Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
As I mentioned in my “new member” post (I don’t know how to do fancy references to other posts yet), I am looking for a notion of “remainder” in category theory.
I’m describing a “system behavior” and a map of that behavior to the behavior of a “component” of the system. The question at hand is: what is the best description I can find of the behavior of the other components?
Because a composition of components (and synchronization between) them, in my category leads to a “jointly monic family” of maps from the system behavior to the behavior of the components, I could therefore say that I’m looking for the “smallest” component that allows me to form a “jointly monic pair” with the map into the “known” component.
I got to this definition at first:
I though it was a rather natural notion, but have not found references to it yet.
Mike Shulman and David Roberts already responded with two questions:
1) can I give an example of whether such a remainder can exist at all? He didn’t think it would exist for Set.
2) Isn’t it a trivial notion? If you take $Y = X$, the identity might do the job.
As to 1)… I’m going to think about it. I think I have an example in the category of prefix orders, on which I just posted a page, but I still need to check it. (Apologies for asking the question before doing my homework… but sometimes you can spend hours figuring out what other people already know ;-)
As to 2)… shoot… that is a suggestion I overlooked. In my own examples there are often “smaller” candidates for $Y$ than $X$, but they do not seem to come out this way. Intuitively, in Set, the map $g$ should not be injective at points where $f$ is already injective. But the definition still allows that apparently. Back to the drawing board… Any suggestions?
David is right, btw, about any category and any $f$ at all: your definition says it is an initial object in some full subcategory of $X/\mathcal{C}$, so as long as that subcategory contains the initial object of $X/\mathcal{C}$ (namely $Z=X$ and $g=id_X$) — and it does — that is also the initial object in the subcategory. I think I was originally trying to solve the problem you actually had in mind of making a map that’s “only injective where $f$ is not”, and that’s what I thought would not exist. Perhaps what you want is a terminal object of the same category? But look at $Set$ first; suppose $f:X\to Y$ is the first projection $2\times 2 \to 2$, where $2=\{0,1\}$. What do you want $g$ to be? Since $f(0,1)=f(0,0)$ and $f(1,1)=f(1,0)$, you need $g$ to separate those pairs. But if you want $g$ to be noninjective “at points where $f$ is already injective” then presumably you want something like $g(0,1) = g(1,1)$ or $g(0,1) = g(1,0)$ — but you can’t have both of those, otherwise $g(1,0)=g(1,1)$ which isn’nt allowed, and there’s no universal way to pick one but not the other.
Thanks Mike, good example. (I’m not a full time category theorist, but after some parsing I can get the first part of your post as well :-)
Indeed, let’s look at Set first. If $f: X \rightarrow Y$ is the first projection of $2 \times 2 \rightarrow 2$, then (intuitively) I would like $g$ to become the second projection. That would at least be my idea of division within Set… Whenever $f$ is not an actual projection, it will be a bit tricky to tell what $g$ should be, but indeed for projections it should work out in any/most categories.
So, what are we doing really then… I think $g$ should split the points that $f$ equates, but it can identify points that $f$ distinguishes. Furthermore, $g$ should identify “as many points as possible”, and that is where the difficulty starts.
Okay, then let’s look at whether terminal objects work…
Second try: given $f : X \rightarrow Y$, and object $Z$ is a remainder of $f$ if there exists a map $g : X \rightarrow Z$ such that $(f,g)$ is jointly monic, and for any $g' : X \rightarrow Z'$ with $(f,g')$ jointly monic there is a map $h : Z' \rightarrow Z$ such that $h g' = g$.
Notice that I have reversed the direction of $h$ but also dropped the uniqueness requirement, because it will give trouble whenever $g'$ is non-surjective.
This seems to work for the simple $2 \times 2$ example. It is a standard result that the two projections form a jointly monic pair. Furthermore, any candidate $Z'$ must at least contain two points in order to distinguish $(0,0)$ from $(0,1)$, otherwise $(f,g')$ cannot be jointly monic. Furthermore, the image of $g'$ will contain at most $4$ points, and the remaining points an be mapped arbitrarily to $2$. Simply enumerating the possibilities for images that contain $2$,$3$ or $4$ points quickly leads to the conclusion that indeed the second projection is a witness to show that $2 \times 2$ divided by $2$ gives $2$.
