[proofplan] The inverse of $F(f)$ in $\mathcal D$ is a morphism $F(Y)\to F(X)$. Fullness lets us lift this inverse to a morphism $g:Y\to X$ in $\mathcal C$. The two inverse identities in $\mathcal D$ then become equalities between the $F$-images of $g\circ f$ and $\operatorname{id}_X$, and between the $F$-images of $f\circ g$ and $\operatorname{id}_Y$. Faithfulness reflects these equalities back to $\mathcal C$, proving that $g$ is a two-sided inverse for $f$. [/proofplan] [step:Lift the inverse of $F(f)$ along fullness] Assume that $F(f):F(X)\to F(Y)$ is an isomorphism in $\mathcal D$. Let $h:F(Y)\to F(X)$ be the inverse morphism to $F(f)$ in $\mathcal D$, so \begin{align*} h\circ F(f) &= \operatorname{id}_{F(X)},& F(f)\circ h &= \operatorname{id}_{F(Y)}. \end{align*} Since $F$ is full, the induced map \begin{align*} F_{Y,X}:\operatorname{Hom}_{\mathcal C}(Y,X)&\to \operatorname{Hom}_{\mathcal D}(F(Y),F(X))\\ u&\mapsto F(u) \end{align*} is surjective. Hence there exists a morphism $g:Y\to X$ in $\mathcal C$ such that $F(g)=h$. [guided] We start with the only datum supplied by the hypothesis beyond fullness and faithfulness: $F(f)$ is an isomorphism in $\mathcal D$. By definition, this means there is a morphism $h:F(Y)\to F(X)$ in $\mathcal D$ satisfying the two inverse identities \begin{align*} h\circ F(f) &= \operatorname{id}_{F(X)},& F(f)\circ h &= \operatorname{id}_{F(Y)}. \end{align*} The morphism $h$ lives in $\mathcal D$, but to prove that $f$ is an isomorphism in $\mathcal C$ we need a candidate inverse in $\mathcal C$. This is precisely where fullness is used. Fullness says that for every pair of objects $A,B\in \operatorname{Ob}(\mathcal C)$, the map \begin{align*} F_{A,B}:\operatorname{Hom}_{\mathcal C}(A,B)&\to \operatorname{Hom}_{\mathcal D}(F(A),F(B))\\ u&\mapsto F(u) \end{align*} is surjective. Applying this with $A=Y$ and $B=X$, the morphism $h\in \operatorname{Hom}_{\mathcal D}(F(Y),F(X))$ has a preimage. Therefore there exists a morphism $g:Y\to X$ in $\mathcal C$ such that $F(g)=h$. [/guided] [/step] [step:Reflect the left inverse identity using faithfulness] Because $F$ is a functor and $F(g)=h$, we have \begin{align*} F(g\circ f) &= F(g)\circ F(f) \\ &= h\circ F(f) \\ &= \operatorname{id}_{F(X)} \\ &= F(\operatorname{id}_X). \end{align*} Since $F$ is faithful, the induced map \begin{align*} F_{X,X}:\operatorname{Hom}_{\mathcal C}(X,X)&\to \operatorname{Hom}_{\mathcal D}(F(X),F(X))\\ u&\mapsto F(u) \end{align*} is injective. Therefore $g\circ f=\operatorname{id}_X$. [guided] We now translate the first inverse identity for $h$ into an inverse identity for $g$. Since $F$ is a functor, it preserves composition and identities. Thus \begin{align*} F(g\circ f) &= F(g)\circ F(f) \\ &= h\circ F(f) \\ &= \operatorname{id}_{F(X)} \\ &= F(\operatorname{id}_X). \end{align*} This equality takes place in $\operatorname{Hom}_{\mathcal D}(F(X),F(X))$. To conclude equality before applying $F$, we use faithfulness. Faithfulness says that for every pair of objects $A,B\in \operatorname{Ob}(\mathcal C)$, the map \begin{align*} F_{A,B}:\operatorname{Hom}_{\mathcal C}(A,B)&\to \operatorname{Hom}_{\mathcal D}(F(A),F(B))\\ u&\mapsto F(u) \end{align*} is injective. Applying this with $A=X$ and $B=X$, the equality $F(g\circ f)=F(\operatorname{id}_X)$ forces \begin{align*} g\circ f=\operatorname{id}_X. \end{align*} So $g$ is a left inverse of $f$. [/guided] [/step] [step:Reflect the right inverse identity using faithfulness] Similarly, using functoriality and $F(g)=h$, we obtain \begin{align*} F(f\circ g) &= F(f)\circ F(g) \\ &= F(f)\circ h \\ &= \operatorname{id}_{F(Y)} \\ &= F(\operatorname{id}_Y). \end{align*} Since $F$ is faithful, the induced map \begin{align*} F_{Y,Y}:\operatorname{Hom}_{\mathcal C}(Y,Y)&\to \operatorname{Hom}_{\mathcal D}(F(Y),F(Y))\\ u&\mapsto F(u) \end{align*} is injective. Therefore $f\circ g=\operatorname{id}_Y$. [guided] The second inverse identity is handled in the same structural way, but now in the endomorphism set of $Y$. Functoriality gives \begin{align*} F(f\circ g) &= F(f)\circ F(g) \\ &= F(f)\circ h \\ &= \operatorname{id}_{F(Y)} \\ &= F(\operatorname{id}_Y). \end{align*} Again, the equality currently lives only after applying $F$. Faithfulness supplies injectivity of the map \begin{align*} F_{Y,Y}:\operatorname{Hom}_{\mathcal C}(Y,Y)&\to \operatorname{Hom}_{\mathcal D}(F(Y),F(Y))\\ u&\mapsto F(u). \end{align*} Therefore the equality $F(f\circ g)=F(\operatorname{id}_Y)$ implies \begin{align*} f\circ g=\operatorname{id}_Y. \end{align*} So $g$ is also a right inverse of $f$. [/guided] [/step] [step:Conclude that $f$ is an isomorphism] We have constructed a morphism $g:Y\to X$ in $\mathcal C$ such that \begin{align*} g\circ f &= \operatorname{id}_X,& f\circ g &= \operatorname{id}_Y. \end{align*} Thus $g$ is a two-sided inverse for $f$, and hence $f:X\to Y$ is an isomorphism in $\mathcal C$. [/step]