# Free Abelian groups satisfy the projective property.

I want to show that if $F$ is a free Abelian group, then $F$ satisfies the projective property.

That is, for any finitely generated Abelian groups $B$ and $C$, given any homomorphism $\alpha: F \to C$ and a surjective homomorphism $\beta: B \to C$, there exists a homomorphism $\gamma$ which makes the diagram below commute - i.e. $\beta \gamma = \alpha$

What I've done so far:

Let $F = \left<x_1, \dots ,x_n \right>$. It is enough to define $\gamma$ on this set of generators. Let $\alpha(x_i) = c_i$. As $\beta$ is a surjection, $\exists b_i \in B$ such that $\beta (b_i) = c_i$. Thus, if we define $\gamma (x_i) = b_i$, we get that $\beta \gamma = \alpha$ on the set of generators for $F$.

I am having trouble showing that the map I've defined is a homomorphism. So far, I've shown:

$\beta (\gamma (x_i + x_j)) = \alpha(x_i + x_j) = \alpha(x_i) + \alpha(x_j) = \beta(\gamma(x_i)) + \beta(\gamma(x_j)) = \beta(\gamma(x_i) + \gamma(x_j))$.

This shows that $\gamma (x_i + x_j)$ and $\gamma(x_i) + \gamma(x_j)$ have the same image under $\beta$, but we don't know that $\beta$ is injective.

Perhaps the map I've defined isn't right - an alternate idea would be to somehow map the generators of $F$ to those of $B$ in such a way that we get commutativity, but how to do so isn't clear to me.

The fact that your map $\gamma$ is a homomorphism follows almost directly from your statement "It is enough to define $\gamma$ on a set of generators," although you want a minimal set of generators, i.e. a basis. When we define a map on a basis and extend linearly, we do so precisely to ensure the resulting map is a homomorphism. That is, if $f\in F$, we write $f$ uniquely (hence why it's important to have a basis) as $$f=a_1x_1+\dots+a_nx_n$$ and then define $$\gamma(f)=a_1\gamma(x_1)+\dots+a_n\gamma(x_n).$$ From this you should be able to check the remaining details.