# Realizing a map between first homology groups of surfaces by a continuous map

Suppose $S$ is a closed, oriented surface of genus $g$ and $\chi$ is some element of $H^1(S; \mathbb{C})$, which I prefer to think of as a map $\chi : H_1(S) \to \mathbb{C}$. Now suppose this $\chi$ is such that its image is a lattice $\Lambda$ in $\mathbb{C}$. Then for each homology class $\gamma$ you could interpret $\chi(\gamma)$ as a closed loop on the complex torus $\mathbb{T}^2 = \mathbb{C} / \Lambda$, and looking at the homology class of that loop you could think of $\chi$ as a map $\chi : H_1(S) \to H_1(\mathbb{T}^2)$. Something I'm reading now makes a remark that this map between homology groups must be induced by a map $f : S \to \mathbb{T}^2$, and I don't see why this needs to be the case. In general maps between homology groups don't need to be induced by maps between the spaces, so is there something special here that promises the existence of such an inducing map in this case?

For a hands-on, low-tech way of seeing this, consider the standard presentation of $S$ as a quotient of a $4g$-gon. The boundary of the $4g$-gon is made up of edges with identifications, with each edge representing a class in $H_1(S)$. Given any $\chi:H_1(S)\to H_1(\mathbb{T}^2)$, there is a map from the boundary of the $4g$-gon to $\mathbb{T}^2$ which agrees with $\chi$ on the homology classes represented by each edge of the boundary of the $4g$-gon. To extend this map over the interior of the $4g$-gon and thus get a map from $S$ to $\mathbb{T}^2$, you just need a nullhomotopy of the map $S^1\to \mathbb{T}^2$ given by going all the way around the boundary of the $4g$-gon. Such a nullhomotopy is guaranteed to exist since this loop represents a product of commutators in $\pi_1(S)$, and $\pi_1(\mathbb{T}^2)$ is abelian (or alternatively, this product of commutators represents $0$ in $H_1(S)$ so $\chi$ sends it to $0$ in $H_1(\mathbb{T}^2)=\pi_1(\mathbb{T}^2)$).
For a higher-tech perspective, note that $\mathbb{T}^2$ is an Eilenberg-Mac Lane space $K(\Lambda,1)$: its only nontrivial homotopy group is $\pi_1$. Thus for any CW-complex $X$, homotopy classes of maps $X\to\mathbb{T}^2$ are naturally in bijection with cohomology classes in $H^1(X;\Lambda)$, or equivalently homomorphisms $H_1(X)\to\Lambda$.
You might as well take the special case $\Lambda=\Bbb Z^2$. Then a map $S\to \Bbb T^2$ is really a pair of maps $S\to\Bbb T= \Bbb R/\Bbb Z$. So basically we want each group homomorphism $H_1(S)\to H_1(\Bbb T)\cong\Bbb Z$ to be induced by a continuous map. But a map $H_1(S)\to \Bbb Z$ is essentially a cohomology class in $H^1(S,\Bbb Z)$. Take a smooth model of $S$; then we can embed $H^1(S,\Bbb Z)$ in $H^1_{dR}(S)$, the first de Rham cohomology group. So our map $H_1(S)\to H_1(\Bbb T)$ comes from a closed $1$-form $\omega$ on $S$. For the cohomology class to be integral, one needs all its period integrals to be integers. In this case $\omega$ lifts to an exact one form on $S^*$, the universal cover of $S$, and this is $df$ where $f:S^*\to\Bbb R$. The condition that periods are integers means $fs$ induces a map $F:S\to \Bbb T$ and this is exactly what we need.