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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?

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user125103
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1 Answers1

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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 $\tilde f$.

Giorgio Mossa
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  • What is meant by $(E,e)$ and $(X,x)$? Are these distinct entities from the fundamental groups $\pi_1(E,e)$ and $\pi_1(X,x)$? – user125103 May 06 '14 at 16:00
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    Oh, I see: http://en.wikipedia.org/wiki/Pointed_space – user125103 May 06 '14 at 16:01
  • So, as in the case of my problem above, if we are forget about the pointed spaces, then $\widetilde{f}$ needn't be unique for each choice of $f$. That is, different choices of $f$ could result in different choices of $\widetilde{f}$. Is this the case? – user125103 May 06 '14 at 16:03
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    @user125103 So what? You need just a lift for the original problem, so why not pick points and take the corresponding lift. – Hagen von Eitzen May 06 '14 at 16:08
  • Ok -- so we don't need uniquness of $\widetilde{f}$ in the problem. Is what I wrote in the "verification that I understand the hint" section above correct? – user125103 May 06 '14 at 16:17
  • @user125103 yes your proof is correct. – Giorgio Mossa May 06 '14 at 16:23