# Homomorphism between free groups induced by group homomorphism of abelian groups.

Suppose we have finitely generated abelian groups $G_1$ and $G_2$. Then there exists free groups $F_1$ and $F_2$ such that $F_i \stackrel{q_i}\rightarrow G_i$ maps the basis elements of $F_i$ to generators of $G_i$. Given a group homomorphism $g: G_1 \rightarrow G_2$, I want to define a map $f: F_1 \rightarrow F_2$ such that the following diagram commutes: \begin{array}{ccccc} F_1 & \stackrel{q_1}\rightarrow & G_1 & \rightarrow &0 \\ \downarrow f & & \downarrow g\\ F_2 & \stackrel{q_2}\rightarrow & G_2 & \rightarrow &0 \end{array} The obvious map is to let $f(x)$ be an/the element $y \in F_2$ such that $q_2(y) = g \circ q_1(x)$. I don't think that this is well defined in general though.

My question is then whether or not this map is ever well defined, or if we can construct a different map that is well defined? I can't think of any possible maps that may work.

Edit: I've thought about this, we need $f(x)$ to be an element of $q_2^{-1}(g\circ q_1(x))$. There may be several elements to choose from, but I don't think there's any danger in just choosing one for each basis element of $F_1$ and defining $f$ using our choices?

## 1 Answer

You want $f$ to be a group morphism. Simply letting $f(x)$ be any element of $q_2^{-1} \left( (g \circ q_1)(x)\right)$ may not produce a group morphism.

For example, let $g$ be the identity $\mathbb{Z}^2 \to \mathbb{Z}^2$. Say $\mathbb{Z}^2$ is generated by $x,y$ and $F_1 = F_2$ by $a,b$. You could make the choices $f(a) = x$, $f(b) = y$ and $f(ab) = ab[a,b]$, but then $f$ is not a morphism.

Edit: it seems you've covered this in your edit by indeed restricting to the basis elements, as I did below.

Instead, let's restrict your idea to the generators of both groups. Let $\{y_1, \dots, y_n\}$ be a minimal generating set of $G_1$. Then $g$ is completely determined by the images of those generators, i.e. $g(y_1), \dots, g(y_n)$.

Now let $F_1$ be the free group of rank $n$ with generators $x_1, \dots, x_n$, and let $q_1: F_1 \to G_1$ be determined by $q_1(x_1) = y_1$.

We can now take your idea, but restrict it to the generators: for each $i \in \{1, \dots, n\}$, define $f(x_1)$ by taking any element from $q_2^{-1} \left( (g \circ q_1)(x_1)\right)$. This uniquely determines $f$ as a map, and free groups have the nice property (see this answer) that any choice of images of the generators will produce a morphism (since there are no relations to be satisfied).