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I have been told the statement in the title, but I don’t see how to construct such a function. An hint was to “collapse the $S^1$”, but I am not sure what that means. I should find a surjective function that is homotope to the identity, maybe?

tommy1996q
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  • This was asked so many times, see questions linked to https://math.stackexchange.com/questions/1782261/degree-1-map-from-torus-to-sphere – Moishe Kohan Jul 12 '19 at 18:03
  • Perhaps a more general question is easier: You can construct a degree 1 map from any compact, orientable surface to the sphere. – Ted Shifrin Jul 12 '19 at 18:06
  • Do you have any hints gor the more general case? Also, I have seen the other question, but the answer given below seems simpler. – tommy1996q Jul 12 '19 at 20:03

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You can think of $S^1\times S^1$ as a closed square with pairs of opposite edges identified. Now identify them to a point. This represents $S^2$ as a quotient of $S^1\times S^1$, and "most" points in $S^2$ have just one point in their preimage, so the degree of the map (subject to orientation) is $1$.

If you like, think of the torus with two circles, going around the torus in the two most natural ways. Collapsing these $S^1$s to a point gives an $S^2$.

Angina Seng
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  • It makes sense, but collapsing things won’t give me something non smooth? The only thing that puzzles me about your approach is the smoothness – tommy1996q Jul 12 '19 at 20:01
  • I'm sure all this can be "smoothed out" but I don't want to work out the details right now.... – Angina Seng Jul 12 '19 at 20:05