I'm trying to solve this exercise:

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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 $A'\to A$).

  • $\begingroup$ Hint: You have identity arrows $A \to A$ and $A' \to A'$. $\endgroup$ – SCappella Mar 1 at 5:28
  • $\begingroup$ @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 $f:A\to A$. $\endgroup$ – user634426 Mar 1 at 5:38
  • $\begingroup$ Don't focus on the commutative square just yet. Only use $f$, not $\phi$. $\endgroup$ – SCappella Mar 1 at 5:59
  • $\begingroup$ @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? $\endgroup$ – user634426 Mar 1 at 6:20
  • $\begingroup$ math.stackexchange.com/questions/704891/… $\endgroup$ – user634426 Mar 2 at 2:08

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 $g$.

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

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