# Show that no covering of the 2-torus has the homotopy type of a figure eight.

I'm not sure how to go about proving this.

I know that the standard cover of the torus is $\mathbb{Z}\times\mathbb{Z}$ and that's coming from $\pi_1(T)\cong \mathbb{Z}\times\mathbb{Z}$, correct?

I think I've also seen somewhere that the figure eight is a subset of the torus? Or there's an inclusion map from the figure eight to the torus? If there is such an inclusion map, then wouldn't it be obvious that the figure eight doesn't cover the torus? A subset of a space can't cover the space.

Or I had another thought and I'm not sure if this will work. If the fundamental group of the torus is $\mathbb{Z}\times\mathbb{Z}$, then by the Galois correspondence wouldn't subgroups of $\mathbb{Z}\times\mathbb{Z}$ correspond to covering spaces of the torus. So, if the figure eight was a covering space of the torus, then the fundamental group of the figure eight would be a subgroup of $\mathbb{Z}\times\mathbb{Z}$. But the fundamental group of the figure eight is the free group on two generators which is not a subgroup of $\mathbb{Z}\times\mathbb{Z}$. (Since $\mathbb{Z}\times\mathbb{Z}$ is abelian and $F(a,b)$ is not abelian. )

• I'd like to add that a subset of a space absolutely can cover the space. The torus has many subsets homeomorphic to $\mathbb{R}^2$, which is its universal cover. Also, the question is not about whether the figure 8 itself covers the torus, rather whether a space homotopy equivalent to the figure 8 does. – Charlie Cifarelli Aug 15 '16 at 1:01
There is an injection $\pi_1(Y) \to \pi_1(X)$ if $Y \to X$ is a covering (with some nice hypothesis on the spaces). But there is no injection : $F_2 \to \mathbb Z^2$ so there is no such covering.
• Any subgroup of $\mathbb{Z}^2$ is abelian. – Pedro Aug 15 '16 at 1:19
The fundamental group of the eight figure is the free group generated by two elements if $p:X\rightarrow T^2$ is a covering map, it is a Serre fibration, you have the exact sequence $\pi_2(T^2)=1\rightarrow \pi_1(X)\rightarrow \pi_1(T^2)$, thus $\pi_1(X)$ is a subgroup of $\pi_1(T^2)$, thus it is commutative. This implies that $X$ does not have the homotopy type of the figure eight since the fundamental group of $X$ and the figure eight are not isomorphic.