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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeMay 3rd 2010
• (edited May 6th 2010)

Any suggestions for something to read on the train ride home? :)

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeMay 3rd 2010

my suggestion: read something basic that allows you to progress with your technical understanding of category theory. It is all the more fun the more one understands it concretely. How is your familiarity with adjoint functors by now?

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 3rd 2010

here is an idea: since you say you are currently enjoying diffeological spaces:

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

• CommentRowNumber4.
• CommentAuthorEric
• CommentTimeMay 3rd 2010

Thanks Urs.

How is your familiarity with adjoint functors by now?

Improving (from a very low base), but still not “familiar” at all. When I read a statement about adjoint functors, I can often nod and pretend I understand, but I’d be stuck cold if I had to actually construct an adjoint functor. if you gave me one, I “might” be able to verify it was one.

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

Thanks. This might occupy more than 1 train ride though, i.e. it might be a train ride or two just to get up to page 38 :) I’ll make a point to give this a try, but I’m frankly scared :)

For today, I chose Andrew’s:

Differential Topology of Loop Spaces

I didn’t get through it all and skimmed a lot, but found it very nice reading :)

I thought it was interested how in Section 4.5 The Enemy of my Enemy is not my Friend, there was the expression

$L[E,F] = [L E,L F]$

which reminded me of

$C([X,Y]) = [C(X),C(Y)].$

Is there a name for a functor that satisfies that condition?

That is, a functor satisfying

$F([X,Y]) = [F(X),F(Y)].$
• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeMay 3rd 2010
• (edited May 3rd 2010)

What do the brackets mean, internal hom? I would call that a closed functor (page does not exist) between closed categories, in analogue with a monoidal functor between monoidal categories. (Although $=$ should become $\cong$ with coherence conditions.)

• CommentRowNumber6.
• CommentAuthorEric
• CommentTimeMay 3rd 2010
• (edited May 3rd 2010)

Thanks Toby.

I would call that a closed functor (page does not exist)

It does now :)

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeMay 3rd 2010

OK, but your definition didn’t include the coherence conditions. I changed $\cong$ back to $=$ and stated it as the definition of strict closed functor. I also mentioned what weak (aka strong) and (op)lax closed functors should be, but without the coherence laws, I don’t know the complete definition.

• CommentRowNumber8.
• CommentAuthorAndrew Stacey
• CommentTimeMay 3rd 2010

For today, I chose Andrew’s:

Differential Topology of Loop Spaces

I didn’t get through it all and skimmed a lot, but found it very nice reading :)

Thanks!

I’ve a vague plan to transfer that to the nLab (when I first wrote it I had vague plans of turning it into a book, but I think that the nLab is better) so any comments as you read through would be greatly appreciated.

Indeed, one could regard manifolds of mapping spaces as part of the transfer.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeMay 3rd 2010

when I first wrote it I had vague plans of turning it into a book, but I think that the nLab is better

+1

• CommentRowNumber10.
• CommentAuthorEric
• CommentTimeMay 5th 2010

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

Thanks. This might occupy more than 1 train ride though, i.e. it might be a train ride or two just to get up to page 38 :) I’ll make a point to give this a try, but I’m frankly scared :)

Today’s train ride got me up to page 17 :)

I’m nervous going forward. I never did manage any intuition when it comes to presheaves.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeMay 5th 2010

I never did manage any intuition when it comes to presheaves.

You once said you liked the motivation for sheaves, cohomology and higher stacks. That’s precisely the kind of motivation and intuition that you need for the article that you are reading.

• CommentRowNumber12.
• CommentAuthorEric
• CommentTimeMay 5th 2010
• (edited May 5th 2010)

Yeah, I’m pretty sure it just hasn’t clicked yet. I’m pretty comfortable with smooth spaces but that level of comfort hasn’t discovered its relation to the word “presheaf” yet. It’s a matter of connecting my intuition to the formal mathematical definitions.

PS: Thanks for reminding me about motivation for sheaves, cohomology and higher stacks.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeMay 5th 2010

Right, so when you hear “presheaf” you should think throughout: rule that tells you for each test object the set of ways (“plots”) of throwing this test object into a given would-be space.

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeMay 6th 2010

By the way, is there any reason that this is in the trash (scratch paper) category?

• CommentRowNumber15.
• CommentAuthorEric
• CommentTimeMay 6th 2010

@Toby: I guess I could have created a new category for “Casual” or something. I take a train ride every day and was hoping to get in the habit of asking for recommendations regularly and didn’t want to clutter up any more serious category.

• CommentRowNumber16.
• CommentAuthorTobyBartels
• CommentTimeMay 6th 2010

I think that’s what the Atrium is for.

I don’t mean to complain, I just don’t want your interesting post to get lost. It fits right in at the Atrium.

• CommentRowNumber17.
• CommentAuthorEric
• CommentTimeMay 6th 2010

Ok. Done :)

• CommentRowNumber18.
• CommentAuthorEric
• CommentTimeMay 7th 2010

Last night’s train ride consisted of me reading and rereading pages 18-20 :)

I realized that I don’t have a good intuition for $D^{op}$. I know it is basically the objects of $D$ with arrows reversed, but how exactly do you reverse arrows? For example, how should I think of $Set^{op}$? Given a function $f:X\to Y$, is $f^{op}:Y\to X$ the preimage? Some explicit examples on opposite category might be nice.

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeMay 7th 2010

You reverse arrows by declaring that they now (i.e. in the new category $D^{op}$ that you are constructing) point in the other direction from the way they used to (i.e. in the old category $D$ that you started with). The arrows themselves are still the same things.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeMay 7th 2010
• (edited May 7th 2010)

That’s right: arrows can be taken the same; you then merely swap the domain and codomain functions. And since arrows are now reversed in orientation, the order of composition gets reversed too.

But all that is just a formal description of opposite. Often it’s of interest to describe opposites of familiar categories of concrete mathematical structures as themselves categories of concrete structures. For example, it turns out that the category opposite to the category of abelian groups is (equivalent to) the category of compact Hausdorff topological abelian groups. The category opposite to the category of finite sets is (equivalent to) the category of finite Boolean algebras. And there are many other such “duality theorems”. Understanding examples of these is an important aspect of one’s categorical training.

We could actually take the time to work through some of these examples, if you wanted.

Edit: I hadn’t been paying attention to the rest of this thread. If the suggestion above is merely distracting, please ignore it.

• CommentRowNumber21.
• CommentAuthorEric
• CommentTimeMay 7th 2010

@Mike: I’m sure your statement is one of those that will seem completely obvious once I understand it, but at the moment, it is still completely mysterious to me :)

Todd said: We could actually take the time to work through some of these examples, if you wanted.

That would be awesome. How about FinSet? How do you get finite Boolean algebras from that? That obviously sounds interesting from a computer science perspective.

Edit: I hadn’t been paying attention to the rest of this thread. If the suggestion above is merely distracting, please ignore it.

Sorry Todd. I will never ignore anything you say :)

• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeMay 7th 2010

I don’t want to give this away too suddenly, but let me say that the picture which you apparently were forming of the opposite of sets and functions was a good start. John Baez had started to talk about this once in connection with the “arrow of time”.

