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• CommentRowNumber1.
• CommentAuthorPeter Heinig
• CommentTimeJun 3rd 2017

Created unnatural isomorphism, with references.

A cleaner working out and linking between the concepts of

• unnatural isomorphism (a structure which can of course also be constructed in situations where there is a natural isomorphism

and

• unnaturally isomorphic (a property)

appears to be a worthwhile thing to do.

1. I changed “it is impossible to construct a natural isomorphism” to “there is no natural isomorphism”, since these could be different if you think that “construct” means working constructively.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJun 3rd 2017

Maybe I’m being dense, but I don’t really “get” the example of Brandenburg that you describe. (His example 2 is clearer in my opinion.)

2. I think what Brandenburg is saying in his MO answer is that two functors $\{\bullet\to\bullet\}\rightarrow \mathsf{C}$ are unnaturally isomorphic if they land on two parallel arrows that aren’t isomorphic.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJun 3rd 2017

Thanks. And right, that’s what Brandenburg says. But that’s not what the article says.

3. I changed it to

Two functors from the interval category, $I$, to a category $C$ are naturally isomorphic only if the morphisms they land on are isomorphic in $C$‘s arrow category. But they will be unnaturally isomorphic if they land on any parallel pair of arrows.

• CommentRowNumber7.
• CommentAuthorPeter Heinig
• CommentTimeJun 3rd 2017

Thanks. The previous version was indeed not clear enough, mostly as a result of not clearly structuring the article along a isomorphism-of-categories-isomorphism-of-functors-distinction. Will be improved further.

• CommentRowNumber8.
• CommentAuthorPeter Heinig
• CommentTimeJun 15th 2017
• (edited Jun 15th 2017)

Both polished and abbridged unnatural isomorphism, to make it reasonably clean. I have a large document, written as an exercise, containing examples in a uniform notation, but I am not satisfied with it and will not include it for the time being. Needless to say, others are free to include examples. Part of the idea of the article is to not just duplicate the many existing examples floating around, but to cleanly and systematically present them. There appear to be more important things to do that that though.

One central exposition-problem is to explain that, in a sense, all examples can be seen as failure of generalized equivariance (w.r.t. monoid actions).

Moreover, keeping track of objectwise isomorphic pairs of assignations seems to call for methods of higher category theory.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeJun 15th 2017

I’m not sure I really get (or agree with) the example of the Yoneda isomorphism $Set^{C^{op}}(C(-, c), F) \cong F c$ being non-canonical.

Of course I agree there can be more than one natural isomorphism of that form, for the reason given: that $1_C$ may have nontrivial automorphisms. However, I would reckon that the identity automorphism is canonical one among all automorphisms. And identities are what we use: the standard isomorphism above takes $\theta: C(-, c) \to F$ to $\theta(1_c)$.

On another matter: I think full citations to MathOverflow posts (author, title, version, date, etc.) should be given. The cite button on MO posts makes this very easy.

• CommentRowNumber10.
• CommentAuthorTim_Porter
• CommentTimeJun 15th 2017
• (edited Jun 15th 2017)

I always taught that $1_c$ is used since it is the ONLY morphism that we know exists that has codomain $c$. It therefore surely is meta-canonical!

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeJun 15th 2017

Also, for the purposes of the introducing a theoretically useful notion, I wonder whether unnatural isomorphism should really be defined as an isomorphism of the form $U(F) \cong U(G)$ where $U: Set^C \to Set^{|C|}$ is the evident forgetful functor ($|C|$ denotes the underlying set/discrete category of $C$). In other words, invoke the red herring principle where for example a “nonassociative algebra” might in fact be associative and “unnatural isomorphism” might in fact be natural, it’s just that they don’t have to be.

For example, under the red herring version, unnatural isomorphisms are closed under composition. Under the current negative version, they aren’t.

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeJun 15th 2017

I agree completely with #11, and that the Yoneda isomorphism is canonical in any sense one might want.

On the other hand I don’t think it is necessary to give full citations to anything on the nLab. The author of the thing cited should certainly be mentioned, by way of giving proper credit. But beyond that, the purpose of the rest of the data in a citation is to enable the reader to find the thing cited, and a hyperlink surely suffices for that.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeJul 6th 2017

I did some major reorganization and rewriting of unnatural isomorphism. In particular I changed the definition along the lines suggested in #11, and removed some things that didn’t make any sense to me.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeJul 6th 2017

I also added the example of the two functors relating vector spaces and affine spaces, which are one of my favorites.

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeJul 6th 2017

Thanks very much, Mike. I added more to the Examples section, and worked on the Reference section (and the links thereto). (One of the links referred to a comment on triangulated categories under a MO question which in turn was just a bare Wikipedia link; I just linked directly to Wikipedia there.)

