Intuition on homotopy group

Question

I just started learning some algebraic topology and trying to make sense of the following identity, $$ \pi_m(S^n) = \begin{cases} \mathbb{Z} , & m = n \\ 0 , & m < n \, . \end{cases} $$ I think of $\pi_1(S^1) = \mathbb{Z}$ in the following way: looping/ winding around a circle twice or once are not equivalent. Or I can not "bend, stretch, shrink" a loop with "winding" = 1 to "winding" = 2 on a circle. Hence $\mathbb{Z}$ is nothing but the winding numbers. First of all is this way of thinking correct? Second, I can not make sense of why should $\pi_1(S^2)$ vanish ?

Answer 1

Yes that way of thinking is correct. When you learn the proof of this fact from the point of view of covering space theory you will see how to make this intuition precise.

For your second question, this is actually a non-trivial theorem. The subtle point is that it is possible for a continuous map from $S^1$ to $S^2$ to be surjective. However, it is always possible to make a homotopy that makes a surjective $f: S^1 \to S^2$ no longer surjective, essentially because $S^1$ is one-dimensional so you can pick a point and push the image off bit by bit (you pick a neighborhood of this point that misses your basepoint, then there is a linear homotopy to the boundary circle that you can write down - this is probably best explained with a picture, not words). Once you have a map $f: S^1 \to S^2$ which is not surjective, it will factor through $R^2$ (since the sphere without a point is homeomorphic to $R^2$ via the stereographic projection), and now you can use the contractictibility of $R^2$ to conclude that any such loop was nullhomotopic.