# Functors between morphism categories

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.)

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

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.

• 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. – Dan Normand May 15 '18 at 1:34
• 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! – Dan Normand May 15 '18 at 2:03
• 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. – Randall May 15 '18 at 2:05
• 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. – Billy May 15 '18 at 12:56