Context:
To introduce some symbols and such, what I'm seeking to prove is this:
Let $F$ be a faithful functor. Suppose $F(f)$ is a monic arrow. Show $f$ is monic.
This came up as part of a class assignment recently, tied to an introduction to category theory as we began to introduce free objects. I have an attempt ready, but I have a slight feeling that something's wrong with it. I can't really pin it down on what, it's just a lurking feeling; alternatively, I could be overthinking.
Any thoughts?
My Attempt at a Proof:
Let $F : \scr{C} \to \scr{D}$ be a faithful functor. Let us also have arrows $g,h : A \to B$ and $f : B \to C$, where $A,B,C,f,g,h \in \scr C$. Further, let $F(f)$ be monic and let $f \circ g = f \circ h$.
Since $F$ is faithful, it is injective on the hom-sets of $\scr C$. Thus,
$$F(f \circ g) = F(f \circ h) \tag 1$$
Since $F$ is a functor, it preserves composition: thus,
$$F(f \circ g) = F(f) \circ F(g) = F(f) \circ F(h) = F(f \circ h) \tag 2$$
Since $F(f)$ is monic, $(2)$ yields
$$F(f) \circ F(g) = F(f) \circ F(h) \implies F(g) = F(h) \tag 3$$
Again since $F$ is faithful, $(3)$ yields
$$F(g) = F(h) \implies g = h \tag 4$$
Thus, since our assumptions $F$ is faithful, $F(f)$ is monic, and $f \circ g = f \circ h$ imply $g = h$ we can conclude $f$ is monic.
As noted in the comments, it seems (at this time) the consensus is that this is indeed correct as an approach (though these days I would probably phrase it a little differently).
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.