There is a smooth map of degree 1 from $S^1 \times S^1$ to $S^2$

Question

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?

Answer 1

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 $S^2$.