2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.