Supposing we have a natural isomorphism $\tau : S \rightarrow T$ between functors $S,T : \mathscr{C} \rightarrow \mathscr{D}$, how exactly do we show that if $S$ is faithful, then so is $T$?
If $S$ is faithful, then it is injective on the hom-sets, i.e. if $f = g$, then $S(f) = S(g)$. I also know that the components of the natural transformation form a commutative square defined by $\tau \circ S(f) = T(f) \circ \tau$, and that the ultimate implication I want to show is (probably along the lines of?)
$$f = g \;\;\; \Rightarrow \;\;\; S(f)=S(g) \;\;\; \Rightarrow \;\;\; T(f)=T(g)$$
(with the first implication being given as stated by $S$ being faithful).
I've tried playing with around with some of the compositions but the closest I've gotten would be $T(\tau(f)) = T(\tau(g))$ (which came from $S(f)=S(g)$, composing on the right by $\tau$ which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure.
Anyhow, sorry for the dumb question, but any ideas?
The point is that $T(f)=\tau_y S(f)\tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!