Defining $f$ and $g$ in $\mathbb{Z} \xrightarrow[]{f} \frac{\mathbb{Z}}{2 \mathbb{Z}} \xrightarrow[]{g} \mathbb{Z}$ such $fg=1_{\mathbb{Z}}$

Question

I already work a counterexample seems to work for answer my own question here:

Homotopy category is not Abelian.

But I need to know if in the following sequence of functions in the category of abelian groups

$$\mathbb{Z} \xrightarrow[]{f} \frac{\mathbb{Z}}{2 \mathbb{Z}} \xrightarrow[]{g} \mathbb{Z}$$

there is some way to define $f$ and $g$ such $gf=1_{\mathbb{Z}}$. My intuition says this is not posible but cannot justify why?

Answer 1

Hint: $g$ cannot be surjective.

Answer 2

If such maps $f$ and $g$ exist then $f$ must be injective, but that is an absurd because $\mathbb{Z}/2\mathbb{Z}$ is a finite group whereas $\mathbb{Z}$ is an infinite group.