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.


A necessary and sufficient condition for a full functor $F\colon C\to D$ to be conservative is that $Fg$ being the identity implies $g$ is an isomorphism (e.g. that there exists $f$ and $h$ such that $fg$ and $gh$ are identities, as then $f=fgh=h$). Since a faithful functor has $Fg$ the identity morphism only if $g$ is already the identity morphism, it follows that full and faithful functors are conservative.
The necessity of the condition is immediate because $F$ being conservative implies that if $Fg$ is the identity morphism, which is an isomorphism, then $g$ is an isomorphism.
For sufficiency: if $Fg$ is an isomorphism, i.e. if there are $\phi$ and $\psi$ such that $\phi Fg$ and $Fg\psi$ are identites, then fullness implies there exists $f$ and $h$ with $Ff=\phi$ and $Fh=\psi$, which then have the property that $FfFg=F(fg)$ and $FgFh=F(gh)$ are identities. Then the desired property implies $fg$ and $gh$ are isomorphisms, whence $((fg)^{-1}f)g$ and $g(h(gh)^{-1})$ are identities, so $g$ is an isomorphism, as desired.