I also tried to replace indirect allusions with more direct descriptions; for example one bit of the mockery referred (I guess) to something Weil said; I just quoted him directly. The other referring to a German textbook: what is the direct quote, please? (If it’s published, then why not?)

• CommentRowNumber16.
• CommentAuthorPeter Heinig
• CommentTimeJul 7th 2017
• (edited Jul 12th 2017)

Many thanks for the edits.

Will not have time to read it carefully today, but made one simple edit (see below).

I also tried to replace indirect allusions with more direct descriptions; for example one bit of the mockery referred (I guess) to something Weil said; I just quoted him directly. The other referring to a German textbook: what is the direct quote, please? (If it’s published, then why not?)

Thanks. Yes, in one instance it was A. Weil who was meant. As for the other instance: it seems better to not mention this. (I have simply removed it.) When writing the draft of the nLab article I thought (and still think) that a dose of such descriptive material (describing the literature) is useful, scientific and engaging. In that particular instance, for various reasons, I will not make that explicit. Giving reasons for why not, even here, would more or less do what not mentioning it is trying to avoid.

To get back to mathematics, the project to make this into an informative nLab article for me mainly stalled because I was (and am, but with low priority) working on a satisfactory treatment of three further examples (in no case did I reach a publishable exposition yet):

(0) One can generalise, and work-out-the-essence-of, the widely known example of the

endofunctor-of-Vect-which-takes-any-vector-space-to-the-“same”-vector-space-but-with-the-“twisted”-ring-action-which-maps-pairs-($\lambda$,$v$)-not-to-($\lambda\cdot v$)-as-usual-but-to-( $(-\lambda)\cdot v$)

for example by generalising it to “twisted” actions of commutative rings on abelian groups (thus making it an example about modules-in-the-commutative-algebra-sense), with an appropriate use of the concept of equivariant monoid-actions.

[EDIT: I’ll leave that uncensored to keep the thread understandable, but note: as it was given in the short-form above, this does not define a vector space.]

(1) There is something of a staple of category theory textbooks, which appears to admit of a fruitful treatment in the context of unnatural isomorphisms: the “unreasonableness” of defining exponential objects “pointwise” (as opposed as indirectly via Yoneda). Some treatments, as something of an expositional ritual, briefly touch upon the pointwise definition of exponential objects, choose one verbal dismissal or another, but not making it clear what the pointwise definition is and what it is not, and then launch into the “correct” definition of $O_0^{O_1}$. It seems to me that one can instructively treat this classic material in the context of unnatural isomorphism. (No time to elaborate on this now.)

(2) There appears to be a precise sense in which existential quantification is a functor unnaturally isomorphic to universal quantification. While the treatment of $\exists_f$ and $\forall_f$ as functors is of course widely known, making that connection, and this kind of exposition, seems to be new. (Again, will try to write more on this soon, but not now.)

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeJul 7th 2017

the “same” vector space but with the “twisted” ring action which maps pairs ($\lambda$,$v$) not to ($\lambda\cdot v$) as usual but to ( $(- \lambda)\cdot v$)

Um, I don’t think that’s a vector space. You need $(1,v)\mapsto v$, for instance.

the “unreasonableness” of defining exponential objects “pointwise” (as opposed as indirectly via Yoneda)

I have no clue what you mean here. What is the “pointwise” definition of exponential objects?

There appears to be a precise sense in which existential quantification is a functor unnaturally isomorphic to universal quantification

I can think of all sorts of reasons that that can’t be true. One of the more obvious is that, at least in the simplest case, both are functors between posets, so that any unnatural isomorphism between them would automatically be natural.

• CommentRowNumber18.
• CommentAuthorPeter Heinig
• CommentTimeJul 7th 2017
• (edited Jul 9th 2017)

Will be forced to be brief and to focus one of them for the time being (hope to come back to the others soon):

Um, I don’t think that’s a vector space. You need (1,v)↦v(1,v)\mapsto v, for instance.

Right, I was writing carelessly, and I was abusing the minus-sign as (wrong, of course) shorthand for a non-identity-ring-automorphism (as you rightly point out, $x\mapsto -x$ is not a ring-automorphism, and I knew that).

