First, I know that any compact connected surface is either $\mathbb S^2$, $\mathbb P^2 \# \cdots \# \mathbb P^2$ ($n$ copies), or $\mathbb T^2 \# \cdots \# \mathbb T^2$ ($n$ copies).
Also, suppose I know the following result:
-Given $X$ is a connected space with a contractible universal covering
space, $f : Y \to X$ is null homotopic if and only if the induced homomorphism $f_* : \pi_1(Y, y) \to \pi_1(X, f(y))$ is trivial for each $y \in Y$.
For the first case, I know that since the fundamental group of $\mathbb S^2$ is trivial, it must be that any continuous $f$ from $\mathbb S^2$ to $\mathbb S^1$ should be trivial.
For the second case, when $n$ is just one, since the fundamental group of the projective plane is just a group of order $2$.
However, I am not sure how to proceed with other cases, and show that there exists/does not exist non-null homotopic maps.
Also, what should I do to prove the fact I am assuming?
Best Answer
(This should be a comment but I dont have enough reputation to make a comment)
The non-orientable surface $\Sigma$ has a contractible covering space hence it is $K(\pi_{1}(\Sigma),1)$ in order to give a map to $S^1$ which is $K(\mathbb{Z},1)$ you can give a map at the level of groups. As you pointed out the fundemental group of $\mathbb{RP}^2$#$\mathbb{RP}^2$ is the infinite dihedral group, hence you get a map to $\mathbb{Z}$.
Even in the case of orientable surfaces you can produce a lot of map by repeating the arguement above, as the universial cover of orientable surfuce of genus $\geq$ 2 is disc (or equivalently the upper half plane). Hence the question boils down to a purely group theoretical question. In particular if you have elements of infinite order in the fundemental group you have a map. Which might have several interesting geometric intepretations.