I’ve tried a simple example in my category of prefix orders as well, and it also seems to work there.
Next, I guess I should prove that in general when $f$ is the first projection of a binary product, $g$ is the second projection. Or try a few more examples. And after that, we still have the question whether/when $Z$ is unique…
And pondering the diagram a bit… what would the consequence be if we weaken the definition to:
Third try: given $f : X \rightarrow Y$, and object $Z$ is a remainder of $f$ if for every $g : X \rightarrow Z'$ with $(f,g)$ jointly monic, there is an $h : Z' \rightarrow Z$ such that $(f,hg)$ is jointly monic.
Whenever the second try holds, the third try holds as well. But of course the third try has fewer properties in general. If the third try still gives a unique choice for $Z$ however, it means that the definition becomes possible in more categories.
My point about $2\times 2$ is that given only a 4-element set and one of its projections to a 2-element set, there is no canonical, hence no universal, way to identify another map as “the second projection”. Maybe it would be clearer to consider the function $f: \{\heartsuit,\spadesuit,\diamondsuit,\clubsuit\}\to \{0,1\}$ defined by $f(\heartsuit)=f(\clubsuit)=0$ and $f(\spadesuit)=f(\diamondsuit)=1$. What would you want $g$ to be in this case?
Theorem (using either definition version 2 or 3): If in our category of choice all pushouts exist and all epi’s have sections (the axiom of choice, I believe), then the natural projections of a binary product are each-others remainder.
Proof: Let $f : X \times Y \rightarrow X$ and $g : X \times Y \rightarrow Y$ be natural projections of $X \times Y$. It is a standard result that $(f,g)$ is jointly monic. Furthermore, let $g' : X \times Y \rightarrow Z$ be such that $(f,g')$ is jointly monic. Then we can create the pushout $g'$ and $g$, given by maps $p : Y \rightarrow P$ and $q : Z \rightarrow P$. It is a standard result that $p$ and $q$ are epimorphisms. Let $p' : P \rightarrow Y$ be the section of $p$, then we define $h = p' q$ and find $h g' = p' q g' = p' p g = g$. So we may conclude that $Y$ is a remainder of $f$, witnessed by $g$.
If I didn’t make any mistakes, in Set the remainders (if they are indeed unique) therefore behave as I would like them to for the time being.
Here’s an analogy that might be helpful: given a subspace of a vector space $U\subseteq V$, there is no canonical way to single out a complementary subspace $W$ such that $U\oplus W = V$. What you can do is consider the quotient space $V/U$, which plays a similar role. Or if $V$ is equipped with an inner product, you can define the orthogonal subspace $U^\perp$.
@Mike. Yes, I see. But I’m not directly concerned with the choice of $g$ exactly. I’m concerned with finding the object $Y$…
From a category-theoretic point of view, you shouldn’t expect to determine an object without the morphisms that structure it.
You may be on to something with the non-uniqueness however; a weakly terminal object in some category could be what you want.
Thanks for the analogy.
Division between objects would be a ’quotient object’. Division between maps would be a ’quotient map’. Division between an object and map would be what?
In my use-case, I have a map from an object representing a system behavior to an object representing a component, and I would like to find a specification of the remaining components. I’m “sideways” interested in finding a map into this specification as well, but am also happy to allow multiple of such maps. So perhaps version 3 of my definition is closest to my intuition at the moment.
Hm. How do you quote in this forum?
Thanks for the remark that I should not expect to determine an object without the morphisms that structure it.
It is “easy” to see that any two remainder are adjoint by definition. So perhaps that is the best I can hope for?
I’ve been reading up on congruence to understand quotient spaces better, but they have the ’opposite’ effect of what I want to achieve, it seems.
What reasoning leads you to dub this a ’remainder’? What is your working example that you want to apply this concept to?
