3
$\begingroup$

this is a problem from Lee's Topological Manifolds, 11-21. It asks the following:

What compact, connected surfaces $M$ admit a non-nullhomotopic map to the circle?

So, we use the classification of surfaces. I can prove that none exist for the sphere since the sphere is simply connected and has trivial fundamental group, so the induced map would be trivial on the fundamental group; then there is a theorem as mentioned here (in the answer) that gives the result:

Continuous map from Projective Plane to Torus

This theorem also works for a single copy of the projective plane $\mathbb{P}^2$, but doesn't immediately work for $\mathbb{P}^2 \# \mathbb{P}^2 \# \cdots \# \mathbb{P}^2$. I can explicitly exhibit non-nullhomotopic maps from the orientable positive genus surfaces. So the part I am stuck on is the sums of projective planes. I definitely can't see one on the Klein Bottle so I tend to believe that there are none, but I can't prove it.

Can anyone give a hint to get it started? Thanks a lot!

$\endgroup$
4
$\begingroup$

There is one from the Klein bottle. The Klein bottle can be defined as follows: let it be $S^1 \times [0,1]/ (x,0)\sim (-x,1).$ This definition comes equipped with a map $K \to S^1$; it's $f(x,t)=t \in [0,1]/(0 \sim 1)$. If you calculate the fundamental group of $K$ (Try can Kampen along with the above definition - you'll essentially be doing van Kampen on two Möbius bands) you can calculate that this homomorphism is nontrivial on $\pi_1$.

To get maps from any $M = K \# n\Bbb{RP}^2$ (meaning the Klein bottle, connected sum $n$ projective planes) homotope the above to be constant on the ball you delete when connect summing, then extend it to be constant on the copies of $\Bbb{RP}^2$ you add.

(In general maps to the circle are in bijection with elements of $\text{Hom}(\pi_1(X),\Bbb Z)=H^1(X;\Bbb Z)$.)

$\endgroup$
2
  • $\begingroup$ Ok thanks, really appreciate it! $\endgroup$ – John Samples Jan 5 '16 at 8:23
  • $\begingroup$ I think the easiest way is to subsequently pinch the identified circular 'ends' of the tube to a point and show you get a "pinched torus" using quotient uniqueness. No SVK needed. Am I right? $\endgroup$ – John Samples Jan 5 '16 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.