Explicitly, here is a simplified (in particular, by restricting to rank one) version of the kind of examples for (0) that I meant:

$R:=$ a non-trivial commutative ring

$\mathcal{C}:=$ the full subcategory of $R$-Mod consisting of the free rank-one $R$-modules only

$U:=$the faithful functor $\mathcal{C}\rightarrow Set$ which sends any morphism of $\mathcal{C}$ to the underlying set-map

$F_0$ $:=$ the identity-endofunctor of $\mathcal{C}$

$t$ $:=$ some non-identity automorphism of $R$

$F_1(t)$ $:=$ the endofunctor of $\mathcal{C}$ which

• sends any $O\in Ob(\mathcal{C})$ to the “twisted” rank-one free $R$-module $F_1(O)\in Ob(\mathcal{C})$ which has the same underlying abelian group $A=A(O)$ as $O$ does, but has the $R$-action $\odot\colon R\times A\rightarrow A$ given by $r \odot v := t(r)\cdot v$, where $\cdot$ denotes the $R$-action of the module $O$,

• sends any module-morphism $O_0\overset{f}{\rightarrow}O_1$ of $\mathcal{C}$ to the morphism $F_1(O_0)\overset{F_1(f)}{\rightarrow}F_1(O_1)$ of $\mathcal{C}$ with

• for each $v\in A$ we have $U(F_1(f))(v) = U(f)(v)$

If I am not mistaken, then $F_0$ and $F_1$ are unnaturally isomorphic but not naturally isomorphic.

Examples of this kind are widely known.

My attempt to document them in a general, presentable and informative form foundered on attempts

• to adapt to the “red herring” definition of “unnatural isomorphism”,

• to write a section analysing the role of the “concretization” functors $U$ in such examples,

• to write generalising sections on when and if an identity endofunctor is unnaturally-but-not-naturally-isomorphic to another endofunctor, using the concepts of monoid-actions and equivariance.

Somewhat curiously, looking through my notes on unnatural isomorphisms, I found, in a portion on the above type of examples, a remark reminding readers that any ring-automorphism must preserve $1$, so $t(1)=1$, and $\odot$ acts as it should, and that $R$ cannot be taken to be $\mathbb{R}$; a comment which, had I remembered it when writing from memory comment #16, would perhaps have prevented me from this slip-of-the-key.

• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeJul 7th 2017

Why so elaborate though? You can regard a ring $R$ as a one-object ($Ab$-enriched if you like) category, and $1_R, t$ as endofunctors on $R$. The identity on $R$ then gives an unnatural isomorphism from $1_R$ to $t$.

Correct me if I’m wrong, but the operation you are describing is then “whiskering” that unnatural isomorphism with modules $A$ regarded as functors:

$R \overset{\overset{\longrightarrow}{\Downarrow}}{\underset{t}{\to}} R \stackrel{A}{\to} Ab.$

where the unlabeled arrow is the identity. This gives an unnatural isomorphism $Ab^R \overset{\overset{\longrightarrow}{\Downarrow}}{\underset{Ab^t}{\to}} Ab^R$. You are then restricting this from $Mod_R = Ab^R$ to $\mathcal{C}$ (why I’m not sure, since the operation extends to the generality indicated here – maybe that was an artifact of working with the wrong “unnatural” before?).

Actually – and this should perhaps be added to the article – the red herring notion of unnatural transformation (called “mere” transformations here, leaving off the word “natural”) arises “naturally” by regarding the 1-category $Cat$ as enriched in itself via the “funny” closed monoidal structure on $Cat$: if $[C, D]$ is the internal hom with respect to this structure, then morphisms in the category $[C, D]$ are exactly mere transformations. The whiskering operation mentioned above can be derived formally from this closed monoidal structure.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeJul 7th 2017

Do we have any material already in the nLab on the funny symmetric monoidal closed structure on $Cat$? A cursory search hasn’t turned up anything so far.

• CommentRowNumber21.
• CommentAuthorMike Shulman
• CommentTimeJul 8th 2017

Todd, I don’t think it actually does give an unnatural isomorphism between endofunctors of $Ab^R$. That would mean any $R$-module $M$ is isomorphic, as an $R$-module, to itself with twisted action, but I don’t think that’s true. It does happen to work for free rank-one modules, but that’s basically just an accident because the category of free rank-one modules has only one isomorphism class of objects. So there’s really nothing going on in this example that has anything to do with rings or modules; it’s just the fact that any two endofunctors of a one-object category are unnaturally isomorphic.

It looks like the only mention of the funny tensor product is a brief paragraph at Gray tensor product.

• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeJul 8th 2017

Mike: thanks; yes, of course. I seem to have made an unconscious slip, writing the ordinary internal hom $Ab^R$ instead of a funny internal hom $[R, Ab]$.

• CommentRowNumber23.
• CommentAuthorPeter Heinig
• CommentTimeJul 9th 2017
• (edited Jul 12th 2017)

Why so elaborate though?