Re: terminology, a “quotient object” is actually technically already a quotient of an object by a map; you can’t divide one object by another without a morphism to relate them. Usually a “quotient map” is the map into a quotient object that gives it its universal property.
What do you mean by “adjoint”?
You could try looking at pushout complements and span rewriting, which are another way to “delete part of an object and replace it with something else”.
How do you quote in this forum?
Prefix by >
, as in
> How do you quote in this forum?
If you want to find out how someone on the nForum does something – quote, link, etc. – you can just find an instance and click on ’Source’.
What reasoning leads you to dub this a ’remainder’? What is your working example that you want to apply this concept to?
I still have some doubts on whether it should be called remainder. Actually ’fraction’ may be a better term.
As I mentioned in the opening of my post, I have a model of behavior of an entire (software) system, and I have a model of one of its components. Categorically, the relation between an entire system and its components can be seen as a jointly monic family. But in my current situation I have only been given a part of that family, and I want to make a “best guess” as to what the “remainder” may be, or the “fraction” of what I’m missing.
The example from Set suggested by Mike shows that pretty well. If you have been given only one projection out of a pair of canonical projections from a product, my intuition tells me that finding the remainder means finding the other projection. In case products exist in a category, you can rephrase my question as: $f : X \rightarrow Y$ has been given, now construct $Z$ and a monomorphism $X \rightarrow Y \times Z$ that agrees with $f$. In other words, given $X$ and $Y$ you are finding the smallest $Z$ such that $X \subseteq Y \times Z$.
Does that make sense?
The ’Source’ button is usefull ;-)
You could try looking at pushout complements and span rewriting
That would require me to know more than ’just’ the map $f : X \rightarrow Y$ to start with.
But the suggestion is helpful in the sense that, in my definition, I could require an epimorphism into the remainder/fraction (I’m starting to prefer the term fraction, thanks for questioning David), rather than just a map. Switching to weak final objects leads to non-uniqueness, but requiring epi restores some of that.
In summary so far, working through the above, how about this:
Definition: given $f : X \rightarrow Y$, a map $g : X \rightarrow F$ is a fraction of $f$ if $(f,g)$ is jointly monic and for any jointly monic pair $(f,h)$ there exists a $k$ such that $k h = g$.
Uniqueness theorem: if $g : X \rightarrow F$ and $g' : X \rightarrow F'$ are epic fractions of $f : X \rightarrow Y$, then $F$ and $F'$ are isomorphic.
Proof: (I guess this is a standard result for weakly final epis, but I am a real newbe, so I need to do this for my own understanding.) Consider $g : X \rightarrow F$ and $g' : X \rightarrow F'$ to be epimorphic fractions of $f : X \rightarrow Y$. Then there exist $k$ and $k'$ such that $k g = g'$ and $k' g' = g$. So $1 g = g = k' g' = k' k g$ and because $g$ is epic we find $k' k = 1$. Similarly, $k k' = 1$. Q.E.D.
Basic intuition theorem: consider the natural projections $f : X \times Y \rightarrow X$ and $g : X \times Y \rightarrow Y$ of a product $X \times Y$, then $f$ is an epic fraction of $g$ and $g$ is an epic fraction of $f$.
Proof: that projections are epic is standard, and that the theorem is symmetric is clear, so we only need to prove that $g$ is a fraction of $f$. How to go about that is not so clear to me. Maybe it is not even true in general, in which case my definition might need more tweaking.
The definition at least works for the $2 \times 2$ example in Set, and also seems to work for my own use-case examples. Next, I think it is worthwhile to see what tweaking is necessary to get the basic intuition theorem to work out…
So here’s an interesting test case: Take for $f\colon X\to Y$ some surjective function. Special cases of this are projections $X = Y\times F \to Y$, for some standard fibre $F$. But how to proceed when all the fibres are not necessarily isomorphic? In a suitable Boolean lextensive category (not sure of the addition extra requirements) one can do things like decompose $X$ into a disjoint sum of the subobject on which $f$ is injective, and the rest.
I’m trying to follow, but I lack a lot of category theoretic ’basics’ I’m afraid.