Functions are “many to one”: a function can send many elements to the same element; it can’t send one element to many. Your apparent picture is correct: a morphism in $Set^{op}$ reverses that, hence is “one-to-many”. But no two of those “ones” can go to the same one of the many. If you have a clear picture of what I meant by that last sentence (and how it follows by taking the opposite), then let’s continue.

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeMay 7th 2010

I don’t want to discourage you from understanding what the opposites of some familiar categories look like (i.e. are equivalent to), but I think that it’s also really important to understand that forming the opposite of a category is, on its own, a purely formal construction. A simpler example is the opposite of a group: if I have a group $(G,\cdot, e)$, then I can form a new group that has the same underlying set as $G$, but a new operation $\star$ defined by $x\star y = y\cdot x$. The elements of this new group $G^{op}$ are the same as the elements of $G$, but they are a different group because they have a different operation on them. The situation with the opposite of a category is completely analogous: we have the same objects and arrows, but a new category because the operations like “source” and “target” and “composition” are different.

• CommentRowNumber24.
• CommentAuthorDavidRoberts
• CommentTimeMay 7th 2010
• (edited May 7th 2010)

(just expanding Mike a bit…)

If you want to think about categories internally (say a topological category arising from a group action or something), then given a category $C$ where the source and target are described by the standard map $(s,t):C_1\to C_0\times C_0$, then $C^{op}$ consists of the same object and arrows, where the source-target map is the composite

$C_1 \to C_0\times C_0 \stackrel{swap}{\to} C_0 \times C_0.$

All that changes is the source and target labels have been switched, but it’s still the same arrow. Composition is then defined as being ’backward’, just as in the group case that Mike just mentioned.

• CommentRowNumber25.
• CommentAuthorEric
• CommentTimeMay 7th 2010

@Todd: Thank you :)

So functions send elements of $X$ to elements of $Y$, but not every element of $Y$ gets something mapped to it and several elements of $X$ can map to the same element of $Y$. The opposite function seems like it sends elements in $Y$ to something like a “set of fibers”, but that can’t be quite right because that would be a map $Y\to P(X)$, i.e. the powerset of $X$. BUT… I think we have a map $P(X)\to X$ which just takes the union of elements in $P(X)$. So I think $f^{op}:Y\to X$ sends each element of $Y$ to its preimage in $P(X)$ and then takes the union of all preimages. Or something…

• CommentRowNumber26.
• CommentAuthorUrs
• CommentTimeMay 7th 2010
• (edited May 7th 2010)

I have expanded opposite category a bit

please add to that entry all further information that you give Eric

• CommentRowNumber27.
• CommentAuthorEric
• CommentTimeMay 7th 2010

PLEASE add all further information that you give Eric there.

This is my job in exchange for help :)

• CommentRowNumber28.
• CommentAuthorTim_Porter
• CommentTimeMay 7th 2010

To return to presheaves, I have always found it useful to remember that the inclusion of open sets in a space naturally results in restriction of functions defined on those sets, so to capture ‘restriction’ as an idea one uses the opposite category of open sets within your $X$. I do not know if that helps at all.

When I used to teach sheaves etc. I always looked at the example of presheaves of local sections of a map, as there the constructions were quite easy to illustrate, and of course, that goes towards sheaves.

• CommentRowNumber29.
• CommentAuthorFinnLawler
• CommentTimeMay 7th 2010

@Eric #25:

Two things:

The opposite function seems like it sends elements in Y to something like …

It sounds like you’re thinking of turning $f \colon X \to Y$ in $Set$ into a function from $Y$ to $X$, but that’s not what’s happening. In passing to $Set^{op}$, you’re just formally reversing the direction of the morphisms, not changing their behaviour. An analogy might involve a bunch of electrical cables with plugs on one end and sockets on the other – you can call either the plug or the socket the ’source’ or the ’target’ of a cable, but that doesn’t change the direction the electrons travel in. (I don’t know if that’s a help or a hindrance, though.)

Also,

we have a map $P(X) \to X$ which just takes the union of elements in $P(X)$

No, we don’t – ’union’ takes a set of subsets of $X$ and gives you back a subset of $X$. So it’s a map $P P X \to P X$. If you think about it, a map $P X \to X$ would have to take a subset of $X$ and give you back an element of $X$, which is too much to ask for every $X$. But, as I said, you don’t need such a map to form $f^{op}$, because it’s exactly $f$, only considered as living in $Set^{op}$.

Does that help, or have I completely misinterpreted your comment?

• CommentRowNumber30.
• CommentAuthorTodd_Trimble
• CommentTimeMay 7th 2010
• (edited May 7th 2010)

Eric #25: let me split this thread to start a new discussion; the presheaf discussion should continue without my hijacking it.

• CommentRowNumber31.
• CommentAuthorTim_van_Beek
• CommentTimeMay 7th 2010

To return to presheaves, I have always found it useful to remember that the inclusion of open sets in a space naturally results in restriction of functions defined on those sets, so to capture ‘restriction’ as an idea one uses the opposite category of open sets within your X. I do not know if that helps at all.

The most concrete picture of a presheaf that I have in mind is the sheaf of holomorphic functions on $\mathbb{C}$.

• CommentRowNumber32.
• CommentAuthorTim_Porter
• CommentTimeMay 7th 2010

Exactly the sort of example that I mean. The origins of sheaves and presheaves are rich in intuition that might help Eric.

• CommentRowNumber33.
• CommentAuthorUrs
• CommentTimeMay 7th 2010
• (edited May 7th 2010)

Eric certainly want to think of presheaves to understand opposite categories – because right now he is reading a text on concrete presheaves that made him want to understant what an opposite category is! :-)

But: I would suggest to think about those concrete presheaves in question, not presheaves on open subsets of a topological space for a moment, in order not to get side-tracked by yet one more detail.

Eric, you have already seen the oppositeness that we are after in action here in this thread, let’s just recall it:

for X a smooth manifold, and U a test space, the collection of plots of $X$ by $U$ is the set $Hom_{Diff}(U,X)$.

Now, given a function of test spaces $\phi : V \to U$ we can take any plot $f : U \to X$ in $Hom_{Diff}(U,X)$ and turn it into a plot $V \stackrel{f}{\to} U \stackrel{\phi}{\to} X$ on $V$ in $Hom_{Diff}(V,X)$. This determines a map of set

$f^* : Hom_{Diff}(U,X) \to Hom_{Diff}(V,X) \,.$

So we have a functorial assignment of the form

$\array{ W && \mapsto && Hom_{Diff}(W,X) \\ \downarrow^g &&&& \uparrow^{g^*} \\ V && \mapsto && Hom_{Diff}(V,X) \\ \downarrow^f &&&& \uparrow^{f^*} \\ U && \mapsto && Hom_{Diff}(U,X) }$

You see, it looks almost like a functor from the category of test spaces to the category of sets. Only that it behaves as if the morphisms on the left were not pointing downwards, but upwards.

So it really is a functor on the opposite of the category of test spaces to sets.

That’s all there is to it.

