Let $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ be two functors such that $\alpha:1_{\mathcal{B}}\cong F\circ G$ and $\beta:G\circ F\cong 1_{\mathcal{C}}$. I want to show that $F$ is full and faithful.
It can be deduced that $F\circ G$ is full and faithful. From this, we can conclude that $G$ is faithful. I am not able to prove that $F$ is full.
Let $C,C'\in\mathcal{C}$ and $b:F(C)\rightarrow F(C')$. We have to find a morphism in $\mathcal{C}(C,C')$ whose image under $F$ is equal to $b$. I thought about using the natural isomorphism $\alpha$: $$\alpha_{F(C')}\circ b=F(G(b))\circ\alpha_{F(C)}.$$ But this seems like a dead-end. I imagine the proof is rather trivial but I am not able to see it. Any hints?
Edit:
Solved: I think the candidate is $\beta_{C'}\circ G(b)\circ\beta^{-1}_C$.
Since $\mathcal{D}(d, d’)\xrightarrow{G}\mathcal{C}(Gd, Gd’)\xrightarrow{F} \mathcal{D}(FGd, FGd’)$ is an isomorphism, the second map is surjective. Now by $G\circ F\cong \mathrm{id}$ we know $G$ is essentially surjective, so for any $c, c’\in \mathcal{C}$ the above observation can be used to prove that $\mathcal{C}(c, c’)\xrightarrow{F} \mathcal{D}(Fc, Fc’)$ is surjective.