Because merely giving the example (which I remembered from somewhere, though I do not recall the source) with $R:=\mathbb{C}$, $t:=$complex conjugation, and $\mathcal{C}:=$ the category of 1-dimensional $\mathbb{C}$-vector spaces, seems unsatisfactory and I wanted to understand what the general reasons are for why a contradiction can be derived from the hypothesis of naturality. This made me consider free modules over commutative rings and “concretization functors”.

It does happen to work for free rank-one modules, but that’s basically just an accident because the category of free rank-one modules has only one isomorphism class of objects. So there’s really nothing going on in this example that has anything to do with rings or modules; it’s just the fact that any two endofunctors of a one-object category are unnaturally isomorphic.

Very illuminating, thanks. And apologies for not having said in #18 why I made this restriction to rank one. Basically, I did not say so because I was not sure whether it was necessary: I only knew it was sufficient, since in my notes I need it in a calculation to get to a contradiction (from the hypothesis of the functors being naturally isomorphic). At one point, one seems to need the lemma that “every endomorphism of the module under consideration is of the form $v\mapsto a\cdot v$”, for some fixed $a$, which is true in the rank-one setting, a property which in my notes I called (toy.Riesz), alluding to the Riesz-representation theorem.

• CommentRowNumber24.
• CommentAuthorMike Shulman
• CommentTimeJul 9th 2017

I added the remark about one-object categories to unnatural isomorphism as an example, with rank-one free modules as a special case.

• CommentRowNumber25.
• CommentAuthorPeter Heinig
• CommentTimeJul 10th 2017
• (edited Jul 10th 2017)

I have no clue what you mean here. What is the “pointwise” definition of exponential objects?

To respond clearly to this still needs some preparation on my part, but I am working on it, on and off, and should soon post something relevant.

Meanwhile, since it could help me to make the presentation better and more informed, permit me to ask the following two questions:

(vague question)

extrapolating from the usual equivalence of categories $\Sets^J\simeq \Sets/J$, for any given discrete category $J$, what do you consider a usual “sensification” of the nonsensical “equivalence” (the right-hand side is non-sensical, since $\mathsf{C}$ is not known to be an object ot $\Sets$)

for any small category, $\Sets^{\mathsf{C^{op}}} \simeq \Sets / \mathsf{C^{op}}\qquad$ (not.in.general.sensical)

Of course, indexed categories come to mind, but seem not to be strictly relevant.

(terminological question) EDIT: the “usual” way seems to be the one discussed in sufficient detail in Section 2 of the current version of cartesian monoidal category: when the abstractly given monoidal product is symmetric and the coherence conditions given there are true. Sorry for asking this too rashly. I’ll leave the question here, but am not asking for more explanations.

Already purely terminologically, every cartesian monoidal category is a monoidal category. But what is the “usual” way of dealing with the cheap converse question “Is every monoidal category equivalent to a cartesian monoidal category?” This question actually splits into two (modulo a perhaps definitional issue): suppose that every “monoidal category” has all binary products. (Whether this is true I did not stop to think about.) Then the facile question splits as follows: Given a monoidcal category $C$.

• Does there necessarily exist any cartesianmonoidal category equivalent to $C$?

• How to characterize when $C$ is equivalent to the monoidal category having $\times$ instead of the abstractly/additionally-given $\otimes$ as its monoidal product?

• CommentRowNumber26.
• CommentAuthorTodd_Trimble
• CommentTimeJul 10th 2017

Regarding the penultimate question, I guess you really mean monoidally equivalent to $C$. There’s a 2-category whose objects are monoidal categories, whose 1-cells are (lax) monoidal functors, and whose 2-cells are monoidal transformations. The notion of equivalence makes sense in any 2-category, and a monoidal equivalence is by definition an equivalence in that 2-category.

Anyway, in order for a monoidal category $(C, \otimes, I)$ to be cartesian, the monoidal unit $I$ must be terminal, and there is a natural transformation $\delta$ from $1_C$ to the composite functor

$C \stackrel{diag}{\to} C \times C \stackrel{\otimes}{\to} C,$

such that for every object $x$ the data $(x, \delta_x: x \to x \otimes x, !_x: x \to 1)$ is a comonoid structure on $x$ (so $\delta$ is coassociative, and there are counit laws relating $!_x$ to $\delta_x$.

A monoidal category is semicartesian if the monoidal unit is terminal. So for example, the category of abelian groups with its usual tensor product is not semicartesian, much less cartesian, because the monoidal unit $\mathbb{Z}$ is not terminal.

With regard to the first question, there are probably several different directions in which one could give answers. But the one which might be the most useful is the equivalence between presheaves on $C$ and discrete opfibrations over $C$ (taking $C$ to be a small category). If $C_0$ is the set of objects, then a discrete opfibration consists of a set $X$ and an function $F: X \to C_0$ and data $M: C_1 \times_{C_0} X \to X, I: C_0 \to X$ which roughly speaking makes $X$ a module over $C$ (as an internal category in $Set$, which can be defined as a monoid in a suitable sense of endospans). This formulation realizes $Set^{C^{op}}$ as monadic over the slice $Set/C_0$.