• CommentRowNumber34.
• CommentAuthorEric
• CommentTimeMay 8th 2010

Thanks Urs. I get the sense that there are two things going on here. On the one hand, we have contravariant functors and on the other hand we have opposite categories in their own right. A contravariant functor can be thought of as a covariant functor from the opposite category. In that sense, I can see how it is just a purely abstract “reversing arrows”.

On the other hand, if we start with FinSet and reverse arrows we get (something equivalent to) FinBooAlg. These are different categories that would be of interest to different people.

For example, my favorite

$\Omega:SmoothSpace\to SmoothSpaceForms$

is contravariant with

$\Omega(f\circ g) = \Omega(g)\circ\Omega(f),$

which is usually written as pullback

$(f\circ g)^* = g^*\circ f^*.$

You could say $\Omega$ is a functor from $SmoothSpace^{op}$ but you never really need to construct $SmoothSpace^{op}$. Instead you could just say you’re dealing with contravariant functors and never need to worry about opposite categories.

But in some cases, you really do want to construct the opposite category and then it seems there is a little more to it then simply reversing arrows. You have to think about what the reverse arrow actually is.

If there was a way the two senses above are equivalent, I’d love to hear about it. It would probably help me understand.

So after writing the above, I now think I have a better shot at understanding presheaves. A presheaf is just a contravariant functor. No need to complicate things by throwing in unnecessary ${}^{op}$s :)

• CommentRowNumber35.
• CommentAuthorMike Shulman
• CommentTimeMay 8th 2010

If there was a way the two senses above are equivalent, I’d love to hear about it. It would probably help me understand.

You can always do a purely abstract reversing of arrows. Sometimes, when you do this, you get something equivalent to a category you already knew. Identifying that category and finding the equivalence can take some work – sometimes a lot of work! But merely constructing the opposite is a purely formal operation. Are you looking for more than that?

By the way, I understand that if opposites are hard to understand, then contravariant functors may be easier. But there is of course a reason that categorists generally do it the other way: if you consider a contravariant functor as simply an ordinary (covariant) functor on a different category, then you don’t need a general theory that includes two different kinds of functors.

• CommentRowNumber36.
• CommentAuthorEric
• CommentTimeMay 8th 2010

Hi Mike,

But merely constructing the opposite is a purely formal operation. Are you looking for more than that?

Maybe I’m thinking about it the wrong way. How would you construct $FinSet^{op}$ and what would $f^{op}:Y\to X$ do to elements of $Y$? Given $y\in Y$, what is $f^{op}(y)$? You can tell from opposite of FinSet that I don’t have a clue what $f^{op}(y)$ should be. It can’t be the preimage because that would be a map $Y\to PX$. So what is it?

• CommentRowNumber37.
• CommentAuthorDavidRoberts
• CommentTimeMay 8th 2010

$f^{op}$ isn’t a function, like the arrows of many categories. Because $f$ is a function, we can talk about what it does to elements, and because of (…something…) we can talk about preimages of elements. But in general, what if I give you a category where the arrows are not functions to begin with? The opposite makes complete sense, but we don’t even have a slight handle to talk about that the opposite arrow ’does’. Sometimes taking (fin)Set as the first example to play with is a misleading exercise, because it has too much structure. A good idea is to have two examples to work on simultaneously, and see what is common to both. I suggest the fundamental groupoid (of a generic space) as a foil to the finite set example. Because there the opposite arrow is literally the same path but traversed in the opposite direction. If you want a non-groupoidal example (and one where $a^{op}$ is not just $a^{-1}$, try the fundamental category of a directed space. Then the opposite category is again just the same thing, but all the orders of the paths are reversed (or take the space with the order reversed first, which gives the same thing - but this is a subtlety that is best not dwelt on too much, because as I said, no one example can give you all the insight. This is why categories were invented in the first place (actually this is a lie, but it’s probably a meta-reason): there are different categories around, and we want to capture all the abstract properties, and not just for those categories that happen to have functions for arrows.

• CommentRowNumber38.
• CommentAuthorEric
• CommentTimeMay 8th 2010

$f^{op}$ isn’t a function

Sure. I know it is not a function, but what is it? It has to do something to $Y$. What does it do? Since it does something to $Y$, I’d think it does something to elements of $Y$. What does it do to elements of $Y$? Are those questions just not allowed?

• CommentRowNumber39.
• CommentAuthorTodd_Trimble
• CommentTimeMay 8th 2010
• (edited May 8th 2010)

Mike’s #35 is completely to the point. You don’t need opposite categories; you can do everything you need with covariant functors and contravariant functors, and never mention an opposite. But the “opposite” is a formal construction that eliminates the need for two types of functors: a contravariant functor $C \to D$ is the same as a covariant functor $C^{op} \to D$, so you need only deal with covariant functors. It’s actually a very handy construction.

As for ’what is $f^{op}$?’ As David just finished saying, not all arrows considered in category theory can be conceived as functions that do things to elements. (Oh, there are generalized elements, but that’s not what I’m talking about right now. I can’t make up my mind whether it would be a good idea to talk about that right now.) For example, this idea is meaningless for the homotopy category of topological spaces.

So in complete generality, you don’t always have morphisms as functions that do things to elements (in the plain vanilla sense of ’element’). So in complete generality, the question is not “allowed”.

That said, you deserve maybe a slightly less austere answer. In the discussion we’re having at “opposite of finite sets”, the $f^{op}$ is a relation. That is to say: an ordinary function $f: X \to Y$ can be regarded as one type of relation (namely, a subset $graph(f) \subseteq X \times Y$). We in fact have a category whose objects are finite sets, and whose morphisms $X \to Y$ are relations = subsets of $X \times Y$. These are composed by defining a relational composite, and the more familiar category of finite sets and functions is a subcategory of the category of finite sets and relations. The opposite $f^{op}$ can be regarded as the opposite relation, where you take the image of the subset $graph(f)$ under the switching bijection $X \times Y \to Y \times X$, to get a subset of $Y \times X$.

There is a large class of categories where you can mimic this trick (of embedding the category $C$ into a bigger category $Rel(C)$ of relations, sending morphisms $f: X \to Y$ to relations $graph(f) = \langle 1, f \rangle: X \hookrightarrow X \times Y$, and then think of $f^{op}$ as the morphism in Rel(C) gotten by applying a switching isomorphism to $graph(f)$). To do this, what you need is an appropriate category structure on $Rel(C)$, an appropriate analogue of relational composition. The categories $C$ where this has a satisfying answer are regular categories. Examples include toposes and categories of algebras for an algebraic theory.

Sometimes these “opposite relations” can be thought of again as “structure-preserving functions” in one way or another, sometimes not. But category theory has come a long way since the time that people generally regarded categories as being about “classes of sets with structure and functions which preserve the structure”. I guess you know that (Eric).

• CommentRowNumber40.
• CommentAuthorMike Shulman
• CommentTimeMay 8th 2010

category theory has come a long way since the time that people generally regarded categories as being about “classes of sets with structure and functions which preserve the structure”.

I want to really stress this point, because your (Eric’s) question

what would $f^{op}:Y\to X$ do to elements of $Y$?

really sounds to me as though you are thinking that every morphism in every category is some sort of “function” that can be “applied” to “elements” of the object that is its domain. But in general, all of that is false.

