Why are there no injective continuous maps from $F_2$ - the free group on two generators - and $\mathbb{Z} \times \mathbb{Z}$? Is it because the free group is uncountable but $\mathbb{Z} \times \mathbb{Z} $ is countable?
(I want to show there is no covering map between the figure 8 and the torus. Showing there is no injection from the free group to the cartesian product would be enough.)
There is an injective map $\phi: F_2\rightarrow\mathbb{Z\times Z}$ as both groups are countably infinite.
However, there is no injective homomorphism from $F_2$ to $\mathbb{Z}\times\mathbb{Z}$ because there are two-generated groups which are non-abelian ($S_3$ is a good example). This proves that the figure-of-eight does not cover the torus.
That is, if there exists an injective homomorphism $\phi:F_2\hookrightarrow\mathbb{Z\times Z}$ then $F_2$ would be abelian (why?). On the other hand, by the universal property of free groups, we have a surjective homomorphism $F_2\twoheadrightarrow S_3$, so $F_2$ cannot be abelian. Hence, there is no injection $\phi$.