Basic question concerning equivalence of categories

Question

Suppose $C,D$ are categories and $F:C \rightarrow D$ covariant functor which is an equivalence together with a functor $G:D \rightarrow C$ . How can I then show that $ G \cdot F :Mor_{C}(X,Y) \rightarrow Mor_{C}( G \cdot F(X), G \cdot F(Y))$ is a bijection? I fail to see it is surjective.

Answer 1

This follows from a more general fact that both $F$ and $G$ are fully faithful.

There is a nice characterization: a functor $F\colon \mathcal{C}\to \mathcal{D}$ is an equivalence of categories if and only if it is 1) fully faithful (meaning that $\operatorname{Hom}_\mathcal{C} (X,Y) \to \operatorname{Hom}_\mathcal{D} (F (X), F (Y))$ is a natural bijection) and 2) essentially surjective (meaning that for every object $A\in \mathcal{D}$ there exists $X\in \mathcal{C}$ such that $F (X) \cong A$).

It is an instuctive exercise to show that any equivalence is fully faithful. Note that you have families of natural isomorphisms $\eta_X\colon X\xrightarrow{\cong} G F (X)$, so by looking at the corresponding naturality squares it is easy to show that

1) $F$ is faithfull: $F (f) = F (f')$ implies $G F (f) = G F (f')$ and then $f = f'$.

2) $F$ is full: for an arrow $F(X) \xrightarrow{g} F(Y)$ in $\mathcal{D}$ you can find an arrow $f$ such that $G F (f) = G (g)$. This implies $F (f) = g$ by 1).

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Edit: in your particular case, to see that $f \mapsto GF (f)$ is surjective, look at the naturality square

$$\require{AMScd} \begin{CD} X @>{\eta_X}>> GF (X) \\ @VV f V @VV GF (f) V\\ Y @>\eta_Y>> GF (Y) \end{CD}$$

$\eta_X$ and $\eta_Y$ are isomorphisms, hence $GF (f) = \eta_Y\circ f\circ \eta_X^{-1}$. So for any given morphism $g\colon GF (X) \to GF (Y)$ you may consider $f = \eta_Y^{-1}\circ g\circ \eta_X$, and then $GF (f) = g$.