What is $f^{op}:Y\to X$? It’s $f:X\to Y$, regarded as a morphism in $FinSet^{op}$ rather than as a morphism in $FinSet$. (Actually, I think putting the $(-)^{op}$ on $f$ is bad notation; it makes you think that $f$ is changing somehow, when really all that changes is what we say that its domain and codomain are.) In particular, $f^{op}:Y\to X$ is still a function from $X$ to $Y$, so that the only thing we can do with it is apply it to elements of $X$ and get elements of $Y$. The point is that by definition, a morphism from $Y$ to $X$ in $FinSet^{op}$ is defined to be a function from $X$ to $Y$.

• CommentRowNumber41.
• CommentAuthorEric
• CommentTimeMay 8th 2010

Mike said:

But in general, all of that is false.

I’m not thinking this for every category. I am trying to think this for $FinSet^{op}$ though, which seems like a reasonable thing to do.

Todd’s explanation in terms of relations seems to make some sense, but I haven’t thought seriously about relations before, so will have to do just that. It sounds like regular categories encompass most of the things I would probably ever care about though.

This exercise is very helpful. Thank you for your efforts.

• CommentRowNumber42.
• CommentAuthorUrs
• CommentTimeMay 8th 2010
• (edited May 8th 2010)

Eric, you write

On the other hand, …

namely:

…if we start with FinSet and reverse arrows we get (something equivalent to) FinBooAlg.

Sure, but maybe you shouldn’t care about that at the moment!

Here is a general observation about your activity here:

• on the one hand, it is great how you are asking basic questions about category theory, make people explain it to you and thus, as a nice side-effect, make the nLab slowly but steadily become better where it is about exposition of basics of category theory;

• on the other hand, it seems to me you are too easily side-tracked. The problem here is that whatever elementrary question you ask, all the experts around will start mentioning all sorts of related things. Then you ask for more explanation of these, then these experts mention other things, and so on and so forth. It’s fun for a while, but it would be even more fun if from time to time we’d be sure we can record some net progress.

Here I am thinking that it would be really handy if we could do things small-step by small-step. I would like to see you get to the point that you can say “now I understand what a presheaf is”. Or even “now I understand the category of presheaves on CartSp”. Then after that is clear, you can start trying to understand why $FinSet^{op}$ is equivalent to $FinBoolAlg$.

Do you see what I mean? I think it is better first to understand what a functor is, and then try to understand what the cograph of a functor is. First to understand what an adjunction is, and then try to reformulate it. First understand how a presheaf is a functor on an opposite category, and then try to understand what opposites categories can be equivalent to. And so on.

Of course, if you tell me now that the notion of presheaf on CartSp is all easy and clear to you and that you feel bored with it and rather think about $FinSet^{op}$ now then that would be great and fine with me. But is it the case?

• CommentRowNumber43.
• CommentAuthorEric
• CommentTimeMay 9th 2010

Hi Urs,

Thank you for your comment. I understand where it is coming from and I understand if I can be frustrating at times. Your suggestion is totally reasonable and well thought out. I’ll do my best to take it to heart.

My questions over the last couple of days are actually traceable and (I think) justified given my goal. My goal is to try my best to answer the challenge you gave me in comment #3 above:

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

This is a very loaded question. Before I can understand the question, I need to have a handle on

I got up to page 18 without much of a hitch until I saw the definition of presheaf. Although I have seen it, I’ve never actually used it. Your challenge represents the first time I’ve ever needed to explicitly deal with presheaves, which made me nervous, but I wanted to at least try to answer your question because I appreciate you taking the time to pose it. Frankly, I don’t have the background needed to tackle it directly, but that’s what I’m working on and making some progress. My first hiccup when looking at the definition of presheaf was that it had an ${}^{op}$ in it. To a certain extent, I can nod and say I agree it is just a formal thing, but I wanted to have a decent understanding of ${}^{op}$. Mostly so I could understand presheaves, which is mostly so that I can answer your question. At a very elementary level, this is an example of what you mentioned recently about something being so completely (even painfully) obvious, once you get it, it only adds to your frustration for not getting it right away. I think “getting” ${}^{op}$ is probably pretty important for understanding lots of things in category theory. Now, I’m forming an opinion that ${}^{op}$ is little more than a excuse for not having to talk about contravariant functors, i.e. if you understand contravariant functors, you understand ${}^{op}$.

Now, for some good news. Now, I believe I have a decent understanding of presheaf. I don’t need to know what ${}^{op}$ really is to understand presheaf. I can just think in terms of contravariant functors, which I’m already ok with because one of my favorite functors $\Omega$ is contravariant. I’ve now added a line to this effect at presheaf, but this only makes me want to modify contravariant functor a bit so it doesn’t feel so circular.

Todd is now helping me to understand $FinSet^{op}$, which I now (I didn’t originally) think is a little peripheral to your original question, but going down that path was motivated by your question and I’d like to continue as long as Todd’s patience with me continues or until I get it. But that doesn’t mean I’ve lost focus of trying to understand your original question.

Another good thing about your question is that it will force me to revisit adjoints. This will give me an excuse to once and for all finish one of your earlier challenges, i.e. understand the adjoint relation between directed graphs and categories, i.e. show that for each graph morphism $G\to U(C)$, there is a unique morphism $F(G)\to C$. I include this as part of my current challenge.

• CommentRowNumber44.
• CommentAuthorUrs
• CommentTimeMay 9th 2010

Thanks, Eric.

I started adding some list of examples of presheaf, but in a haste. Have to run now….

• CommentRowNumber45.
• CommentAuthorEric
• CommentTimeMay 9th 2010

Another train ride and I’m still on page 18 :)

I’m looking at Definition 14:

Given a covering family $(f_i: D_i\to D|i\in I)$ in $\mathbf{D}$ and a presheaf $X:\mathbf{D}^{op}\to Set$, a collection of plots $\{\phi\in X(D_i)|i\in I\}$ is called compatible if whenever $g:C\to D_i$ and $h:C\to D_j$ make this diagram commute:

$\array{ C & \stackrel{h}{\rightarrow} & D_j \\ \mathllap{g\;}{\downarrow} & & \mathrlap{\downarrow}{\quad f_j} \\ D_i & \stackrel{f_i}{\rightarrow} & D }$

then $X(g)(\phi_i) = X(h)(\phi_j)$.

However, since functors preserve commuting diagrams, contravariant functors preserve commuting diagrams with arrows reversed, so

$f_i\circ g = f_j\circ h \mapsto X(g)\circ X(f_i) = X(h)\circ X(f_j).$

So what is the relation between $\phi_i$ and $X(f_i)$?

• CommentRowNumber46.
• CommentAuthorUrs
• CommentTimeMay 9th 2010
• (edited May 9th 2010)

That’s right Eric, but the point here is that we do not consider plots on $D$ in this case, hence do not just apply the contravariant functor to the full diagram. You are of course right that if we did this, we would be guaranteed to get another commuting diagram.

Here the point is rather that we want to discuss descent (don’t look that that page! :-): this means: we want to consider plots on all the $(D_i)$ that as a collection have the property that we can build a plot on $D$ from them.

