Let $f\colon \Sigma \rightarrow S^1$ be a map from a genus $2$ surface to $S^1$. If $y\in S^1$ is a regular value of $f$ and $f^{-1}(y)$ is a nonseparating circle of $\Sigma$. How can I prove that $f$ is not a nullhomotopic map?
Assuming that $f$ is nullhomotopic, we can factor the map through $\mathbb{R}$, how can we conclude from here?
If $f$ factorizes through the universal covering $\exp2\pi i\colon \mathbf R \rightarrow S^1$ of $S^1$, then there is a map $\tilde f\colon \Sigma\rightarrow \mathbf R$ such that $\exp2\pi i \tilde f=f$. Since $f^{-1}(y)$ is a circle, it is connected. Its $\tilde f$-image in $\mathbf R$ is, therefore, a singleton $\{\tilde y\}$, and $$ f^{-1}(y)=\tilde f^{-1}(\tilde y). $$ One obtains a disjoint union $$ \Sigma\setminus f^{-1}(y)=\tilde f^{-1}(-\infty,\tilde y)\coprod\tilde f^{-1}(\tilde y,+\infty) $$ which contradicts the fact that $f^{-1}(y)$ is a nonseparating circle. Note that both subsets of the disjoint union are indeed nonempty as $\tilde y$ is a regular value of $\tilde f$.