Why are fully faithful functors conservative? (Or, why are isomorphisms reflected?) Say $F$ is a functor from category $C$ to $D$. By "fully faithful", I mean $f \mapsto Ff$ is injective ("faithful") and surjective ("full") in $C(X, Y) \to D(FX, FY)$.
My question is: when $F$ is fully faithful, why does $FX \cong FY$ imply that $X \cong Y$ for objects $X, Y$ in $C$?
The best I am able to come up with is:
Since $FX \cong FY$, there exists an isomorphism $h \in D(FX, FY)$. And, because $F$ is full, there exists a morphism $f \in C(X, Y)$ such that $F f = h$, and also a morphism $g \in C(Y, X)$ such that $F g$ is the inverse of $h$.
but... I don't know how to prove that $f \in C(X, Y)$ is an isomorphism. I think I need to show that $f g = \text{id}_Y$ and $g f = \text{id}_X$, but I don't know how.
Notation background: I'm using the text

Bradley, T. D., Bryson, T., & Terilla, J. (2020). Topology: A Categorical Approach. MIT Press.


You're nearly there! You've used the fact that $F$ is full, now you need to use the fact that it's faithful...
Suppose $f\in C(X,Y)$ and $g\in C(Y,X)$, such that $Ff\in D(FX,FY)$ and $Fg\in D(FY,FX)$ are inverses. Then $F(f\circ g) = Ff\circ Fg = \text{id}_{FY} = F(\text{id}_Y)$, and since $F$ is faithful, $f\circ g = \text{id}_Y$. The same argument shows $g\circ f = \text{id}_X$.
