Showing that $H_A\simeq H_{A'}$ implies $A\simeq A'$

Question

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 $A'\to A$).

Answer 1

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 $g$.