This is immediate from the definition of equivalence.

Assume we have $G: \mathcal{D} \to \mathcal{C}$ with natural isomorphisms $\eta: \text{id}_\mathcal{C} \Rightarrow G \circ F$ and $\epsilon: F \circ G \Rightarrow \text{id}_\mathcal{D}$.

Essential surjectivity: For any $D$ in $\mathcal{D}$, we have $F(G(D)) \cong D$ via $\epsilon_D: F(G(D)) \to D$.

Fully faithful: Let $A, B$ be objects in $\mathcal{C}$. We show $F_{A,B}: \text{Hom}_\mathcal{C}(A, B) \to \text{Hom}_\mathcal{D}(F(A), F(B))$ is bijective.

Injectivity: If $F(f) = F(g)$, then $G(F(f)) = G(F(g))$. Using naturality of $\eta$: $f = \eta_B^{-1} \circ G(F(f)) \circ \eta_A = \eta_B^{-1} \circ G(F(g)) \circ \eta_A = g$

Surjectivity: For $h: F(A) \to F(B)$, define $f := \eta_B^{-1} \circ G(h) \circ \eta_A: A \to B$. Using triangle identities and naturality: $F(f) = F(\eta_B^{-1}) \circ F(G(h)) \circ F(\eta_A) = h$

Assume $F$ is fully faithful and essentially surjective. We construct quasi-inverse $G: \mathcal{D} \to \mathcal{C}$.

On objects: For each $D$ in $\mathcal{D}$, choose $G(D)$ in $\mathcal{C}$ and isomorphism $\epsilon_D: F(G(D)) \to D$.

On morphisms: For $h: D \to D'$, consider $\epsilon_{D'}^{-1} \circ h \circ \epsilon_D: F(G(D)) \to F(G(D'))$.

Since $F$ is fully faithful, there exists unique $G(h): G(D) \to G(D')$ with: $F(G(h)) = \epsilon_{D'}^{-1} \circ h \circ \epsilon_D$

$G$ is a functor: For identity: $F(G(\text{id}_D)) = \epsilon_D^{-1} \circ \text{id}_D \circ \epsilon_D = \text{id}_{F(G(D))} = F(\text{id}_{G(D)})$

By faithfulness, $G(\text{id}_D) = \text{id}_{G(D)}$.

For composition: Similar diagram chasing shows $G(h' \circ h) = G(h') \circ G(h)$.

Natural isomorphisms: $\epsilon = \{\epsilon_D\}$ gives $F \circ G \Rightarrow \text{id}_\mathcal{D}$ by construction.

For $\eta_A: A \to G(F(A))$, define as unique morphism with $F(\eta_A) = \epsilon_{F(A)}^{-1}$ (exists by full faithfulness). This forms natural isomorphism $\text{id}_\mathcal{C} \Rightarrow G \circ F$. $\square$