Fullness of composition implies fullness of the second functor?

Question

I encountered a statement left as an exerciser in a category theory textbook:

Let F: C→D and G: D→C be functors. Prove that if G◦F is full, then G is full.

My confusion is should there be objects X and Y in D, which are out of the range of the object mapping of F, then
G(hom_D(X, Y)) ≠ hom_C(GX, GY)
can happen (which makes G not full), even when G◦F is full.

I can think of a counterexample:

Update #1: as I keep reading, it seems the author perhaps actually have in mind that

If F◦G is full and F is faithful, then G is full.

Indeed, for any f in hom_C(GX, GY), there's a g in hom_D(X, Y) such that F◦G(g) = F(G(g)) = F(f). The faithfulness of F then implies G(g) = f.

Answer 1

This is not true in general. Counterexample: take $C=\mathbb{N}$ (monoid of naturals by multiplication), $D=\mathbb{N}\sqcup\mathbb{R}$ (coproduct of monoids), $F=i_1\colon\mathbb{N}\to\mathbb{N}\sqcup\mathbb{R}$ (canonical inclusion of the coproduct), $G=I_{\mathbb{N}}\sqcup\Delta_{*}\colon\mathbb{N}\sqcup\mathbb{R}\to\mathbb{N}$ (where $I_{\mathbb{N}}$ is the identity functor and $\Delta_{*}$ is the trivial functor). Then $G\circ F=I_{\mathbb{N}}$ (which is trivially full), but $G$ is not.