So we let $S,T : \mathscr{C} \rightarrow \mathscr{D}$ be naturally isomorphic functors. We seek to show that if $S$ is a full functor, then so is $T$.
As given, we have a natural isomorphism $\tau : S \rightarrow T$, which also means that each component of it $\tau_A : S(A) \rightarrow T(A)$ is invertible.
From here, though, I get pretty lost and have no clue where to go. I'm not even 100% sure how to best show surjectivity in the context of category theory. I feel like that I would have to use the inverse of the natural transformation, $\tau^{-1}$, and that surjectivity implies right-cancellativity, i.e. $S$'s arrow function has a right-sided inverse $S^{-1}$. But I'm honestly just lost.
Does anyone have a potential nudge in the right direction?
Recall that a natural isomorphism $η : F →̣ G$ means we have $η ∘ F f = G f ∘ η$ for all $f$, moreover the $\eta$ are invertible.
Now the required proof progresses as follows :-)
$$\def\room{\qquad\qquad\qquad\qquad\qquad}$$
\begin{align*} & F \; \mathsf{full} \\ ≡\; & \color{green}{\{\text{ Definition of full }\}} \\ & ∀x,y • ∀ f′ : F x → F • ∃ f : x → y \;• \room F f = f′ \\ ≡\; & \color{green}{\{\text{ Using natural isomorphism }\}} \\ & ∀x,y • ∀ f′ : F x → F • ∃ f : x → y \;•\room η⁻¹ ∘ G f ∘ η = f′ \\ ≡\; & \color{green}{\{\text{ Using natural isomorphism }\}} \\ & ∀x,y • ∀ f′ : F x → F • ∃ f : x → y \;•\room G f = η ∘ f′ ∘ η⁻¹ \\ ≡\; & \color{green}{\{\text{ Local declaration ---aka `one point rule' }\}} \\ & ∀x,y • ∀ f′ : F x → F • ∀ f″ : G x → G y • ∃ f : x → y \;• \room G f = f″ \quad ∧ \quad f″ = η ∘ f′ ∘ η⁻¹ \\ ⇒\; & \color{green}{\{\text{ Weaken by discarding a conjunct }\}} \\ & ∀ x,y • ∀ f′ : F x → F y • ∀ f″ : G x → G y • ∃ f : x → y \;• \qquad G f = f″ \\ ⇒ \; & \color{green}{\{\text{ Remove superfluous $∀ f′$ }\}} \\ & ∀ x,y • ∀ f″ : G x → G y • ∃ f : x → y \;•\room G f = f″ \\ ≡ \; & \color{green}{\{\text{ Definition of full }\}} \\ & G \; \mathsf{full} & & \end{align*}