• CommentRowNumber27.
• CommentAuthorPeter Heinig
• CommentTimeJul 11th 2017
• (edited Jul 11th 2017)

Many thanks for the detailed answers in #26. It will take some time to digest all of them. One small terminological question, related to your use of

$C \stackrel{diag}{\to} C \times C \stackrel{\otimes}{\to} C,$

is this (it seems I will need the following variant to respond to Mike’s request for clarification precisely):

• Is there a usual way (or even a place where it appears on the nLab?the page on constant funtors seems not to mention it) to refer to the evident “twisted” diagonal functor [EDIT: caveat lector: this is not in general a functor; thanks to Mike Shulman for pointing out]
$C \stackrel{diag^{\mathrm{twist}}}{\to} C^{\mathrm{op}} \times C$

which with regard to what it does is of course identical to $diag$. (Since, of course, $C^{op}$ differs from $C$ only in the $dom$ and $cod$ functions.) If a functor is to be named after what it does, then one might even be justified in calling $diag^{twist}$ by the name $diag$ again. If one takes the stance that a functor’s name should precisely tell the reader what the source and target category are, then this naming would not be justified.There is probably not much to be said here, and relying on a name would of course be somewhat dangerous; one should probably just spell it out as above and then, since the target has been given, using the name $diag$ seems fine.

• CommentRowNumber28.
• CommentAuthorPeter Heinig
• CommentTimeJul 11th 2017
• (edited Jul 11th 2017)

Another mainly stylistic issue closely related to #27: suppose in an exposition one is trying to give one has started with

“Suppose $\mathsf{S}$ is a closed monoidal category. Suppose $\mathsf{C}$ is any small category. Suppose $F_1\colon \mathsf{C}^{op}\rightarrow\mathsf{S}$ is any functor.”

Then in the text follow some other considerations, which need $F_1$ to be as it was introduced above, but which eventually lead one to having to discuss the following, with $[,]$ denoting the internal hom-functor of $\mathsf{S}$:

[EDIT: caveat lector: some variances in the following are wrong. Thanks to Mike Shulman for pointing out. For corrections, see later comments.] “We note that for any fixed object $s_0$ of $\mathsf{S}$ there is a functor $\mathsf{C}^{op}\stackrel{[F_1(-),s_0]}{\longrightarrow}\mathsf{S}$ which is the composite $\mathsf{C}^{op}\stackrel{F_1}{\longrightarrow}\mathsf{S}^{op}\stackrel{[-,s_0]}{\longrightarrow}\mathsf{S}$.”

Here, the you-may-op-the-target-of-a-functor-which-has-been-introduced-to-readers-with-the-unoped-target-issue appears again. In the above, it would be confusing, or even wrong, to introduce $F_1$ to readers immediately having $\mathsf{S}^{op}$ for its target, and yet, in mid-stream, one suddenly needs the functor $\mathsf{C}^{op}\stackrel{F_1}{\longrightarrow}\mathsf{S}^{op}$, and one should perhaps not swap variances in mid-stream.

Do you think it reasonably acceptable to just write it like it was written above, just giving the target-category $\mathsf{S}^{op}$? Or is there something systematic one should do about this in an exposition?

• CommentRowNumber29.
• CommentAuthorMike Shulman
• CommentTimeJul 11th 2017

Your “$diag^{twist}$” is not a functor; it cannot be defined to preserve domains and codomains of morphisms. Similarly, if $F_1 : C^{op} \to S$ then it is not true that $F_1 : C^{op} \to S^{op}$; what we can say instead is $F_1^{op} : C \to S^{op}$ using the fact that $(-)^{op}$ is itself an endofunctor of $Cat$.

• CommentRowNumber30.
• CommentAuthorPeter Heinig
• CommentTimeJul 11th 2017

at #29: Thanks. Then this was the reason why it seemed wrong.

Does it then seem correct (locally I mean, with $F_1$ having been introduced to the readers in the form $F_1\colon\mathsf{C}^{op}\rightarrow\mathsf{S}$) to write

“We note that for any fixed object $s_0$ of $\mathsf{S}$, there is a functor $\mathsf{C}\stackrel{[F_1^{op}(-),s_0]}{\longrightarrow}\mathsf{S}$ which is the composite $\mathsf{C}\stackrel{F_1^{op}}{\longrightarrow}\mathsf{S}^{op}\stackrel{[-,s_0]}{\longrightarrow}\mathsf{S}$.” ?

