[proofplan] The proof is a direct verification of the two inverse identities. We take the inverse morphism $f^{-1}$ in $\mathcal C$, apply the functor $F$ to the equations defining it, and use the functorial preservation of identities and composition. The resulting equations say exactly that $F(f^{-1})$ is a two-sided inverse of $F(f)$ in $\mathcal D$. [/proofplan] [step:Apply the functor to the inverse identities in $\mathcal C$] Let $g: Y \to X$ denote the inverse morphism $f^{-1}$ in $\mathcal C$. Since $f$ is an isomorphism, the following identities hold in $\mathcal C$: \begin{align*} g \circ f &= \operatorname{id}_X, & f \circ g &= \operatorname{id}_Y. \end{align*} Because $F: \mathcal C \to \mathcal D$ is a covariant functor, it sends objects of $\mathcal C$ to objects of $\mathcal D$, sends each morphism $h: A \to B$ in $\mathcal C$ to a morphism $F(h): F(A) \to F(B)$ in $\mathcal D$, preserves identity morphisms, and preserves composition. Applying $F$ to the two displayed identities gives \begin{align*} F(g \circ f) &= F(\operatorname{id}_X), & F(f \circ g) &= F(\operatorname{id}_Y). \end{align*} Using preservation of composition and identities, this becomes \begin{align*} F(g) \circ F(f) &= \operatorname{id}_{F(X)}, & F(f) \circ F(g) &= \operatorname{id}_{F(Y)}. \end{align*} [/step] [step:Identify $F(f^{-1})$ as the inverse of $F(f)$] The morphism $F(f): F(X) \to F(Y)$ has the morphism $F(g): F(Y) \to F(X)$ as both a left inverse and a right inverse, because \begin{align*} F(g) \circ F(f) &= \operatorname{id}_{F(X)}, & F(f) \circ F(g) &= \operatorname{id}_{F(Y)}. \end{align*} Therefore $F(f)$ is an isomorphism in $\mathcal D$. Since $g = f^{-1}$ by definition, its inverse is precisely \begin{align*} F(g) = F(f^{-1}). \end{align*} This proves the theorem. [/step]