Endomorphisms of an equivalence of categories

Question

For $F:C^{op}\rightarrow C$ an isomorphism of categories, it is easy to see that $End(F)\cong End(\operatorname{id}_C)$ and $Aut(F)\cong Aut(\operatorname{id}_C)$ as sets, where we consider the hom-spaces in the respective functor categories. I try to prove the same thing if $F$ is an equivalence of categories with weak inverse $F'$ and fail to do so. Is it true? Or what is a counterexample?

For the isomorphism $\phi:F'F\Rightarrow \operatorname{id}$, the assignments $(\eta: F\Rightarrow F) \mapsto \phi \circ F'(\eta)\circ \phi^{-1}$ and $(\eta: \operatorname{id}\Rightarrow \operatorname{id}) \mapsto F(\eta)$ are not inverses.

Answer 1

As MPos shows, you can just check this on elements, but you can also argue ''one categorical level higher'', if you will (and this argument works for all fully faithful functors, not only the equivalences of categories). If $F$ is an equivalence of categories, then in particular $F$ is fully faithful. One can see that, for a fully faithful functor $i\colon \mathcal{D}\hookrightarrow\mathcal{E}$ and another category $\mathcal{C}$, it holds that the postcomposition functor $i\circ -\colon\mathrm{Fun}(\mathcal{C},\mathcal{D})\to\mathrm{Fun}(\mathcal{C},\mathcal{E})$ is again fully faithful. Applying this to $F$, we see that the functor $F\circ -\colon\mathrm{Fun}(\mathcal{C},\mathcal{C})\to\mathrm{Fun}(\mathcal{C},\mathcal{C})$ is fully faithful (actually, it is an equivalence of categories also, but we don't need this). Note that I have decided to let $F$ be a functor $\mathcal{C}\to\mathcal{C}$ rather than $\mathcal{C}^\mathrm{op}\to\mathcal{C}$. Now, if $F\circ-$ is fully faithful, it holds that $\mathrm{End}(\mathrm{id}_\mathcal{C})=\mathrm{Nat}(\mathrm{id}_\mathcal{C},\mathrm{id}_\mathcal{C})\cong\mathrm{Nat}(F,F)=\mathrm{End}(F)$. Note that we didn't need $F$ to be an equivalence of categories: we only needed $F$ to be fully faithful.

Running the same argument with the core subcategory $\mathrm{Fun}(\mathcal{C},\mathcal{C})^\cong$ (i.e. the subcategory on all objects but only the isomorphisms) gives you $\mathrm{Aut}(\mathrm{id}_\mathcal{C})\cong\mathrm{Aut}(F)$. Again, you only need that $F$ is fully faithful for this.

Answer 2

The statement is still true, given an equivalence $F$.

As it is pointed out, let $\phi: F'F \Rightarrow \operatorname{id}$ be the counit and $\psi: \operatorname{id} \Rightarrow F F'$ be the unit of the equivalence. In particular, $\phi$ and $\psi$ are natural isomorphisms. Wlog we can assume that $\phi$ and $\psi$ form an adjunction.

Recall that we have the unit-counit equation:

$$ \operatorname{id}_{FX} = F(\phi_X)\circ \psi_{FX} $$

Now, the assignments $$ \operatorname{End}(F) \rightarrow \operatorname{End}(\operatorname{id}), \quad \eta \mapsto \phi \circ F'(\eta) \circ \phi^{-1}$$ and $$ \operatorname{End}(\operatorname{id}) \rightarrow \operatorname{End}(F), \quad \mu \mapsto F(\mu)$$ are in fact inverse to each other, since: \begin{align*} F(\phi\circ F'(\eta)\circ \phi^{-1}) &\overset{(1)}{=} F(\phi)\circ FF'(\eta)\circ F(\phi)^{-1}\\ &\overset{(2)}{=} F(\phi)\circ \psi_{F} \circ \eta \circ \psi_{F}^{-1} \circ F(\phi)^{-1}\\ &\overset{(3)}{=} \eta. \end{align*} Equation (1) is functioriality of $F$, (2) uses naturility of $\psi:\operatorname\Rightarrow FF'$ and (3) follows from the unit-counit equation. The other direction is
\begin{equation*} \phi \circ F'F(\mu) \circ \phi^{-1} = \mu \end{equation*} is immediate from naturality of $\phi$.

Hence, $\operatorname{End}(F)\cong \operatorname{End}(\operatorname{id})$ as monoids.