Special cases of this are projections X=Z×F→Y, for some standard fibre F.
I needed to read-up on fiber for this, but it doesn’t give me enough info on what you mean by this sentence. Can you spell your construction out in smaller steps, please?
David just means a fixed object $F$, which then happens to be the fiber of the projection $Y\times F\to Y$.
As an example of his question, consider $f:\{\heartsuit,\spadesuit,\clubsuit\}\to \{0,1\}$ where $f(\heartsuit)=f(\spadesuit)=0$ and $f(\clubsuit)=1$. Here it doesn’t seem to me that any fraction exists, since both $g:\{\heartsuit,\spadesuit,\clubsuit\}\to \{0,1\}$ where $g(\heartsuit)=f(\clubsuit)=0$ and $g(\spadesuit)=1$ and $h:\{\heartsuit,\spadesuit,\clubsuit\}\to \{0,1\}$ where $h(\clubsuit)=f(\spadesuit)=0$ and $h(\heartsuit)=1$ are jointly monic with $f$, and “maximally” so in that they don’t remain so after composing with any non-injective function out of their codomains; but neither factors through the other.
Thank you! That is a very nice example to chew on. It actually works in my category of behaviors of components as well. But I do notice that, in your example, $g$ and $h$ simply treat $\heartsuit$ and $\spadesuit$ in a ’reverse’ fashion, undistinguishable by $f$. So my feeling is that both should be considered equivalent in some way, perhaps by allowing an auto-isomorphism on $\{ \heartsuit, \spadesuit, \clubsuit \}$.
I.e. fractions should disregard symmetries on $X$. I wouldn’t have found that on my own :-)
Thanks, Mike. That was indeed a typo, should have had $Z$ be $Y$ instead, so I’ve edited.
Here’s another thought: surjective functions with domain $X$ are the same as equivalence relations on $X$, and although your $f$ isn’t assumed surjective, it only matters up to its image, hence up to its kernel pair which is the equivalence relation it determines. On this side, I think that what you are looking for is, given an equivalence relation $R$ on $X$ (the kernel of $f$), you want another equivalence relation $S$ on $X$ such that $R\cap S = \Delta_X$, the diagonal relation, and which is maximal with this property.
Yes, that might work too, I suppose. (According to kernel you actually mean kernel pair, right? The pullback of $f$ along itself.)
After a nights sleep, I figured out how to adapt my own definition a bit as well, so now I have two definitions which both seem reasonable (for the time being).
Theorem 1: Fractions are unique up-to isomorphism.
Proof: let $F$ and $F'$ both be fractions of $f$. Observe that $(f,1)$ is jointly monic. Applying the definition of fraction gives us maps $g : X \rightarrow F$ and $g' : X \rightarrow F'$ such that $(f,g1)$ and $(f,g'1)$ are jointly monic, and $g = g1$ and $g' = g'1$ are epimorphisms. Applying the definition of fraction again gives us maps $h$ and $h'$ such that $hg = g'$ and $h'g' = g$. From this we obtain $1g = g = h'g' = h' h g$, and because $g$ is epic, $1 = h' h$. Similarly, $1 = h h'$. So $h$ and $h'$ are isomorphisms between $F$ and $F'$.
Conjecture: Given a natural projection $f : X \times Y \rightarrow X$, if $f$ has a fraction, then it is isomorphic to $Y$.
Conjecture: Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.
And my first quick attempt to formalize Mike’s suggestion, while keeping it ’comparable’ to the first definition. (I’m not sure if I got the translation into arrows entirely right, please have a look).
Definition 2 : Given a map $f : X \rightarrow Y$. Let $k,k' : R \rightarrow X$ be the kernel pair of $f$ (the pullback of $f$ over itself). An object $F$ is a fraction of $f$ if there exists an epimorphism $g : X \rightarrow F$ for which the kernel pair $h,h' : S \rightarrow X$ (the pullback of $g$ over itself) has the property that every $m : X \rightarrow Z$ with $hm = h'm$ and $km = k'm$ is a monomorphism.
Conjecture: Fractions are unique up-to isomorphism.
