Theorem: Let $F : \mathbf{A} \rightarrow \mathbf{B}$ be a functor.

$$F$$ is an equivalence of categories if and only if $$F$$ is full, faithful, and essentially surjective on objects. *Proof*: Assume that $$F$$ is an equivalence. Then, we have a functor $$G : \mathbf{B} \rightarrow \mathbf{A}$$, and two natural isomorphisms $$\eta : 1_{\mathbf{A}} \rightarrow G \circ F$$ and $$\varepsilon : F \circ G \rightarrow 1_{\mathbf{B}}$$. We first show that $$F$$ is faithful. Let $$s : A \rightarrow A'$$ be an arrow in $$\mathbf{A}$$. We have a component $$\eta_A : 1_{\mathbf{A}}(A) \rightarrow (G \circ F)(A)$$. The naturality condition says that $$\eta_{A'} \circ 1_{\mathbf{A}}(s) = (G \circ F)(s) \circ \eta_A$$. Of course $$1_{\mathbf{A}}(A) = A$$ and $$1_{\mathbf{A}}(s) = s$$. So simplifying a bit, we have $$\eta_{A'} \circ s = (G \circ F)(s) \circ \eta_A$$. If there were another arrow $$t : A \rightarrow A'$$ with $$F(s) = F(t)$$, it would follow that $$(G \circ F)(s) = (G \circ F)(t)$$. But then, we obtain by combining this with the respective naturality conditions:$$

\eta{A'} \circ s= (G \circ F)(s) \circ \eta_A = (G \circ F)(t) \circ \eta_A = \eta{A'} \circ t

$$Since $$\eta$$ is a natural isomorphism, $$\eta_{A'}$$ is an isomorphism, thus the outer-most inequality implies $$s = t$$. Now we show that $$F$$ is full. Consider $$g : F(A) \rightarrow F(A')$$ an arrow in $$\mathbf{B}$$. Take $$f = \eta_{A'}^{-1} \circ G(g) \circ \eta_A$$. This is well-defined since $$\eta$$ is a natural isomorphism and thus $$\eta_{A'}$$ is an isomorphism. By naturality we have $$\eta_{A'} \circ f = (G \circ F)(f) \circ \eta_A$$. Then $$f = \eta_{A'}^{-1} \circ (G \circ F)(f) \circ \eta_A$$. Equating this with the previous expression for $$f$$:$$

\eta{A'}^{-1} \circ G(g) \circ \eta_A = \eta{A'}^{-1} \circ (G \circ F)(f) \circ \eta_A

$$and since $$\eta_{A'}$$ and $$\eta_A$$ are both isomorphisms we can conclude that $$(G \circ F)(f) = G(g)$$. This really means that $$G(F(f)) = G(g)$$. But, $$G$$ is an equivalence -- we showed that equivalences are faithful in the preceding paragraph. Thus, we can conclude that $$F(f) = g$$. Finally, we show that $$F$$ is essentially surjective on objects. Let $$B \in \mathbf{B}$$. We claim that $$G(B)$$ is the desired element so that $$F(G(B)) \cong B$$. In fact, $$\varepsilon_B : (F \circ G)(B) \rightarrow 1_{\mathbf{B}}(B)$$ is an isomorphism between $$F(G(B))$$ and $$B$$. Now assume that $$F$$ is full, faithful, and essentially surjective on objects. We will define the functor $$G$$. For a $$B \in \mathbf{B}$$, define $$G(B)$$ as any particular object in $$\mathbf{A}$$ whose image under $$F$$ is isomorphic to $$B$$, say via a particular isomorphism $$\varepsilon_B$$ (there is at least one since $$F$$ is essentially surjective on objects). That is, we have $$\varepsilon_B : F(G(B)) \rightarrow B$$. We now need to define $$G$$ on arrows. For an arrow $$s : B \rightarrow B'$$ in $$\mathbf{B}$$, note that $$F$$ gives a bijection $$\Hom(G(B), G(B')) \leftrightarrow \Hom(F(G(B)), F(G(B'))$$, since it is full and faithful. Then let $$G(s) = F^{-1}(\varepsilon_{B'}^{-1} \circ s \circ \varepsilon_B)$$ -- the inner term is a map $$F(G(B)) \rightarrow F(G(B'))$$ and we take the inverse image under the bijection given by $$F$$. We now claim that we have a natural isomorphism $$\varepsilon : F \circ G \rightarrow 1_{\mathbf{B}}$$, with components $$\varepsilon_B$$ as chosen above. Note that since every $$\varepsilon_B$$ is an isomorphism, $$\varepsilon$$ will be a natural isomorphism as long as it is a natural transformation -- so we need only show the latter, which amounts to checking the naturality condition, which is straightforward: For a map $$s : B \rightarrow B'$$ in $$\mathbf{B}$$, we have:$$

\varepsilon{B'} \circ (F \circ G)(s) = \varepsilon{B'} \circ (\varepsilon{B'}^{-1} \circ s \circ \varepsilon_B) = s \circ \varepsilon_B = 1{\mathbf{B}}(s) \circ \varepsilon_B

$$and $$\varepsilon$$ is a natural isomorphism. Finally, we define $$\eta : 1_{\mathbf{A}} \rightarrow G \circ F$$. For $$A \in \mathbf{A}$$, we need to define $$\eta_A : A \rightarrow G(F(A))$$. Note that $$\varepsilon_{F(A)}^{-1}$$ is a map $$F(A) \rightarrow FGF(A)$$. Since $$F$$ is full and faithful, it gives an bijection $$\Hom(A, GF(A)) \leftrightarrow \Hom(F(A), FGF(A))$$. So, let $$\eta_A = F^{-1}(\varepsilon_{F(A)}^{-1})$$. It is straightforward to check that since $$\varepsilon_{F(A)}^{-1}$$ is an isomorphism, its preimage under the bijection given by $$F$$ will be as well. So each $$\eta_A$$ is an isomorphism and we need only check naturality. So, consider a $$t : A \rightarrow A'$$ in $$\mathbf{A}$$. Considering $$F(t) : F(A) \rightarrow F(A')$$ an arrow in $$\mathbf{B}$$, we know the following is true:$$

FGF(t) \circ \varepsilon{F(A)}^{-1} = \varepsilon{F(A')}^{-1} \circ F(t)

$$(since the inverse of $$\varepsilon$$ is also a natural isomorphism $$1_{\mathbf{B}} \rightarrow F \circ G$$). But, both sides can be rewritten to yield:$$

F(GF(t) \circ \etaA) = F(\eta{A'} \circ t)

$$and since $$F$$ is faithful, we can conclude that$$

GF(t) \circ \etaA = \eta{A'} \circ t

$$

which is precisely the naturality condition for $\eta$. Thus, $\eta$ is a natural isomorphism, completing the proof.