• CommentRowNumber31.
• CommentAuthorTodd_Trimble
• CommentTimeJul 11th 2017
• (edited Jul 11th 2017)

Re #30: I understand, and think that I might have struggled over this once myself long ago, but nevertheless I write it simply as $[F-, s]: C^{op} \to Set$ (for a given $F: C \to Set$), and I think I practically never see $[F^{op}-, s]$. I can’t swear that there are no established authors who write that, but if I ever saw that, I would consider it somewhat iconoclastic.

Let’s step back a bit. Suppose we have a morphism $f: a \to b$ in $C$, and an object $c$ of $C$. There is an induced map $[b, c] \to [a, c]$. How do we denote it? I think everyone denotes it by $[f, c]: [b, c] \to [a, c]$. It may seem to some struggling with such issues that $f$ in $[f, c]$ should be adorned with something, maybe an $^{op}$, to remind us that it occupies a $C^{op}$ argument, not a $C$ argument. But nobody does that, and I would venture to say that more importantly, no one gets confused by that lack of extra adornment. Probably most people just think of this as a contravariant action of $C$, pure and simple.

We can think of $C$ and $C^{op}$ as having the same objects and morphisms, only composed differently. And correspondingly, for a functor $F: C \to D$, we can think of $F$ and $F^{op}$ as connoting the same pair of functions $F_{Ob}: Ob(C) \to Ob(D), F_{Mor}: Mor(C) \to Mor(D)$. Hence we write $[F-, s]$ for the functor that takes an object $c$ to $[F c, s]$ and a morphism $f$ to $[F f, s]$ in accordance with the last paragraph. Just as there is no confusion in the last paragraph, so there is no confusion here.

• CommentRowNumber32.
• CommentAuthorPeter Heinig
• CommentTimeJul 12th 2017
• (edited Jul 12th 2017)

For various reasons I decided to change my plan to answer Mike Shulman’s request for clarification on (1) in #16 by a polished expostion, and rather make this a more modular and communicative process.

I do think that there is something of substance to be discussed, in particular, something which can result in some small improvements to the nLab.

There are more or less two distinct aspects, so I will try to structure this by writing two separate comments, see below.

• CommentRowNumber33.
• CommentAuthorPeter Heinig
• CommentTimeJul 12th 2017
• (edited Jul 12th 2017)

Comment clarifying what I meant in #16.

I have no clue what you mean here. What is the “pointwise” definition of exponential objects?

Briefly, I mean the treatment of the closed monoidal structure on presheaves in some references (which will not be given).

Preliminary grammatical comment: my writing

but not making it clear what the pointwise definition is and what it is not,

was not the clearest use of English, I’m afraid. This was not meant in the sense “I know what the pointwise definition is, but the texts do not tell readers what it is”. It was meant in the sense “several texts do give a pointwise definition, which henceforth I will denote $(F_0^{F_1})_{unn}$, but then do not make it clear (to me), what $(F_0^{F_1})_{unn}$ is, and what it is not.

Now the mathematics.

Let $\mathsf{C}$ be any small category. Let $\mathsf{S}:=Set$, for brevity. (One can, and perhaps should, treat it for general closed monoidal categories, but let us stick to presheaves in the traditional sense for the time being.)

Then the functor category $[\mathsf{C}^{op},\mathsf{S}]$ is cartesian closed. The most interesting aspect of this (to me) is the existence of internal homs. Let us recall that, by definition, this means the existence, for any presheaf $F_1\colon\mathsf{C}^{op}\rightarrow\mathsf{S}$, of a right-adjoint to the endofunctor $(-)\times F_1\colon [\mathsf{C}^{op},\mathsf{S}]\rightarrow [\mathsf{C}^{op},\mathsf{S}]$. I understand (or so I think) how to prove this. Whether one motivates the definition beforehand (see below) or not, one just gives the usual definiton via a composite natural transformation, checks that this is a right-adjoint, and since adjoints are unique, this proves the existence and uniqueness of internal homs in the presheaf category.

What I would like to discuss are expository/pedagogical/stylistic issues.

Let $F_0^{F_1}$ denote the exponential object in $[C^{op},\mathsf{S}]$ in the standard sense.

Specifically, two things are rather customary in expositions of closed monoidal structure on presheaves:

• (0) Mentioning the class-function

$(F_0^{F_1})_{unn}\colon Ob(\mathsf{C}^{op})\ni O \mapsto [F_1(O),F_0(O)]\in Ob(Set)$,

which is defined since $Set$ has internal homs (so to speak) and dismissing it in various ways before launching into the usual treatment.

It is (more or less), the function $(F_0^{F_1})_{unn}$ what I meant by “pointwise definition of exponential objects” in #16.