Conjecture: Given a natural projection $f : X \times Y \rightarrow X$, if $f$ has a fraction, then it is isomorphic to $Y$.
Conjecture: Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.
I’ll get back later to seeing whether I can prove any of the conjectures, or find conditions under which they can be proven.
Definition 2 : Given a map $f : X \rightarrow Y$. Let $k,k' : R \rightarrow X$ be the kernel pair of $f$ (the pullback of $f$ over itself). An object $F$ is a fraction of $f$ if there exists an epimorphism $g : X \rightarrow F$ for which the kernel pair $h,h' : S \rightarrow X$ (the pullback of $g$ over itself) has the property that every $m : X \rightarrow Z$ with $h m = h'm$ and $k m = k'm$ is a monomorphism.
Close! But I think the inclusions go the other direction: $h m = h' m$ means that the kernel pair of $m$ contains $(h,h')$, not conversely. Also there should be some kind of maximality condition, e.g. that their union in the poset of equivalence relations is the full relation, i.e. they are complements in that poset. Finally, an object in category theory should be equipped with a morphism rather than just be such that a morphism exists (e.g. a cartesian product is equipped with projections $A\leftarrow A\times B \to B$). So if we turn all the equivalence relations into epimorphisms, I would state the definition something like this:
Definition 2: Given a map $f : X \rightarrow Y$, a fraction of $f$ is an epimorphism $g:X\to F$ such that (1) if $e:X\to Z$ is an epimorphism such that $f$ and $g$ both factor through it, then $e$ is an isomorphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).
However, with this definition as stated, your first conjecture
Fractions are unique up to isomorphism.
is false in complete generality. Consider any lattice $P$ as a thin category and let $x\in P$ be an object that has more than one distinct complement, i.e. $y,y'$ with $x\wedge y = x\wedge y' = 0$ and $x\vee y = x\vee y' = 1$. Then since every morphism in a thin category is epi, $0\to y$ and $0\to y'$ are nonisomorphic fractions of $0\to x$. This is also a counterexample to your second conjecture, since $0 = x\times y = x\times y'$.
However, a more reasonable formulation would be to add conditions on the category and/or restrict to some well-behaved subclass of epimorphisms like strong, extremal, or regular ones. Of course it’s still false if you ask for the isomorphism to respect the morphisms from $X$, but absent that condition I don’t yet know the answer (though there could still be an easy one, I haven’t thought about it very hard).
Moreover, your third conjecture:
Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.
is almost true, in good cases. It’s not quite true even in $Set$ because $X\times Y \to Y$ may not be an epimorphism (consider $X=\emptyset$), but it is true in $Set$ if both $X$ and $Y$ are nonempty. More generally, I believe it is true in any regular category if instead of “epimorphism” we say “regular epimorphism” (or assume that all epis are regular, as in a pretopos) and assume that $X$ and $Y$ are well-supported, i.e. $X\to 1$ and $Y\to 1$ are regular epi. (The proof I have in mind uses the internal language, though, and I haven’t tried to compile it out in terms of arrows.)
I feel like the structure of the lattice of equivalence relations (or quotients) of an object in a regular category must have been studied before, but I don’t know where to find it. Perhaps stackexchange might help.
Classically, the lattice of equivalence relations on a set forms a geometric lattice or a matroid lattice. I don’t think I’ve ever considered the situation for more general regular (or exact) categories.
Definition 2: Given a map f:X→Y, a fraction of f is an epimorphism g:X→F such that (1) if e:X→Z is an epimorphism such that f and g both factor through it, then e is an isomorphism; and (2) the pushout of f and g is constant (factors through the terminal object).
Thanks Mike, I already felt actually creating the kernel pairs was cumbersome. I’m still chewing on the ’constant’ part, which is a new trick for me.
My thinking was that if $f$ is also epi, then the pushout should be the terminal object, since that would correspond to the union of the kernel pairs of $f$ and $g$ in the lattice of equivalence relations being the full relation. If $f$ is not epi then whatever extra stuff is in its codomain will still be there in the pushout, but the map from $X$ to the pushout factors through the image of the image of $f$, hence through the terminal object.
