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I am in the process of trying to use the general lifting lemma (Lemma 79.1) Munkres to show that every continuous map $f$ from $S^n$ to $S^1$ is null-homotopic for $n>1$. Now I know this question has been asked before on this site but my question is related to a small detail in the proof of it.

Because the fundamental group of $\pi_1(S^n)= 0$ I know that my $f$ can be lifted to a map $\tilde{f}$ from $S^n$ to $\Bbb{R}$ that is a covering space for $S^1$. This map is defined as follows: Suppose the basepoint of $S^n$ is $y_0$, the basepoint of $S^1$ is $x_0$ and the basepoint of $\Bbb{R}$ is $z_0$. For any $y_1 \in S^n$, choose a path $\alpha$ from $y_0 $ to $y_1$. Then $f \circ \alpha$ is a path in $S^1$ from $x_0$ to $f(y_1)$. I now lift this to a path $\gamma \in \Bbb{R}$ with startpoint $z_0$ and endpoint

$$p^{-1}(f(y_1)).$$

I now say that $\tilde{f}(y_1) = p^{-1}(f(y_1))$. $p$ is the usual covering map from $\Bbb{R}$ to $S^1$ given by the exponential function: $p(x) = e^{ix}$.

I want to define a null homotopy between $\tilde{f}$ and the constant map at $z_0$. I want it to be a straight line homotopy and I try with $$\begin{eqnarray*} F :&S^n \times I& \longrightarrow \Bbb{R} \\ &(y_1,t)& \longmapsto z_0(1-t) + tp^{-1}(f(y_1)). \end{eqnarray*}$$ The problem now is that when I take the inverse image under $p$ I don't have a well defined map? How do I get around this? Can I just say we pick a point in the inverse image and it will do the job?

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Why can't you just take $(y,t)\mapsto z_0(1-t)+t\tilde f(y)$? It seems that your lift is already guaranteed to exist. –  Andrew Aug 19 '12 at 7:04

1 Answer 1

up vote 4 down vote accepted

What you're actually having trouble with is that you can lift the map $S^n \to S^1$ to a map $S^n \to \mathbb{R}$. This is the map you're calling $p^{-1}(f(y))$, and is not a multi-valued map.

I would advise rereading over the section where they showed you can lift maps to covering spaces, since after that it is trivial (set $g(x)=p^{-1}(f(x))$, which is a continuous map from $S^n\to \mathbb{R}$, hence has an obvious null-homotopy, what you wrote with $p^{-1}(f(y_1))$ replaced with $g(y_1)$).

Cheers,

Rofler

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Yes I realised that I missed that bit. Indeed the map is not multivalued. I forgot we checked that in proving the lemma. So the null homotopy is well-defined and composing it with $p$ we get a null homotopy between $f$ and the constant map at the basepoint of $S^1$. –  user38268 Aug 19 '12 at 7:11

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