• (1) Motivating the definition of $F_0^{F_1}$ on objects by a hypothetical reasoning, before checking that the full definition of $F_0^{F_1}$ satisfies (one of the) definition(s) of a right-adjoint.

I write one comment for each of (0) and (1), respectively.

• CommentRowNumber34.
• CommentAuthorPeter Heinig
• CommentTimeJul 12th 2017
• (edited Jul 12th 2017)

Re (0) in #33:

First of all, a suggestion to make the definition of unnatural isomorphism even more permissive than it currently is in unnatural isomorphism:

• Wouldn’t it be useful to be able to permissibly say that a functor is unnaturally isomorphic to a class function? ${}\qquad$ (permissive)

At least when it comes to explain/describe/comment mathematical practice, then this would free one from the obligation to always first “wrap” one mathematician’s usage of unnatural isomorphic into a functor, and then explain. Arguably, often, in particular if one of the mathematicians is not categorically inclined, they tend to not have any assignation of morphisms to morphisms in mind, or maybe not even any category except $Set$.

Adopting (permissive) for the sake of discussion, it follows terminologically that $(F_0^{F_1})_{unn}$ is unnaturally isomorphic to the functor $F_0^{F_1}\colon\mathsf{C}^{op}\rightarrow\mathsf{S}$.

With the current definition of unnatural isomorphism, $F_0^{F_1}$ and $(F_0^{F_1})_{unn}$ are incomparable, of different types, so to speak.

My main (expository) motivation was to make precise sense of various, often purely verbal dismissals of this class function when it comes to defining $F_0^{F_1}$. One treatment I once saw said something to the effect that this cannot be made into a functor “in any reasonable way”.

In a way, the goal is to write “the correct” dismissal of $(F_0^{F_1})_{unn}$.

What somewhat puzzles me in the treatments (probably there are good mathematical reasons for this that I am not aware of) are two things:

(0.0) Why no treatment that I know of ever discusses if and why the value of $F_0^{F_1}$ on an object of $\mathsf{C}$, namely

• $F_0^{F_1}(O) = [\mathsf{C}^{op},\mathsf{S}] ( \mathsf{C}( - , O ) \times F_1(-) , F_0(-) )$ is isomorphic as an object of $\mathsf{S}$ to the pointwise exponentioal $[ F_1(O) , F_0(O) ] = \mathsf{S} ( F_1(O) , F_0(O) )$.

(0.1) Why no treatment that I know of discusses what seems to me the obvious functorial contender to the usual definition of $F_0^{F_1}$, namely the functor $(F_0^{F_1})_{var}\colon \mathsf{C}^{op}\rightarrow\mathsf{S}$ which is the composite of functors

$\mathsf{C}^{op} \overset{diag}{\longrightarrow} \mathsf{C}^{op}\times \mathsf{C}^{op} \overset{1_{\mathsf{C}^{op}}\times F_0}{\longrightarrow} \mathsf{C}^{op}\times \mathsf{S} \overset{ F_1\times 1_S }{\longrightarrow} \mathsf{S}\times\mathsf{S} \overset{ (-)^{op}\times 1_{\mathsf{S}} }{\longrightarrow} \mathsf{S}^{op}\times\mathsf{S}\overset{internal\quad hom\quad [-,-]}{\longrightarrow}\mathsf{S}$
• CommentRowNumber35.
• CommentAuthorPeter Heinig
• CommentTimeJul 12th 2017
• (edited Jul 12th 2017)

Re (1) in #33:

I do not think it necessary to motivate the definition of $F_0^{F_1}$ (one should get to know a concept by working with it, not by trusting one linear narrative) but if one decides to do give a motivation an linear narrative, shouldn’t one then motivate the usual definition of $F_0^{F_1}$ on morphisms of $\mathsf{C}^{op}$, and not only on objects (which is what some of the texts do). One rather old treatment I know, resolves to give a motivation of the definition, but then only gives the usual suppose-there-were-a-right-adjoint-then-by-the-very-definition-of-adjoints-combined-with-one-standard-application-of-the-Yoneda-lemma-we-are-forced-to-define-the-exponential-on-objects-in-the-usual-way, but then suddenly gives free rein to the expositional horse by just stating the usual definition of $F_0^{F_1}(h)$ as the set map which maps any natural transformation $\theta\colon \mathsf{C}(-,dom(h))\times F_1(-) \Rightarrow F_0(-)$ to the composite

$[\mathsf{C}^{op},\mathsf{S}]( \mathsf{C}(-,dom(h))\times F_1(-), F_0(-) ) \overset{ \mathsf{C}(-,h)\times 1_{F_0} }{\longrightarrow} [\mathsf{C}^{op},\mathsf{S}]( \mathsf{C}(-,cod(h))\times F_1(-) , F_0(-) ) \overset{\theta}{\longrightarrow} F_0(-)$

and then checks that this gives the desired right-adjoint. In a sense, the motivational section never gets off the ground. A motivation of a definition on objects only seems more or less not a defintion of a functor at all, from a category theoretic point of view.