@Todd, thanks! Do you know if anything is known about the relation of “having isomorphic quotients” on that lattice?
Mike, I haven’t been following this thread, and so I may be misunderstanding or maybe I’m being dense, but: what do you mean? I consider two quotients $q: X \to Y$ and $q': X \to Y'$ to be isomorphic if there is an isomorphism $Y \to Y'$ making the obvious triangle commute. In that case, $q$ and $q'$ correspond to the same equivalence relation. So I don’t know what you’re asking.
I mean an isomorphism $Y\cong Y'$ not making the obvious triangle commute.
Trying to relate your definition to mine, I discovered the following (which is probably already obvious to you).
Theorem: in a category in which all products exist $(f,g)$ is jointly monic if and only if any map $m$ that factors through $f$ and $g$ is monic.
That any epi that factors through $(f,g)$ is iso would then be true in any category where mono+epi implies iso.
in a category in which all products exist $(f,g)$ is jointly monic if and only if any map $m$ that factors through $f$ and $g$ is monic.
That doesn’t sound right. The map to the terminal object always factors through $f$ and $g$, but isn’t generally monic. Maybe I misunderstood what you mean?
Ah, or maybe I misunderstood what you mean by “factors through”. I couldn’t find a formal definition… I’ll elaborate on what I found and make it self-contained, including the standard definitions. (Is there a way to put diagrams in posts?)
Definition: a product of $Y$ and $Z$ is an object $Y \times Z$ with projections $k : Y \times Z \rightarrow Y$ and $l : Y \times Z \rightarrow Z$ such that for any pair $f : X \rightarrow Y$, $g : X \rightarrow Z$ there is a unique map $h : X \rightarrow Y \times Z$ such that $kh = f$ and $lh = g$.
Theorem 1: given a pair $(f,g)$ of jointly monic maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$, let $h : X \rightarrow B$ be such that there exist $k : B \rightarrow X$ and $l : B \rightarrow Y$ with $f = kh$ and $g = lh$, then $h$ is monic.
Proof: assume two $i,j : A \rightarrow X$ such that $hi = hj$, then $fi = khi = khj = fj$ and $gi = lhi = lhj = gj$, so by joint monicity $i = j$. Therefore, $h$ is monic.
Theorem 2: assuming products exist, let $(f,g)$ be a pair of maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$. Furthermore, assume that for any $h : X \rightarrow B$ with $k : B \rightarrow X$ and $l : B \rightarrow Y$ such that $f = kh$ and $g = lh$ we find $h$ is monic. Then $(f,g)$ is jointly monic.
Proof: Observe that in particular $h : X \rightarrow Y \times Z$ is assumed to be monic, with $k : X \times Y \rightarrow X$ and $l : X \times Y \rightarrow Y$ the natural projections from the product. Now, let $i,j : A \rightarrow X$ be a pair of maps such that $fi = fj$ and $gi = gj$. Then by definition of product there is a unique arrow $u : A \rightarrow Y \times Z$ such that $ku = fi$ and $lu = gi$, as well as a unique arrow $u' : A \rightarrow Y \times Z$ such that $ku' = fj$ and $lu' = gj$. Because $fi = fj$ and $gi = gj$ we find $u = u'$, and because $khi = fi$ and $lhi = gi$ we find $u = hi$ and similarly $u' = hj$. Therefore, $hi = hj$. We assumed $h$ to be monic, therefore $i = j$, which concludes the proof that $(f,g)$ is jointly monic.
Ah, I see the confusion. When $f = h k$ we say that $f$ factors through $h$ (since writing $f = h k$ is “factoring” $f$, analogously to $6 = 2\cdot 3$), not that $h$ factors through $f$.
