Why is the 2nd homotopy group of the circle trivial?

Question

As stated in the title I am trying to prove $\pi_{2} (S^1) \cong 0$ . I have no clue where to start. I know that the meaning is: there is no way to transform a sphere into a circle.

Can anyone help me on elaborating the proof? I would be so glad.

Answer 1

Let $f: (S^2, x_0) \to (S^1, y_0)$ be a continuous map. Our goal is to show that $f$ is homotopic to the trivial map sending every point in $S^2$ to $y_0$.

Recall that $\mathbb R$ is the universal cover for $S^1$. Let $p : (\mathbb R ,z_0) \to (S^1, y_0)$ be the standard projection map. Since $\pi_1(S^2, x_0) = 0$, it is clear that $f_\star(\pi_1(S^2, x_0)) \subset p_\star (\pi_1(\mathbb R, z_0))$. Therefore, by the lifting criterion, our map $f : (S^2, x_0) \to (S^1, y_0)$ lifts to a map $\widetilde f : (S^2, x_0) \to (\mathbb R, z_0)$ obeying $p \circ \widetilde f = f$.

But $\mathbb R$ is convex, so $\widetilde f$ is clearly homotopic to a trivial map sending every point in $S^2$ to $z_0$. Composing this homotopy with $p$, we see that $f$ is homotopic to the trivial map sending every point in $S^2$ to $y_0$, as required.