Cfrgtkky

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 :-)

$$defroom{qquadqquadqquadqquadqquad}$$

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*}

Hint: You have to relate the maps $hom(c,d)to hom(S(c),S(d))$ and $hom(c,d)tohom(T(c),T(d))$ for any $c,dinmathscr C$. This can be done by using the natural isomorphism in order to produce a bijective map $hom(S(c),S(d))to hom(T(c), T(d))$ that commutes with the above maps.