Showing that every map $f : S^2 \rightarrow S^1$ is homotopic to the trivial map

Question

Problem: Prove that every map $f : S^2 \rightarrow S^1$ is homotopic to the trivial map.

Hint: Use the covering space $E: \mathbb{R} \rightarrow S^1$. If you can show that every map $f: S^2\rightarrow S^1$ lifts to a map $\tilde{f}: S^2 \rightarrow \mathbb{R}$ then you can conclude $f$ is nullhomotopic because $\mathbb{R}$ is contractible.

I'm having some trouble understanding the big picture reasoning the hint is providing. So I want to map it out step by step to make sure my understanding is correct.

Verification that I understand the hint:

1. To say that every map $f: S^2\rightarrow S^1$ is homotopic to the trivial map is to say that $f$ is nullhomotopic. So what we're really trying to show is that $f$ is nullhomotopic.

2. I'm assuming also that all of the maps $f$, $\widetilde{f}$, and $E$ are continuous.

3. Now suppose we did establish that every $f: S^2\rightarrow S^1$ lifted to a map $\widetilde{f}: S^2 \rightarrow \mathbb{R}$ so that $E \circ \widetilde{f} = f$.

4. Then $\widetilde{f}$ is homotopic to a constant map since $\mathbb{R}$ is contractible.

5. Then $f = E \circ \widetilde{f}$ is homotopic to a constant map from (4).

6. Then $f$ is nullhomotopic as desired.

Question: So it seems like like all we need to show is that $\forall f$, we have that $\exists \widetilde{f}$ s.t. $f$ lifts to $\widetilde{f}$. (Then, from (1)-(6), we would have the desired result). How do we show this?

Answer 1

That's a theorem, actually a really important one about covering spaces.

This theorem states that

For every covering space $p \colon (E,e) \to (X,x)$ a function $f \colon (Y,y) \to (X,x)$, where $Y$ is connected and locally path connected, then $f$ has a lifting $\tilde f\colon (Y,y) \to (E,e)$, i.e. a continuous map such that $p \circ \tilde f=f$, if and only if the induced map $f_* \colon \pi_1(Y,y) \to \pi_1(X,x)$ has image contained into the image of the map $p_* \colon \pi_1(E,e) \to \pi_1(X,x)$.

note: the theorem deals with pointed spaces because of the uniqueness requirement if you forget about fixed points the the result implies the existance of a lifting but you lose the uniqueness requirement.

You can apply this theorem since $\pi_1(S^2) \cong \pi_1(\mathbb R)=0$ are the trivial group, so the image of the $f_*$ and $p_*$ are the trivial group. From that you can get a lift $\tilde f$.