# Is there a map from the torus to the genus 2 surface which is injective on homology?

Let $T$ denote the torus and $M_2$ the genus 2 surface. Specifically, I am wondering if there is a map $f\colon T\to M_2$ such that $f_*\colon H_1(T)\to H_1(M_2)$ is injective. By thinking about the explicit generators of each of these groups, it seems to me that no such map exists, but I'm not sure how to show it. By considering the fundamental polygons of these surfaces and assuming $f$ is cellular, it looks like if such a map exists, then it would have to "wrap" $T$ completely around $M_2$ and send each generator in $H_1(T)$ to a sum in $H_1(M_2)$.

Note that not all maps $T\to M_2$ are null-homotopic: there is a map which sends one generator of $H_1(T)$ to a generator of $H_1(M_2)$, and the other to $0$, by projecting $T$ onto an annulus and mapping that onto the desired generator.

Let $M_g$ be a genus $g$ surface, $g>1$.

If you had some map $f: T \to M_g$ inducing an injection on first homology, it would also induce an injection on fundamental groups. This is because the map on first homology is the abelianization of the map on fundamental groups, and $\pi_1(T)$ is already abelian.

So if this were the case, $\pi_1(M_g)$ would have a subgroup isomorphic to $\Bbb Z^2$. So by general covering space theory $M_g$ has a covering space with fundamental group $\Bbb Z^2$; as all covering spaces of a surface are surfaces, and (by classification of compact surfaces, and because fundamental groups of noncompact surfaces are free) the only surface with fundamental group $\Bbb Z^2$ is the torus, this means there is a covering map $T \to M_g$. Because $T$ is compact, this must be a finite-sheeted cover. But if $X \to Y$ is an $n$-sheeted cover of finite CW complexes (or manifolds, if you like), then $\chi(X) = n\chi(Y)$; and $\chi(T) = 0, \chi(M_g) = 2-2g$. This is a contradiction.

So no map $T \to M_g$ for $g>1$ induces an injection on first homology.

As a result of this, we can classify all maps $T \to M_g$ up to homotopy.

Because there is a unique map up to homotopy $K(\pi, 1) \to K(\pi',1)$ for every homomorphism $\pi \to \pi'$, and the universal cover of $M_g, g>0$ is homeomorphic to $\Bbb R^2$, $M_g$ is a $K(\pi,1)$. Because we know now that no map $\Bbb Z^2 \to \pi_1(M_g)$ is injective (this is the only part we couldn't have done beforehand!), $\pi_1(M_g)$ has no torsion (if it did, there's a surface quotient of $\Bbb R^2$ with finite fundamental group; but this surface must be noncompact so as above this is not possible), and the only torsion-free quotients (that aren't the 'trivial quotient') of $\Bbb Z^2$ are $\Bbb Z$ and the trivial group, the image of any homomorphism $\Bbb Z^2 \to \pi_1(M_g)$ is cyclic or trivial.

In the first case, we may factor the homomorphism as $\Bbb Z^2 \to \Bbb Z \to \pi_1(M_g)$, where the first map is the quotient by $\langle (a,b)\rangle$, where $\text{gcd}(a,b) = 1$. Such a map is represented by a map $T \to T \to S^1$ that sends $(a,b)$ to $(0,1)$ and then projection onto the first factor of $T = S^1 \times S^1$. The map $S^1 \to M_g$, then, is just the map representing some element of $\pi_1(M_g)$.

So the homotopy classes of maps $T \to M_g$ are the trivial map, and those that factor as above.

• I am not particularly satisfied by my invoking classification of surfaces; the only way I see that point is by using uniformization and noting that $\Bbb Z^2$ is not a free group of isometries on the hyperbolic plane, which one can find a proof of out there in the wild. I don't really want to invoke a geometric argument here. – user98602 Apr 12 '15 at 0:40
• You only need the weaker result that a noncompact surface has free fundamental group. See, for example, mathoverflow.net/questions/18454/…. – Qiaochu Yuan Apr 12 '15 at 0:47
• @QiaochuYuan Thanks for reminding me of that. I've updated the answer. – user98602 Apr 12 '15 at 0:50
• Very nice proof! – Jason DeVito Apr 12 '15 at 2:04