# Showing that $H_A\simeq H_{A'}$ implies $A\simeq A'$ [duplicate]

I'm trying to solve this exercise: So we are given that for all objects $$B$$, there is a natural isomorphism $$H_A(B)\simeq H_{A'}(B)$$ Let's write this isomorphism as $$f\mapsto \bar f$$. The commutativity of the square tells us $$\overline{f\circ\phi}=\bar f\circ\phi$$ where $$\phi:B'\to B$$ is an arrow in $$\mathscr A$$. I don't see how to use that. I don't even see how to show that there exists an arrow $$A\to A'$$ (or $$A'\to A$$).

• Hint: You have identity arrows $A \to A$ and $A' \to A'$. – SCappella Mar 1 '20 at 5:28
• @SCappella The only way of using this that I can think of is to set $B=B'=A,\phi=id$. But in that case the commutativity of the square gives trivial results like $\bar f=\bar f$ for all $f:A\to A$. – user634426 Mar 1 '20 at 5:38
• Don't focus on the commutative square just yet. Only use $f$, not $\phi$. – SCappella Mar 1 '20 at 5:59
• @SCappella Then if $\psi_B:H_A(B)\to H_{A'}(B)$ denotes an isomorphism, then $\psi_A(1_A)$ is an arrow $A\to A'$ and $\psi^{-1}_{A'}(1_{A'})$ is an arrow $A'\to A$. I guess we need to prove that they are inverses of each other. I tried taking $\phi=\psi^{-1}_{A'}(1_{A'})$; the commutativity told that $\psi_A(f)\circ \phi=\psi_{A'}(f\circ \phi)$. I don't see how to simplify this with my $\phi$. Should I try something different? – user634426 Mar 1 '20 at 6:20
• math.stackexchange.com/questions/704891/… – user634426 Mar 2 '20 at 2:08

## 1 Answer

You do need the definition that $$H_A(B) = \operatorname{Hom}(B, A)$$. In particular, this means that $$Id_A \in H_A(A)$$, so this corresponds to some $$f: A \to A'$$ under the isomorphism $$H_A(A) \cong H_{A'}(A)$$. Similarly, $$Id_{A'} \in H_{A'}(A')$$ corresponds to some $$g: A' \to A$$ in $$H_A(A')$$. The claim is now that $$f$$ is an isomorphism, with inverse $$g$$.

To prove this, the following diagram may be helpful: The details of this can be found in the spoiler block below, but it is a good exercise to write these out yourself.

It just comes down to chasing through the diagram. If we start in the top left with $$Id_A \in H_A(A)$$, then by definition this corresponds to $$f \in H_{A'}(A)$$, which gets sent to $$fg = H_{A'}(g)(f) \in H_{A'}(A')$$. Going the other way, $$Id_A \in H_A(A)$$ is first sent to $$g = H_A(g)(Id_A) \in H_A(A')$$. and by definition $$g$$ corresponds to $$Id_{A'} \in H_{A'}(A')$$ under the isomorphism. By naturality of the isomorphism, these two should be the same, so $$fg = Id_{A'}$$. A similar reasoning starting at $$Id_{A'} \in H_{A'}(A')$$ and using naturality with respect to $$f$$ instead, we also conclude $$gf = Id_A$$. So indeed $$f$$ is an isomorphism with inverse $$g$$.

• The diagram made it unnecessary for me to open the spoiler :) – user634426 Mar 1 '20 at 17:20