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!


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 $\text{Hom}(\pi_1(X),\Bbb Z)=H^1(X;\Bbb Z)$.)

  • $\begingroup$ 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

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