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.

  • $\begingroup$ 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. $\endgroup$ Dec 6 '20 at 22:20
  • $\begingroup$ 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. $\endgroup$ Dec 7 '20 at 0:28
  • $\begingroup$ 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. $\endgroup$ Dec 7 '20 at 0:54
  • $\begingroup$ 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? $\endgroup$ 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 ?

  • $\begingroup$ 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 $\mathbb T^2 \# ... \# \mathbb T^2 \to \mathbb T^2 \to \mathbb S^1$? $\endgroup$ Dec 5 '20 at 22:57
  • $\begingroup$ 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 $\endgroup$ Dec 5 '20 at 23:02
  • $\begingroup$ Shouldn't the fundamental group be $\mathbb Z / 2 * ... * \mathbb Z / 2$? $\endgroup$ Dec 5 '20 at 23:17
  • $\begingroup$ 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. $\endgroup$ Dec 6 '20 at 5:11
  • $\begingroup$ 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 $\endgroup$ Dec 6 '20 at 9:48

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