5
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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 $\tilde f$.

$\endgroup$
6
  • $\begingroup$ What is meant by $(E,e)$ and $(X,x)$? Are these distinct entities from the fundamental groups $\pi_1(E,e)$ and $\pi_1(X,x)$? $\endgroup$
    – user125103
    May 6 '14 at 16:00
  • 1
    $\begingroup$ Oh, I see: en.wikipedia.org/wiki/Pointed_space $\endgroup$
    – user125103
    May 6 '14 at 16:01
  • $\begingroup$ 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? $\endgroup$
    – user125103
    May 6 '14 at 16:03
  • 2
    $\begingroup$ @user125103 So what? You need just a lift for the original problem, so why not pick points and take the corresponding lift. $\endgroup$ May 6 '14 at 16:08
  • $\begingroup$ 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? $\endgroup$
    – user125103
    May 6 '14 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.