The question was arisen from doing Hatcher 1.3.24 (b). I would like to know what is wrong with the following.

Let $X$ be a space that is path-connected and locally path-connected and $G$ be a group that makes $X \twoheadrightarrow X/G$ a normal covering space. Let $H_{1}, H_{2} \leqslant G$ be subgroups. Then $X \twoheadrightarrow X/H_{1}$ and $X \twoheadrightarrow X/H_{2}$ are normal covering spaces as well with $H_{j} = \text{Aut}(X \twoheadrightarrow X/H_{j})$.

Now, fix any $\phi_{1} \in H_{1}$. Consider any $x_{0} \in X$ and by normality, we have $\phi_{2} \in H$ such that $\phi_{2}(x_{0}) = \phi_{1}(x_{0})$. But then $\phi_{1}, \phi_{2} \in G$ and since $X$ is path connected, the deck transformations are uniquely determined by the image of a fixed point. Therefore, we have $\phi_{1} = \phi_{2} \in H_{2}$, showing $H_{1} \leqslant H_{2}$.

The argument can be rewritten by switching $1$ and $2$, so $H_{1} = H_{2}$.

Problem. I was supposed to show that if $X/H_{1}$ and $X/H_{2}$ are isomorphic as covering spaces of $X/G$, then $H_{1}$ and $H_{2}$ are conjugate subgroups in $G$. Clearly, I haven't used the hypothesis of the statement and the conclusion is stronger as well.

  • $\begingroup$ Wy are the coverings $X\to X/H_i$ normal? $\endgroup$ – Andreas Blass Aug 11 '13 at 15:31

Always good to make a diagram:

$$\begin{matrix} X & \xrightarrow{\pi_1} & X/H_1 & \xrightarrow{\psi_1} & X/G\\ & & \downarrow\iota & & \downarrow\\ X & \xrightarrow[\pi_2]{} & X/H_2 & \xrightarrow[\psi_2]{} & X/G \end{matrix}$$

So the task is to show that if there exists a homeomorphism $\iota \colon X/H_1 \to X/H_2$ with $\psi_2 \circ \iota = \psi_1$, then $H_1$ and $H_2$ are conjugate in $G$. (Note: except if $H_1 = H_2$, we do not have $\iota\circ \pi_1 =\pi_2$.)

Now, fix any $\phi_{1} \in H_{1}$. Consider any $x_{0} \in X$ and by normality, we have $\phi_{2} \in H$ such that $\phi_{2}(x_{0}) = \phi_{1}(x_{0})$.

No. The normality of $\pi_i$ says that $H_i$ operates transitively on the fibres of $\pi_i$, not on the fibres of $\pi \colon X \to X/G$. $x_0$ and $\phi_1(x_0)$ generally lie in different orbits under the action of $H_2$, and that means in different fibres of $\pi_2$, i.e. in general, there is no $\phi_2 \in H_2$ with $\phi_2(x_0) = \phi_1(x_0)$. If there always is, then $H_2 \supset H_1$.

So let's prove that if $\iota$ exists, then $H_1$ and $H_2$ are conjugate (the converse holds too, if $H_1$ and $H_2$ are conjugate, then there is such an $\iota$).

Pick any $x \in X$. Choose any $y \in X$ with $\pi_2(y) = \iota(\pi_1(x))$.

Then $\pi_G(y) = \psi_2(\pi_2(y)) = \psi_2(\iota(\pi_1(x))) = \psi_1(\pi_1(x)) = \pi_G(x)$, therefore there is a $g \in G$ such that $gx = y$.

Claim: $\pi_2 \circ g = \iota \circ \pi_1$, i.e. $g$ is a lift of $\iota$.

