2
$\begingroup$

Let ${\cal{C}}$ and ${\cal{D}}$ be categories, and let $\textsf{Mor}({\cal{C}})$, $\textsf{Mor}({\cal{D}})$ be their morphism categories (objects are morphisms, morphisms are pairs of morphisms making the appropriate square commute). Let $\textsf{Mor}({\cal{C}})\xrightarrow{F}\textsf{Mor}({\cal{D}})$ be a (covariant) functor, and suppose moreover that it is an isomorphism of categories.


My question is if $F$'s map on objects necessarily sends identity morphisms of $\cal{C}$ to identity morphisms of ${\cal{D}}$; if it is true, could I perhaps have a hint as to how to approach proving this?


Here's my background. Obviously an isomorphism of categories ${\cal{C}}\cong{\cal{D}}$ induces an isomorphism of categories $\textsf{Mor}({\cal{C}})\cong\textsf{Mor}({\cal{D}})$, and I was wondering if the converse is true (in fact, I was wondering this about the category of isomorphisms first, but after little progress decided to switch to all morphisms). My gut says that it should be true; in a very loose sense, $\textsf{Mor}({\cal{C}})$ should "fully characterize" ${\cal{C}}$ (and certainly contains an embedded copy of ${\cal{C}}$). The purpose for my question is because I have managed to show if $F$ maps identity morphisms to identity morphisms, then I can construct the (obvious) isomorphism ${\cal{C}}\cong{\cal{D}}$.

Thank you for any insight/help!

(As an aside, does MSE have any native support for a commutative diagram package? I am quite proficient with tikz/tikz-cd but can only use them here by compiling locally, clipping the pdf, and then embedding the clipped image.)

$\endgroup$
  • 1
    $\begingroup$ For the aside part : you can draw diagram with MathJax, although it's quite limited compared to tikz-cd. $\endgroup$ – Arnaud D. May 15 '18 at 8:16
  • $\begingroup$ @ArnaudD. That is just what I was looking for, and good enough, thank you! $\endgroup$ – Dan Normand May 15 '18 at 18:04
1
$\begingroup$

I don't have good intuition for this, so I might be wrong, but I don't see why this should be true: Mor(C) doesn't capture much about the composition in C.

Here's my attempt at a counterexample. Let C = D be the cyclic group {1,g} of order 2, thought of as a category with a single element x with two automorphisms 1 and g. Then, if I understand correctly, Mor(C) = Mor(D) is the following category:

  • the elements are 1 and g;
  • there are 8 morphisms:
    1. $\mathbf{id}_1 = (1,1): 1\to 1$
    2. $\mathbf{g}_1 = (g,g): 1\to 1$
    3. $\mathbf{id}_g = (1,1): g\to g$
    4. $\mathbf{g}_g = (g,g): g\to g$
    5. $\alpha_{1,g} = (1,g): 1\to g$
    6. $\alpha_{g,1} = (1,g): g\to 1$
    7. $\beta_{1,g} = (g,1): 1\to g$
    8. $\beta_{g,1} = (g,1): g\to 1$.

(It would be much easier if I could draw this, but I'll let you do it.) Now let F be the functor

  • swapping the elements 1 and g;
  • sending
    1. $\mathbf{id}_1 \leftrightarrow \mathbf{id}_g$
    2. $\mathbf{g}_1 \leftrightarrow \mathbf{g}_g$
    3. $\alpha_{1,g} \leftrightarrow \alpha_{g,1}$
    4. $\beta_{1,g} \leftrightarrow \beta_{g,1}$.

I think F is an isomorphism from Mor(C) to Mor(D), but as it sends 1 to g, it can't come from a morphism from C to D.

$\endgroup$
  • $\begingroup$ I believe this works. I will have to sit down with a piece of paper to double check everything; interestingly your example gives a functor $F$ which does map identity morphisms to identity morphisms, which means my proof that this implies an isomorphism ${\cal{C}}\cong{\cal{D}}$ is, well, not a proof. Let me check where I went wrong and get back. $\endgroup$ – Dan Normand May 15 '18 at 1:34
  • $\begingroup$ Yes this works, and my previous comment is incorrect ($F$ does not send the identity morphism of $\cal{C}$ to an identity morphism - forgive me, I've been staring at far too many arrows today trying to work this out). Thank you very much for your input, it helped immensely. If I could bother you slightly further, do you have any intuition as to why "$\textsf{Mor}({\cal{C}})$ doesn't capture much about the composition in ${\cal{C}}$"? Doesn't $\textsf{Mor}({\cal{C}})$ have an embedded copy of ${\cal{C}}$ inside it (objects are identities, morphisms are pairs of themselves)? Thanks again! $\endgroup$ – Dan Normand May 15 '18 at 2:03
  • $\begingroup$ Yeah, I think $\mathrm{Mor}(\mathcal{C})$ contains everything about composition in $\mathcal{C}$. A composition of two squares with vertical identities will contain the composition in $\mathcal{C}$ across the horizontals. $\endgroup$ – Randall May 15 '18 at 2:05
  • $\begingroup$ I don't think I have a precise idea in mind of what I mean, so it might be nonsense. But it seems to me that, if Mor(C) is somehow able to capture the statements in C that $g^2 = 1$ and $1^2 = 1$, then functoriality should preserve those statements. Maybe a better way to say what I meant is as follows: a functor from C to D preserves C as a group, whereas a functor from Mor(C) to Mor(D) only seems to preserve C as a C-set. $\endgroup$ – Billy May 15 '18 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.