This is the crucial step when we go from presheaves to sheaves. Sheaves are those special presheaves for which it is possible to reconstruct a plot on $D$ from a compatible family of plots on $D_i$s, as soon as the latter cover the former.

For the discussion of diffeological spaces you should think of it this way: if you have a space $X$ that you can test by throwing test spaces $D$ into it, then it should also be possible to break up $D$ into two smaller test spaces, $D_1$ and $D_2$ that overlap a bit, then see how we can throw $D_1$ into $X$ and $D_2$ into $X$ separately. Then if among all those ways of throwing these two spaces sperately you look at these pairs where the image in $X$ coincides on the overlap of $D_1$ with $D_2$ in $D$, then it should be possible to build one probe of $X$ by $D$ from this information.

This is what the above formalizes: a compatible family of plots on the $D_i$ is a collection of probes that coincides on the joint overlaps of the $D_i$ in the given $D$.

• CommentRowNumber47.
• CommentAuthorEric
• CommentTimeMay 10th 2010

Thanks Urs!

Quick comment and I’ll be sucked into a vortex…

That’s right Eric, but the point here is that we do not consider plots on $D$ in this case, hence do not just apply the contravariant functor to the full diagram. You are of course right that if we did this, we would be guaranteed to get another commuting diagram.

But is it possible that this diagram tells the same story as what you just said? It seems that if you squint, you can see the same story emerge from the simple commuting diagram.

[sucking sound]

• CommentRowNumber48.
• CommentAuthorHarry Gindi
• CommentTimeMay 10th 2010
• (edited May 10th 2010)

I don't get the "sucking sound" reference. Is it because I'm throwing you out of the submarine because you're talking about really simple things and making them really complicated again? The sucking sound comes from the water rushing into the airlock as you sit there waiting for your demise hahahahaha.

Only kidding.

Anyway, in all seriousness, the thing about plots is not relevant to presheaves (this has to do with descent as Urs noted). For it to be a presheaf if literally has to be a contravariant functor into sets and nothing more. If you actually want to read about descent (which is what it seems like you care about), you should check out the beginning of chapter 2 of Hartshorne for a pretty good treatment of the material (albeit they are sheaves of abelian groups). I would also suggest Vistoli's wonderful notes on descent (look on google) if you want a good understanding of descent w/r/t a grothendieck topology.

• CommentRowNumber49.
• CommentAuthorEric
• CommentTimeMay 10th 2010

Hi Harry,

For your reference, Urs gave me something of a homework problem:

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

I’m currently working through that paper and am on page 18. That is where these questions are coming from.

• CommentRowNumber50.
• CommentAuthorUrs
• CommentTimeMay 10th 2010

But is it possible that this diagram tells the same story as what you just said? It seems that if you squint, you can see the same story emerge from the simple commuting diagram.

You are right, what you are seeing is the reverse of the story I told: given a plot on $D$, you can restrict it to plots on each of the $D_i$ and the family of plots on the $D_i$s obtained this way is compatible.

That’s right, that’s what you get from just applying the presheaf/functor to that commuting diagram.

But the nontrivial piece now is going in the reverse direction: we want to find those presheaves for which it is true that all compatible families of plots on the $D_i$s arise this way.

And, with an eye towards Harry’s remark, more traditionally where I say “plot” here other sources would say “section”. it is the same thing. The term “section” is more suggestive when we are thinking petit topos while the term “plot” is more suggestive when it comes to gros toposes. Anyway, “plot” is what people in diffeologicval spaces say, which is what we are talking about here.

• CommentRowNumber51.
• CommentAuthorHarry Gindi
• CommentTimeMay 10th 2010
• (edited May 10th 2010)

I don't really see how "plot" is more suggestive for Gros toposes.

@Eric:

Anyway, I am reasonably certain that you stand to gain a lot more from Vistoli's notes than pretty much anything else you'll read (skip the specific algebraic geometry stuff in it, it is not relevant and will do nothing more than confuse you). There is no handwaving or anything. It's totally rigorous and starts from the basics. I don't know if Urs will agree with me, but it's also not clear to me whether or not Urs has read the notes in question.

• CommentRowNumber52.
• CommentAuthorUrs
• CommentTimeMay 10th 2010
• (edited May 10th 2010)

I don’t really see how “plot” is more suggestive for Gros toposes.

Here is the reason:

in a petit topos (sheaves on open subsets of some $X$), you may think of each object as sitting over a given object $X$. Therefore the value of the sheaf on some open subset is literally the collection of sections of this projection.

For a gros topos however, such as sheaves on CartSp, there is no good sense in which each of the objects is sitting over some fixed object. What the sheaves assign to an object in the site is of course still the collection of all ways to map that object of the site into the given object in the topos. So this is no longer so much to be thought of as a section, but rather as a “probe”. Or a plot. It’s just words, of course, for the same math.

• CommentRowNumber53.
• CommentAuthorHarry Gindi
• CommentTimeMay 10th 2010
• (edited May 10th 2010)

It's a section of the corresponding Discrete fibration, no?

Edit: Eh, maybe not. The above might be stupid, but I don't care enough to check, so I've put this disclaimer.

• CommentRowNumber54.
• CommentAuthorUrs
• CommentTimeMay 10th 2010
• (edited May 10th 2010)

It’s a section of the corresponding Discrete fibration, no?

That’s true of course, of the presheaf’s category of elements over the given object of the site, yes. But this does not quite have the same geometric feel to it. In fact, here its more like we would say that the collection of $V$-plots is the fiber of the category of elements of the presheaf over $V$.

But it’s just words. Whatever you feel like saying, I am fine with. It’s just a historical fact that for presheaves on CartSp there is some traditon to call the value of the presheaf on some object the collection of plots.

• CommentRowNumber55.
• CommentAuthorAlexHoffnung
• CommentTimeMay 10th 2010

@Harry

I have taken a look and the notes look great. I hope to have time to give them a good read sometime soon. There are plenty of things in these notes that I would like to understand better.

Without worrying too much about what one can gain from a particular set of notes, I think Eric has initiated a fantastic way to learn a small topic by consistently being a part of this community, tackling a small project, and graciously asking and accepting the help of the local experts. I hope to follow in his footsteps in the future.

My reason for revisiting the topic which you initiated is that I think it takes a lot of practice and care to guide someone through mathematics. In particular, Urs seems to be working really hard to keep Eric on track, and Eric seems to be treading carefully, only picking up side topics as needed. I think suddenly jumping into another long document seems particularly unhelpful. I find that ignoring large parts of a paper is not easy. It continually gives one the sense of missing some big picture, not too mention the urge to dive into each new exciting word or concept one encounters. In particular, the last few comments between Urs and yourself have made a reasonably big jump from slowly working through details to less relevant, but seemingly more scary territory.

• CommentRowNumber56.
• CommentAuthorHarry Gindi
• CommentTimeMay 10th 2010

I was only suggesting that Eric read those notes because it seemed like the paper he was reading (granted, I skimmed only a part of page 38) uses the ideas of descent anyway (a rose by any other name...). Descent is probably the most important idea in geometry, so I was just suggesting that he should just get to the heart of the matter. Vistoli's notes are great because they are self-contained (at least on the category-theory side, that is, they go from Yoneda to stacks, covering everything in between).

