# Injective maps from free group on two generators

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.)

• Regardless, $F_2$ is not uncountable. Commented Apr 30, 2021 at 15:05
• What is "continuous" here? What topologies are you putting? Judging by the lack of topology tag, did you mean "homomorphism", by any chance? Commented Apr 30, 2021 at 15:07
• Right for some reason I was allowing words of infinite length. Commented Apr 30, 2021 at 15:07
• 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. Commented Apr 30, 2021 at 15:08
• But there is an injection from $F_2$ to $\mathbb Z \times \mathbb Z$, in fact there is a bijection, because both of them are countably infinite, i.e. both have bijections to $\mathbb Z$. Commented Apr 30, 2021 at 15:12

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 $$\phi$$.