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In this theorem, to prove that $\chi$ is onto they construct the function $\phi$ such that $\chi(\phi)=[\sigma]$. They have to prove $\phi \in G$, so they prove $\phi$ is continuous. But in order to have $\phi \in G$ we also need that it's an homeomorphism, but they don't prove it, and they don't even mention that it's easy to prove it. Maybe it's too obvious but I really don't see it. Is it very obvious? I've tried to prove it's onto but if I take $\tilde{e} \in E$, I don't know which $e \in E$ I should take so that $\phi(e)=\tilde{e}$, $\tau'$ depends on $e$ but not on $\tilde{e}$, so I don't know how to relate $e$ and $\tilde{e}$.

Also, when they prove that $\phi$ es contiuous, they only use the fact that $p \phi = p$, but they don't use the definition of $\phi$. Does that mean that any $\gamma$ such that $p \gamma = p$ is continuous?

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user642796
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user29232
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1 Answers1

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  1. To see that $\phi$ is a homeomorphism, construct its inverse. (I can elaborate if you want, but I hope this is a good nudge.)

  2. To see that $\chi$ is onto, let $[\sigma] \in \pi_1(X,x_0)$. Then $\sigma$ lifts to a path in $E$, based at $e_0$, from $e_0$ to $e_\sigma$, where $e_\sigma$ is some element of $p^{-1}(x_0)$.

    So what is $\phi$? It needs to rearrange the elements of $E$.

    Let $e \in E$. Then $\phi$ is the self-map of $E$ which lifts the loop $\tau^{-1} \sigma \tau$ in $X$ to a path in $E$ based at $e$, and then takes the endpoint. It's just like the situation above, where we set $e=e_0$.

  3. Does $p \gamma = p$ imply $\gamma$ is continuous?

    Definitely not! Consider the universal covering map $p:\mathbb{R} \to S^1$ and $\gamma:\mathbb{R} \to \mathbb{R}$ defined by the identity on non-integer points, but $n \mapsto n+1$ on integer points. Observe that $p \gamma = p$, but $\gamma$ is far from continuous.

  4. So why is $\phi$ continuous?

    Let $e_1 \in E$. There is an open pathwise connected neighborhood $U$ of $p(e_1)$ that is evenly covered; let $S_1$ denote the sheet which is mapped homeomorphically onto $U$ by $p$ and contains $e_1$. Now if $e$ is in $S_1$, we can join $e_1$ to $e$ by a path $\tau'$ which is in $S_1$. Now we map the whole picture by $\phi$, and we get two points $\phi(e_1)$ and $\phi(e)$ sitting in a sheet $S_1'$. In particular, $\phi$ restricts to a well defined map $\phi |_{S_1}:S_1 \to S_1'$, but even better, we have a formula for this restriction in terms of restrictions of $p $ to $S_1$ and $S_1'$, which are homeomorphisms: $\phi|_{S_1}=(p |_{S_1'})^{-1}\circ p|_{S_1}$. In particular, $\phi$ is continuous.

Let me know if you have any questions :)

Max
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  • Yes, that is the inverse; moreover $e \mapsto e \cdot [\tau^{-1} \sigma \tau]^{-1}$ is the self-map associated to $\sigma^{-1}$, so you will know that it is continuous once you know that $\phi$ is continuous. And yes, they certainly do use the definition of $\phi$ to prove it's continuous. I will edit my answer shortly so we can digest that part of the proof a little more easily. – Max Nov 30 '15 at 06:12
  • Ok, here's a bit of intuition: The key idea is that $\phi$ does not jump between sheets; $p\phi=p$ not just on all of $E$, but even on small sheets, where $p$ is a homeomorphism. – Max Nov 30 '15 at 08:06
  • It is not because $p \phi = p$, it's because $p|{S_1'} \circ \phi |{S_1} = p |_{S_1}$, which is a stronger statement. – Max Nov 30 '15 at 18:43