• CommentRowNumber57.
• CommentAuthorEric
• CommentTimeMay 11th 2010

Here are Vistoli’s notes:

• CommentRowNumber58.
• CommentAuthorEric
• CommentTimeMay 11th 2010

I just realized Harry is trying to get me to learn algebraic geometry. Good luck! I had a copy of Hartshorne setting on my desk at UIUC for years hoping something would penetrate my skull through osmosis, but never happened. Hartshorne is not for the faint of heart :)

• CommentRowNumber59.
• CommentAuthorDavidRoberts
• CommentTimeMay 11th 2010

Descent is a bit more than alg. geom. - any time someone mentions clutching functions or transition functions they are talking descent. Without you even knowing it, working with manifolds implicitly uses descent a lot, if one has to use charts.

• CommentRowNumber60.
• CommentAuthorEric
• CommentTimeMay 11th 2010

Fair enough :) The word “descent” is thrown around enough around here, I can tell it is important, so I’ll put it high on this list of stuff I need to know.

• CommentRowNumber61.
• CommentAuthorHarry Gindi
• CommentTimeMay 11th 2010
• (edited May 11th 2010)

@David: One uses descent when a manifold is a locally ringed space, or even when a manifold is a variety in the differentiable geometric context (a sheaf on CartSp (with the local diffeo topology) with a jointly surjective cover by open immersions (this is the terminology from algebraic geometry, which I would guess correspond to injective submersions because in AG, one uses smooth injections) of representables).

• CommentRowNumber62.
• CommentAuthorUrs
• CommentTimeMay 11th 2010
• (edited May 11th 2010)

My suggestion: Eric, don’t try to read Vistoli now.

What would help eventually for your exercise with diffeological spaces is that you under stand the sheaf condition. We already talked about it above. The descent condition is the oo-categorification of the sheaf condition. My suggestion: let’s not get into that here before you are well familiar with the sheaf condition maybe in general, but in particular for the case of sheaves on CartSp.

• CommentRowNumber63.
• CommentAuthorEric
• CommentTimeMay 11th 2010

My suggestion: Eric, don’t try to read Vistoli now.

Didn’t plan on it. I already have my train reading for the foreseeable future :)

• CommentRowNumber64.
• CommentAuthorHarry Gindi
• CommentTimeMay 11th 2010

I feel like saying that it's the oo-categorification of the sheaf condition is overstating things a bit (and making them seem harder than they actually are. The word "oo-categorification" makes things sound scary, even to me). Homotopical and cohomological descent (which seem like the real oo-categorification of the sheaf condition) are significantly more complicated than the classical approach using grothendieck pretopologies (for which the abstraction of the sheaf condition on spaces is immediate by replacing the intersection with the fiber product). Chapter 2 of Vistoli covers precisely this abstraction of the sheaf condition.

• CommentRowNumber65.
• CommentAuthorUrs
• CommentTimeMay 11th 2010
• (edited May 11th 2010)

Harry, I guess you are thinking just of descent in the context of stacks. That’s of course just the 1-fold categorification of the sheaf condition.

But I think there is a 1-2-3-infinity thing going on here: beyond already very low values of n, it is easier to handle the full notion of oo-descent than to explicitly do n-descent. But whatever works for you is fine with me. I just don’t think that at the current point our discussion with Eric will benefit from bringing stuff beyond sheaves into the game.

• CommentRowNumber66.
• CommentAuthorHarry Gindi
• CommentTimeMay 11th 2010
• (edited May 11th 2010)

That's what I'm saying. Vistoli chapter 2 covers precisely 0-descent (descent for sheaves). It is the best piece of exposition I know for that topic. Chapter 3 introduces fibered categories and chapter 4 introduces 1-descent for stacks. I am only suggesting that Eric read chapters 1 and 2 of Vistoli. Chapters 3 and 4 complicate the matter, but chapters 1 and 2 are precisely on topic.

• CommentRowNumber67.
• CommentAuthorEric
• CommentTimeMay 12th 2010

I spent another train ride on page 18 staring at Definition 14 again:

Given a covering family $(f_i: D_i\to D|i\in I)$ in $\mathbf{D}$ and a presheaf $X:\mathbf{D}^{op}\to Set$, a collection of plots $\{\phi\in X(D_i)|i\in I\}$ is called compatible if whenever $g:C\to D_i$ and $h:C\to D_j$ make this diagram commute:

$\array{ C & \stackrel{h}{\rightarrow} & D_j \\ \mathllap{g\;}{\downarrow} & & \mathrlap{\downarrow}{\quad f_j} \\ D_i & \stackrel{f_i}{\rightarrow} & D }$

then $X(g)(\phi_i) = X(h)(\phi_j)$.

Now this is right before the definition of a sheaf, so I’m not quite at the sheaf condition yet. This is just the compatibility condition. As I mentioned last time, contravariant functors preserve commuting diagrams with arrows reversed, so we’ve got two expressions relating to compatible plots:

1. $X(g)\circ X(f_i) = X(h)\circ X(f_j)$
2. $X(g)(\phi_i) = X(h)(\phi_j)$

When I look at this I’m tempted to say the compatibility condition is related to simply stating plots are compatible with covering families if

$\phi_i = X(f_i).$

How far from the truth is that?

• CommentRowNumber68.
• CommentAuthorUrs
• CommentTimeMay 12th 2010
• (edited May 12th 2010)

How far from the truth is that?

Not sure how to measure the distance. If it is permutation distance of symbols, then we are close! ;-)

So: as I said before, what you are getting at is that we obtain a compatible family of plots, one for each $D_i$, by taking a single plot $\phi \in X(D)$ on $D$ and pulling it to each $D_i$ by setting $\phi_i := X(f_i)(\phi)$.

clear?

Then here the punchline: we say that $X$ is a sheaf if every compatible family of plots on the $D_i$ arises from pulling back a plot on $D$ this way.

• CommentRowNumber69.
• CommentAuthorEric
• CommentTimeMay 12th 2010

But you read ahead! That is Definition 15! :)

I’m not clear on what compatible means yet (Definition 14) :)

So $X(g)$ is a function from the set $X(D_i)$ to the set $X(C)$. So far, we only know a “plot” $\phi_i$ is an element of the set $X(D_i)$. We don’t know that $\phi_i$ itself is a map yet. Or do we?

From the commuting diagram, we know that for any $\phi\in X(D)$ that

$X(g)\circ X(f_i)(\phi) = X(h)\circ X(f_j)(\phi).$

This is true merely from the fact that $X$ is a contravariant functor and

$f_i\circ g = f_j\circ h.$

Compatibility means

$X(g)(\phi_i) = X(h)(\phi_j).$

So in terms of permutation distance we are now even closer :)

But just because

$X(g)\circ X(f_i)(\phi) = X(h)\circ X(f_j)(\phi)$

and

$X(g)(\phi_i) = X(h)(\phi_j),$

it doesn’t necessarily mean that

$\phi_i = X(f_i)(\phi)$

because there could be some cancellable function thrown in there. So the sheaf condition says there is no cancellable function thrown in there and the relation holds on the nose. Is that right?