Some expository questions suggest themselves (which brings me to the promised nLab-technical aspects):

• What is a good narrative to smoothly summon the full usual definition of $F_0^{F_1}(f)$ from the hypothesis that there were a right-adjoint to $(-)\times F_1$?
• Should one make the inchoate treatment in Section 2 in the current version of closed monoidal structure on presheaves of the construction of $F_0^{F_1}$ complete? (I will gladly do so, after some conceptual discussion here.) Currently, calling it “Proof” is somewhat of a misnomer. It currently only gives the usual motivation of the definition on objects, then refers to a book.
• Should we cross-reference the very relevant exposition Todd Trimble gives here, at least in the direction of having said exposition reachable via hyperlink from within closed monoidal structure on presheaves? I think this would be useful to readers.
• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeJul 12th 2017

$(F_0^{F_1})_{unn}$ is unnaturally isomorphic to the functor $F_0^{F_1}\colon\mathsf{C}^{op}\rightarrow\mathsf{S}$.

Assuming I’m parsing your notation correctly, no it is not.

• CommentRowNumber37.
• CommentAuthorTodd_Trimble
• CommentTimeJul 13th 2017
• (edited Jul 13th 2017)

Various things to sort out here.

Pedagogically, it’s probably best to start out understanding the cartesian closed hom. But I’ll say now there’s another closed monoidal structure which I’ve alluded to in this discussion, where the monoidal product has been called the “funny product” (it might also be called the “white product” on $Cat$, or the “black product” – I can never keep these things straight). The “pointwise exponential” denoted by Peter as $(F_1^{F_0})_{unn}$, and defined by $(F_1^{F_0})_{unn}(c) = F_1(c)^{F_0(c)}$, does turns out to be the funny internal hom adjoint to the funny product. So while Mike is right that this $(F_1^{F_0})_{unn}$ is in no way even unnaturally isomorphic to the cartesian closed hom $F_1^{F_0}$, the pointwise exponential in Peter’s words is the correct hom in the “funny” sense.

But let’s consider the cartesian closed hom before getting into the funny structures. I’ll answer one of Peter’s questions about that here.

(I’d much prefer to write $F^G$ instead of $F_0^{F_1}$, because it requires fewer keystrokes, and because I also believe people overuse subscripts where they are not actually needed. Eilenberg is famous for saying “if you define it right, you won’t need subscripts”, and often I think there’s a point to that.)

Peter asked for a motivated and expositional treatment of $F^G(f)$ where $f: u \to v$ is a morphism in $C$ and $F, G$ are objects of $Set^{C^{op}}$, and $F^G$ denotes the cartesian hom. I guess there are various classroom approaches one could try.

But the bottom line is that if the formula $F^G(v) \cong Nat(C(-, v) \times G, F)$, or in other notation $F^G(v) \cong Set^{C^{op}}(C(-, v) \times G, F)$, is to be natural in $v$, then this already implicitly tells you how $F^G(f)$ must be defined. Stare at the naturality diagram

$\array{ F^G(v) & \stackrel{\widehat{(-)}}{\to} & Nat(C(-, v) \times G, F) \\ \mathllap{F^G(f)} \downarrow & & \downarrow \mathrlap{Nat(C(-, f) \times 1_G, 1_F)} \\ F^G(u) & \underset{\widehat{(-)}}{\to} & Nat(C(-, u) \times G, F) }$

and chase an arbitrary element $x: 1 \to F^G(v)$ around it. Thus $F^G(f)$ is defined by the formula

$\widehat{F^G(f)(x)} = (C(-, u) \times G \stackrel{C(-, f) \times 1_G}{\to} C(-, v) \times G \stackrel{\widehat{x}}{\to} F).$

Put a little differently: if you define the exponential $F^G$ by the formula $F^G(-) = Nat(C(?, -) \times G(?), F(?))$, where the naturality of $Nat$ refers to the $?$ argument, then this definition is literally for both objects and morphisms in the $-$ argument.

• CommentRowNumber38.
• CommentAuthorTodd_Trimble
• CommentTimeJul 13th 2017

Sorry – I was conflating some things in my last post. That there are two closed monoidal structures on $Cat$ was one topic. The other topic was the cartesian closed structure of $Set^{C^{op}}$. I was getting the two mixed up in the second paragraph. (I blame it on insomnia.)