# Degree of the continuous map $S^1 \times S^1 \rightarrow S^2$.

In my course “Introduction to the algebraic topology” I’ve got the following exam problem:

Construct continuous map $S^1 \times S^1 \rightarrow S^2$, which degree is non-zero. Calculate the degree of this map.

I believe that my problem is understanding the concept of the degree. $S^1 \times S^1$ and $S^2$ are obviously orientable manifolds, so there top homology groups are isomorphaic to $\mathbb Z$. Any continuous map $f$ induces homomorphism of this groups $f_*$. By definition, degree of $f$ is just an image of $[1]$ (generator) under $f_*$.

So I need some hints how degree can be calculated in practice, i.e. some examples.

Thanks!

• I think you're putting the cart before the horse in this problem. I would construct a map that I think would work, and only then calculate its degree. Or do you already have a map? – Arthur Jan 22 '18 at 15:27
• Look at the section "differential topology" in en.wikipedia.org/wiki/Degree_of_a_continuous_mapping – Mariano Suárez-Álvarez Jan 22 '18 at 15:28

Look at the quotient map $S^1 \times S^1$ to $S^2$ by collapsing the subspace $S^1 \vee S^1$ to a point( basically taking smash product of circle with circle). This quotient map induces isomorphism on top homology groups hence degree is non zero.
• because it sends the fundamental class of $S^1\times S^1$ to the fundamental class of $S^2$. – Tsemo Aristide Jan 22 '18 at 15:58