Are these two functors, full functors?

Question

Ler $\mathcal{G}$, $\mathcal{G_0}$ and $\mathcal{G_{\infty}}$ be the category of abelian groyps, torsion abelian groups and infinite abelian groups, respectively. The morphism of the first and second categories are group homomorphisms and the morphisms of the third category are injective group homomorphisms.

Are the following two functors, full functors?

$F_1: \mathcal{G} \to \mathcal{G_0}$ with $F_1(G)=Tor(G)$ the torsion subgroup.

$F_2:\mathcal{G_{\infty}} \to \mathcal{G}$ with $F_2(G)= Thin(G)$, the Thin subgroup of $G$. Thin subgroup of $G$ consist of all $g\in G$ such that the cyclic group $(g)$ has infinite index in $G$.

Answer 1

For the torsion subgroup functor, suppose it were full. Then, for any abelian group $G$, there would be a map $G \to Tor(G)$ with image equal to the identity map $Tor(G) \to Tor(Tor(G)) = Tor(G)$. This map would necessarily induce a splitting of $G$ as a direct sum of $Tor(G)$ and another subgroup. However, not every abelian group $G$ has $Tor(G)$ as a direct summand - according to https://en.wikipedia.org/wiki/Torsion_subgroup (although I can't come up with any counterexample off the top of my head, and a quick Google search didn't turn up anything immediately either).

However, the restriction to the full subcategory of finitely generated abelian groups would be full.

Answer 2

I can answer half of this. $F_2$ is not full, since one can just take $G$ uncountable. Then $Thin(G)=G$, and $G$ in general will have non-injective endomorphisms, but these will not be in the image of $F_2$ mapping the $\mathcal{G_{\infty}}$ endomorphisms of $G$ to the ordinary ones.