• CommentRowNumber70.
• CommentAuthorUrs
• CommentTimeMay 12th 2010
• (edited May 12th 2010)

Ah. You write

I’m not clear on what compatible means yet (Definition 14) :)

But a few lines below you write

Compatibiliyt means $X(g)(\phi_i) = X(h)(\phi_j)$.

So there! (… if we demand this for all commuting diagrams as shown ).

So a collection of elements $\{ \phi_i \in X(D_i) | i \in I\}$ is a compatible family, if for all commuting diagrams under discussion, this relation holds. Period.

Not sure, you say you are mixed up abbout something, but it is hard for me to tell about what. You say

We don’t know that $\phi_i$ itself is a map yet.

That sentence does not quite make sense to me. By decree, $\phi_i$ is some element in the set $X(D_i)$.

Let’s see if we can agree up to this point – definition of a compatible family of plots. When we have this saved, I’ll try to say something about your “cancellable functions” (hoping that then it won’t be necessary anymore ;-)

• CommentRowNumber71.
• CommentAuthorTodd_Trimble
• CommentTimeMay 12th 2010

I have to ask: Eric, are you comfortable with the notion of sheaf on a topological space? Because from a pedagogical standpoint, that seems to be the most reasonable place to start. The general notion of sheaf on a site won’t be all that meaningful unless this more vanilla notion of sheaf is familiar and comfortable.

• CommentRowNumber72.
• CommentAuthorUrs
• CommentTimeMay 12th 2010
• (edited May 12th 2010)

I have this feeling that sheaves on topological spaces are an unnecessary distraction here. But maybe I am wrong.

• CommentRowNumber73.
• CommentAuthorTodd_Trimble
• CommentTimeMay 12th 2010

Whether it’s a distraction would depend on who’s learning it. The more I observe Eric’s style, the more I think he’s a very visual concrete learner and that abstract formulations are not the best first approach to take in his case. Witness the discussion about opposite of a category. (Eric, hope you don’t mind this, or speaking of you in third person.)

• CommentRowNumber74.
• CommentAuthorUrs
• CommentTimeMay 12th 2010

Sure, yes, whatever works. I am just wary of opening too many cans of worms at the same time. We should focus on small items by small items.

• CommentRowNumber75.
• CommentAuthorEric
• CommentTimeMay 13th 2010

So there!

This is similar to the case of functor. I can state the compatibility condition from memory, but it doesn’t mean I REALLY understand it until I can draw a picture of it or state it in an alternative way. At this point, I can merely quote the definition (from memory even) :)

Does a coverage of families $(f_i:D_i\to D)$ consist of ALL morphisms with codomain $D$?

Is there a way to state the compatibility condition without reference to elements? If given family members $f_i:D_i\to D$ and $f_j:D_j\to D$ there was a map $f_{ij}:D_i\to D_j$, then would compatibility mean the diagram containing this map commutes?

Or if $\mathbf{D}$ had a terminal object 1, could we stick it into the original commuting square and claim the plots were compatible if the diagram containing 1 commutes? In this case, we could interpret plots as morphisms, i.e. identity $\phi_i\in X(D_i)$ with a function $\phi_i:X(1)\to X(D_i)$ so that compatibility means $X(g)\circ \phi_i = X(h)\circ\phi_j$.

What would be an example of an incompatible plot? I guess it could happen if the image of $X(f_i)$ does not cover all of $X(D_i)$ so an element in $X(D_i)$ that was not in $X(f_i)(X(D_i))$ could correspond to an incompatible plot.

I have to ask: Eric, are you comfortable with the notion of sheaf on a topological space?

I haven’t yet gotten to the definition of sheaf yet. That is Definition 15. I’m still on Definition 14 :)

But once I move on from compatibility condition to sheaf condition, being able to think of topological spaces might help, but I think the actual category itself is less important than my ability to draw diagrams of it. One reason I could never penetrate Hartshorne is that (if I remember correctly) it contains very few (if any) pictures. I am simply unable to absorb purely abstract definitions. I need to be able to draw it and see it or there is very little use for me to even try to think about it.

• CommentRowNumber76.
• CommentAuthorDavidRoberts
• CommentTimeMay 13th 2010

OK, I get it now. $C$ is like the intersection (or pullback, more generally) of $D_i$ and $D_j$. The compatibility condition is saying that the plots $\phi_i$ and $\phi_j$, when restricted along the maps $C \to D_i$ and $C \to D_j$, give the same map. The map $X(g)$ is like ’precomposition with $g$’, if we pretend that a plot $\phi_i$ is a map $D_i \to thing$, where $thing$ is the object that the sheaf is trying to represent. Then $X(g)(\phi_i) '=' \phi_i\circ g : C \to D_i \to thing$. The sheaf condition (which comes next) tells us that if there are a family of plots that agree on the intersections of their domains, then they paste together to form a plot on the union of their domains. (this statement is not really precise, but that’s the idea that is being gotten at) The tricky thing is that it is not assumed that pullbacks exist, so that every square that looks like a pullback, but is not necessarily universal, has to play the part.

• CommentRowNumber77.
• CommentAuthorEric
• CommentTimeMay 13th 2010
• (edited May 13th 2010)

Thanks for your comment David! Seeing other people see the light is encouraging :)

Since $C$ is like the intersection of $D_i$ and $D_j$, I’d think $X(D)$ really is the intersection of $X(D_i)$ and $X(D_j)$.

Starting with the commuting diagram

$\array{ C & \stackrel{h}{\rightarrow} & D_j \\ \mathllap{g\;}{\downarrow} & & \mathrlap{\downarrow}{\quad f_j} \\ D_i & \stackrel{f_i}{\rightarrow} & D }$

we end up with the commuting diagram

$\array{ X(C) & \stackrel{\scriptsize{X(h)}}{\leftarrow} & X(D_j) \\ \mathllap{\scriptsize{X(g)}\;}{\uparrow} & & \mathrlap{\uparrow}{\quad \scriptsize{X(f_j)}} \\ X(D_i) & \stackrel{\scriptsize{X(f_i)}}{\leftarrow} & X(D) }$

More to say, but the vortex (i.e. real life) is pulling me…

• CommentRowNumber78.
• CommentAuthorDavidRoberts
• CommentTimeMay 13th 2010

I’d think $X(D)$ really is the intersection of $X(D_i)$ and $X(D_j)$.

Sadly, no. $X(D)$ is to be thought of as the collection of maps $D \to thing$, so the relationship between the sets in the second square is a bit more complicated. For a start, the maps $X(D) \to X(D_i)$ are neither injective nor surjective (and the same with those involving $X(C)$ etc).

• CommentRowNumber79.
• CommentAuthorEric
• CommentTimeMay 18th 2010

Ok! So I haven’t lost sight of Urs’ suggested trainwork assignment (the train is the only spare time I have to work on this)

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

but it became obvious when I got stuck on page 18 that I did not have the foundations to continue. Harry suggest that to understand the material on sheaves, it would help to understand descent and suggested reading Chapter 2 of Vistoli. I like Vistoli very much. It is written in a style I can resonate with. As I was reading Vistoli, I got to the part about Yoneda lemma.

