$S^2$ cannot cover $T$, the torus

Solution 1:

Observe that the universal cover of $T$ is $\mathbb R^2$, via the map $\mathbb R^2\to \mathbb R^2/\mathbb Z^2\cong T$. If $S^2$ also covered $T$, then since it is simply connected it would also be a universal cover of $T$. Since universal covers are unique, this would imply $S^2\cong \mathbb R^2$, which is clearly false as $S^2$ is compact and $\mathbb R^2$ is not.

Solution 2:

First observe that if $\pi: Y \rightarrow X$ is a covering map and $Y$ is compact and connected, then the degree must be finite.

Second: yes, use Euler characteristics! Remember that if $\pi: Y \rightarrow X$ is a finite degree covering map, then $\chi(Y) = (\operatorname{\deg} \pi) \chi(X)$. I think you'll find that you're done.

Solution 3:

The other answers are great and so I thought I'd just add a 'high powered'* method for getting the same result. As Alex said in his answer, if $S^2$ covered $T$ then $S^2$ would be the universal cover of $T$. Note though, that if $p\colon X\rightarrow Y$ is a covering map, then the induced map $p_*\colon\pi_n(X)\rightarrow\pi_n(Y)$ on homotopy groups is an isomorphism for $n\geq 2$.

The second homotopy group of $S^2$ is cyclic abelian on one generator but the second homotopy group of $T$ is trivial as $T$ is covered by the plane $\mathbb{R}^2$ which has trivial homotopy groups at all levels. That is, $\pi_2(T)\cong\pi_2(\mathbb{R}^2)=0$ and $\pi_2(S^2)\cong\mathbb{Z}$ and so no isomorphism exists between homotopy groups. It follows that no such covering map can exist.

*high powered only in the sense that it uses the higher homotopy groups of a space and results about induced maps on them from covering space theory.