Thanks… But then, when $f = hk$, do you also say $f$ factors through $k$? In that case, using the sentence “factors through” without an indication of left or right is ambiguous…
Now I’ll reread the posts above to see if they make sense again/still to me ;-)
The little theorem is what it is, of course. But I guess it is also pretty standard for people working with joint morphisms often. I guess so standard that there is no reference for it… Does something like that deserve a wiki-page or comment according to you? There is value in such ’trivia’, but there is also a big risk of a wiki become just a collection of such trivia then…
Back to your definition:
Definition 2: Given a map $f: X \rightarrow Y$, a fraction of f is an epimorphism $g : X \rightarrow F$ such that (1) if $e : X \rightarrow Z$ is an epimorphism such that $f$ and $g$ both factor through it, then $e$ is an isomorphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).
In a category where epi+mono implies iso, this is equivalent to saying
Which in a category where products exist comes down to
That definition is closer to my goal of treating compositions as relations between objects, and under mild conditions coincides with your intuition according to my previous post. Nice.
Now, I’ll start studying the factoring through a terminal object as a way to obtain isomorphism of fractions under reasonable conditions :-)
using the sentence “factors through” without an indication of left or right is ambiguous
Technically, yes. Although that would be true regardless of which of $f$ or $h$ is said to “factor through” the other. But if $f$ and $h$ share a domain but not a codomain, or a codomain but not a domain, as is usually the case, then the ambiguity is resolved as only one choice is possible.
Does something like that deserve a wiki-page or comment according to you?
Yes, I would probably generalize it to a statement about one family of morphisms factoring through another jointly-monic family, and add it in a “Properties” section to jointly monic family. The special case when both families are single morphisms is already recorded at monomorphism#properties: if $g f$ is a monomorphism, so is $f$.
I’ll give it a shot as soon as I have the time :-)
Now, I’ll start studying the factoring through a terminal object as a way to obtain isomorphism of fractions under reasonable conditions :-)
Firstly, if I’m not mistaken, using the existence of pushouts is not necessary, because the terminal object is supposed to be the pushout… so
Following your idea (I think) the extension of $f$ and $g$ into a commuting square represents the “union, closed under symmetry and transitivity” of the equivalences represented by the original maps $f$ and $g$. That sounds okay to me. If this union is constant, i.e. factors through the terminal object, that implies that “all information is thrown away” by the new equivalence, i.e. everything is equated.
That last part is close, but not the same I think, as my intuition that: whenever $f$ makes a distinction, then $g$ should try to avoid making this distinction.
The tricky part is that in some categories, $g$ may not be able to avoid making distinctions. In my category of choice, a terminal object exists, but nonetheless I’m not sure that $g$ is free to avoid making distinctions (okay perhaps that just means the distinctions will show up in the terminal object then… true… but this is subtle). I’d have to think about quite examples for a while to make this more concrete though. Furthermore, I don’t like to have to use the terminal object as part of my definition, because I think that having a constant map implies that all information is thrown away, rather than that it is the same as saying all information is thrown away. In a sense, using the terminal object makes the definition stronger than I intend it to be.
Some more thought…
So given maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$ the intuition is that whenever $e = hf = kg$, the map $e$ equates whatever is equated by $f$ or $g$. If we want to state that $e$ equates ’everything’, we could state that as $e$ is constant, but this requires the existence of a terminal object. Alternatively, we could perhaps also state that $X \times X$ is the kernel of $e$?
Pity I still have to assume existence of the product $X \times X$, but imho better than using a terminal object (is there a difference whenever both exist?). Also a pity that $g$ being epic does not simply follow from a “smallest solution” condition, and that I do not immediately see under which conditions we get uniqueness of fractions. I agree with you, however, that this may in general be unreasonable to assume, but should work out in reasonable conditions.
What do you have against terminal objects?
In general, $X\times X$ being the kernel of $e$ is the same as saying that for any two morphisms $u,v$ with codomain $X$ we have $e u = e v$, i.e. that $e$ is a constant morphism in the weak sense. If $e$ has an image, this is the same as saying that that image is subterminal, and I think it would be better for it to be terminal. But note that at constant morphism there is given a way to phrase the strong sort of “constancy” without assuming the existence of a terminal object, so if you really happen to be in a category without a terminal object (but with pullbacks and products? I can’t think of too many categories like that) then you could use that.
1 to 49 of 49