Given that the Yoneda lemma is so important, I hope it doesn’t seem like too much of a distraction from the original task (which I still remember!) to make an effort to understand Yoneda lemma. For this, I might spend at least a train ride or two on Tom Leinster’s notes.

Once I get this and get descent and get sheaves, I’ll return to the original problem.

I know it seems like I’m all over the place, but there is a method to the madness.

• CommentRowNumber80.
• CommentAuthorTim_van_Beek
• CommentTimeMay 18th 2010

There was a question on mathoverflow about the philosophical meaning of the Yoneda Lemma

• CommentRowNumber81.
• CommentAuthorEric
• CommentTimeMay 24th 2010

I feel like the time I can afford to dedicate to maths is dwindling. Any spare time I have soon, should be spent learning Mandarin :)

I hope I can reach some goals before switching gears.

open Baez-Hoffnung on page 38 and convince yourself in full detail of the proof of the claim that concrete presheaves form a reflective subcategory of all presheaves (i.e. that the inclusion functor has a left adjoint).

I’m feeling a bit bummed thinking I might not be able to accomplish this task. I’d like to understand sheaves before switching gears. Is there a reference I might have a chance of understanding that might help me understand sheaves?

• CommentRowNumber82.
• CommentAuthorEric
• CommentTimeMay 24th 2010

PS: Before heading to the train, I printed a copy of Motivation for Sheaves, Cohomology, and Higher Stacks (ericforgy) (I removed the floating TOCs to print it). This is the PERFECT introduction to the ideal reference, but there is a significant gap between this motivating introduction and the main nLab material. Is there a good reference that fills the gap?

• CommentRowNumber83.
• CommentAuthorUrs
• CommentTimeMay 24th 2010
• (edited May 24th 2010)

The description by Dugger at the beginning of his “Sheaves and homotopy theory” gives a good intuitive account of what’s going on with sheaves. You once turned this into a pdf. It is linked to at several places on the nLab. In its role as a pedagogical introduction to sheaves at the end of free cocompletion, for instance.

• CommentRowNumber84.
• CommentAuthorUrs
• CommentTimeMay 24th 2010

here http://ncatlab.org/nlab/files/cech.pdf

• CommentRowNumber85.
• CommentAuthorEric
• CommentTimeMay 24th 2010
• (edited May 24th 2010)
• CommentRowNumber86.
• CommentAuthorUrs
• CommentTimeMay 24th 2010
• (edited May 24th 2010)

Thanks, i was typing while walking and had trouble putting in links.

By the way: be sure not to try to read the whole thing. But just the introductory section on ordinary presheaves and sheaves might be useful.

• CommentRowNumber87.
• CommentAuthorEric
• CommentTimeMay 24th 2010

Looking now. Thanks :)

• CommentRowNumber88.
• CommentAuthorEric
• CommentTimeMay 26th 2010
• (edited May 27th 2010)

Dugger is awesome. I don’t remember promoting it before. I surely did not read it very thoroughly (I couldn’t have been able to at the time). But I felt a small penetration into my thick skull finally. I think I’m starting to understand sheaves a little bit. Still a work in progress though.

• CommentRowNumber89.
• CommentAuthorEric
• CommentTimeJul 28th 2010

Last night on the train, I had the pleasure of reading the introduction of

Smooth Functors vs. Differential Forms

Very very pleasant reading. One of those things that gives you hope for humanity :)

I got hung up on thinking about continuum vs discrete stuff though (as usual). I think the material here could also be used in lattice gauge theory in a very natural way.

• CommentRowNumber90.
• CommentAuthorUrs
• CommentTimeJul 28th 2010

I think the material here could also be used in lattice gauge theory in a very natural way.

That’s where it was first discussed. See this.

• CommentRowNumber91.
• CommentAuthorUrs
• CommentTimeJul 28th 2010
• (edited Jul 28th 2010)

That’s where it was first discussed. See this.

A commented list of this and related references I have compiled – or am compiling – at Differential cohomology in an (oo,1)-topos – References in the section “Connections – In terms of parallel transport”.

• CommentRowNumber92.
• CommentAuthorEric
• CommentTimeJul 28th 2010

Oh thank you! More train reading material. Abstract looks very interesting.

Higher gauge theory – differential versus integral formulation Authors: Florian Girelli, Hendryk Pfeiffer (Submitted on 17 Sep 2003 (v1), last revised 28 Jan 2004 (this version, v2))

Abstract: The term higher gauge theory refers to the generalization of gauge theory to a theory of connections at two levels, essentially given by 1- and 2-forms. So far, there have been two approaches to this subject. The differential picture uses non-Abelian 1- and 2-forms in order to generalize the connection 1-form of a conventional gauge theory to the next level. The integral picture makes use of curves and surfaces labeled with elements of non-Abelian groups and generalizes the formulation of gauge theory in terms of parallel transports. We recall how to circumvent the classic no-go theorems in order to define non-Abelian surface ordered products in the integral picture. We then derive the differential picture from the integral formulation under the assumption that the curve and surface labels depend smoothly on the position of the curves and surfaces. We show that some aspects of the no-go theorems are still present in the differential (but not in the integral) picture. This implies a substantial structural difference between non-perturbative and perturbative approaches to higher gauge theory. We finally demonstrate that higher gauge theory provides a geometrical explanation for the extended topological symmetry of BF-theory in both pictures.

• CommentRowNumber93.
• CommentAuthorEric
• CommentTimeJul 28th 2010
• (edited Jul 28th 2010)

Whoa. Didn’t notice your second post with the link to references. Very cool page. I remember when you were tracking down references. Nice to see everything summarized like that.

Edit: Found the old nCafe discussion.

• CommentRowNumber94.
• CommentAuthorUrs
• CommentTimeJul 28th 2010

Very cool page. I remember when you were tracking down references. Nice to see everything summarized like that.

I should add more references on various aspects of the story. But need to find the time.

• CommentRowNumber95.
• CommentAuthorEric
• CommentTimeJul 29th 2010

Speaking of other references, would it be relevant to include a link to the extended discussion here between you and Domenico?

This is beyond my knowledge, but it seems you were talking about similar ideas.

• CommentRowNumber96.
• CommentAuthorEric
• CommentTimeOct 25th 2010
• (edited Oct 25th 2010)

I’ve gone back and re-collected some of the papers discussed in this thread:

I’ve also gone back to one of my classic references

It is fun to reread Nakahara now that I have a smidgen of category theory under my belt. One thing I think would be neat/helpful to me would be to understand “transition functions” in purely arrow theoretic terms. If I look back at one of David’s comments:

Descent is a bit more than alg. geom. - any time someone mentions clutching functions or transition functions they are talking descent. Without you even knowing it, working with manifolds implicitly uses descent a lot, if one has to use charts.

This statement motivates me to learn some basics about descent (maybe not the full-fledged $\infty$ version though).

I think it might help (?) if I tried to re-understand some basic stuff about manifolds explicitly using some category theory. I never really studied bundles as well as I should have either, so I hope to maybe revisit that from an arrow theoretic perspective as well.

Edit: One more reference: