# Compact connected surfaces such that non-null homotopic map exists from itself to $\mathbb S^1$.

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?

(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.

• Why is having a non-trivial homomorphism between fundamental groups sufficient? I think it would be sufficient only if such homomorphism guarantees that there is a continuous map that induces it. Dec 6 '20 at 22:20
• Because if you have two spaces which are $K(G,1)$ it is enough to give a map at the level of groups. If you have a group homomophism between two groups say H to G then you can make a principle G bundle over any space a principle H bundle via the homomorphism. But principle G bundles for discrete groups are classified by $K(G,1)$ hence it lift to the level of spaces. Hope this helps. Dec 7 '20 at 0:28
• To prove the fact that the covering space, of compact surfaces of higher genus is contractible you may take a look at John Lee's book called Geometry of Surfaces. The basic idea is to get a tesselation of the disc or upper half plane with the $4g$-gon identification of the surface. Dec 7 '20 at 0:54
• I am studying using John Lee’s introduction to topological manifold (this is the problem 11-21) and not sure if I ever saw words like $K(G, 1)$ or bundles yet. Is there a way to get the result you are mentioning with what is in the book I mentioned? Dec 7 '20 at 1:21

You can use the Van Kampen theorem to determine the fundamental group of a connected sum, this should answer your question for the projective planes.

For the torus, what do you think, if $$n=1$$, of the projection $$\mathbb T^2 \cong S^1\times S^1\to S^1$$ ? What does it tell you for $$n\geq 2$$ ?

To prove the result you are assuming, you should use the lifting criterion: suppose $$\tilde X\to X$$ is a covering space, with say both $$\tilde X$$ and $$X$$ connected; then what is a criterion for when a map $$Y\to X$$ can be lifted to a map $$Y\to \tilde X$$ ?

What does it tell you when $$\tilde X$$ is contractible ?

• Could you elaborate bit more? I think I see thee case for $\mathbb P^2$, but not sure for $\mathbb P^2 \# ... \# \mathbb P^2$ in general. Also, for the torus case, do I look at $\mathbb T^2 \# ... \# \mathbb T^2 \to \mathbb T^2 \to \mathbb S^1$? Dec 5 '20 at 22:57
• For the torus case yes, that is a possible example. For the projective plane, do you know how to compute the fundamental group of this connected sum ? Start with two projective planes Dec 5 '20 at 23:02
• Shouldn't the fundamental group be $\mathbb Z / 2 * ... * \mathbb Z / 2$? Dec 5 '20 at 23:17
• What should be the map from a connected sum of T^2 to T^2? Also, I am not sure how to show any continuous map from P^2 # ... # P^2 should be null-homotopic from the result I am using; for example, P^2 # P^2 has a fundamental group isomorphic to an infinite dihedral group, which has non-trivial homomorphism to Z. Dec 6 '20 at 5:11
• So you figured out yourself that the answer was not the free product of Z/2's. In fact, for a connected sum of more than 1 projective planes there will be non nullhomotopic maps to $S^1$, using the fact that those come from maps on $\pi_1$. For tori, note that you can kill a copy of T^2 in the connected sum, and this leave you with a T^2 where a small disk has been shrunk to a point, but this is homotopy equivalent to T^2 Dec 6 '20 at 9:48