We know that $\pi_2(gx) = \iota(\pi_1(x))$. Let $z \in X$ arbitrary, and choose a path $\gamma$ connecting $x$ and $z$. Since paths can always be lifted, there is a path $\beta$ in $X$ starting at $y$ such that $\pi_2 \circ \beta = \iota \circ \pi_1 \circ \gamma$. Now,

$$\pi_G \circ \beta = \psi_2\circ \pi_2 \circ \beta = \psi_2 \circ \iota \circ \pi_1 \circ \gamma = \psi_1 \circ \pi_1 \circ \gamma = \pi_G \circ \gamma,$$

that is, $\beta$ is a lift of $\pi_G \circ \gamma$ starting in $y$. But $g\cdot \gamma$ is also such a lift of $\pi_G \circ \gamma$, and hence $\beta = g\cdot\gamma$ by the uniqueness of lifts, and thus $\beta(1) = g\gamma(1) = gz$, whence $\pi_2(gz) = \pi_2(\beta(1)) = \iota(\pi_1(\gamma(1))) = \iota(\pi_1(z))$.

$\pi_2\circ g = \iota \circ \pi_1$ is a different way of writing $H_2g \supset gH_1$ or $gH_1g^{-1} \subset H_2$.

We have not yet used that $\iota$ was supposed to be a homeomorphism, just that it was a continuous map with $\psi_2\circ \iota = \psi_1$. The same construction for $\iota^{-1}$ choosing $y = gx$ and $x$ as starting points shows $\pi_1 \circ g^{-1} = \iota^{-1}\circ \pi_2$ or $g^{-1}H_2 \subset H_1g^{-1}$. Together $H_2 = gH_1g^{-1}$.

  • $\begingroup$ Thank you. I have followed different books from Hatcher since when I posted this question and now this started to make sense (maybe Hatcher was too difficult for me to start with this subject). $\endgroup$ – Gil Oct 15 '13 at 14:54
  • $\begingroup$ @DanielFischer Can you please explain why is $\pi_2\circ g=i\circ \pi_1$ is same as saying that $H_2g\supseteq gH_1$? Thank you. $\endgroup$ – caffeinemachine May 29 '16 at 11:18
  • $\begingroup$ @caffeinemachine Pick $\xi \in X$ and $h_1 \in H_1$. Then we have $\pi_1(h_1\xi) = \pi_1(\xi)$ by definition of $\pi_1$, and consequently $\iota(\pi_1(h_1\xi)) = \iota(\pi_1(\xi))$. Now using $\iota\circ\pi_1 = \pi_2(g)$ we obtain $\pi_2(gh_1\xi) = \pi_2(g\xi)$, that is, $g\xi$ and $gh_1\xi$ lie in the same fibre of $\pi_2$, i.e. there is a $h_2\in H_2$ with $h_2(g\xi) = gh_1\xi$. The deck transformations $h_2g$ and $gh_1$ agree at one point ($\xi$), so they are the same deck transformation (since $X$ is connected), i.e. $gh_1 = h_2g$. That means $gh_1 \in H_2g$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22
  • $\begingroup$ Since $h_1\in H_1$ was arbitrary, we have $gH_1\subseteq H_2g$. This is the direction I use, and I didn't really intend to state equivalence. But well, I sort of did, so: Conversely, if we have $gH_1\subseteq H_2g$, the covering $\pi_2\circ g\colon X\to X/H_2$ factors through $X/H_1$, i.e. there is a unique covering map $\theta\colon X/H_1\to X/H_2$ with $\pi_2\circ g=\theta\circ\pi_1$, and one verifies that $\psi_2\circ\theta=\psi_1$. By definition of $g$, we have $\iota(\pi_1(x)) = \pi_2(gx) = \theta(\pi_1(x))$, so the two covering maps $\theta$ and $\iota$ agree at the point $\pi_1(x)$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22
  • $\begingroup$ Locally, we can write both of $\iota$ and $\theta$ in the form $\psi_2^{-1}\circ \psi_1$ (with an appropriate branch of $\psi_2^{-1}$), hence [since the $\psi_k$ are local homeomorphisms], the sets $A = \{p \in X/H_1 : \theta(p) = \iota(p)\}$ and $B = \{ p \in X/H_1 : \theta(p) \neq \iota(p)\}$ are both open. By connectedness, one of the two is empty. Since $\pi_1(x) \in A$, it follows that $B = \varnothing$, i.e. $\theta = \iota$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22

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