Let $X$ be a path-connected and locally path-connected topological space. The action of a topolgical group $G$ on $X$ is a covering space action. For any subgroup $H < G$, we have a composition of covering space $X \rightarrow X/H \rightarrow X/G$.

  1. Prove that any covering space of $X/G$ between $X$ and $X/G$ is isomorphic to $X/H$, for a subgroup $H$ of $G$.
  2. $X/H_1 \rightarrow X/G$ and $X/H_2 \rightarrow X/G$ are isomorphic as covering spaces if and only if $H_1$ and $H_2$ are conjugate subgroups of $G$.
  3. The covering space $X/H \rightarrow X/G$ is normal if and only if $H$ is a normal subgroup of $G$, in which case there is an automorphism $$\text{Aut} (X/H \rightarrow X/G) \cong G/H.$$

I have no idea as to (1). The sufficiency parts of (2) and (3) seems to be easy. As to the necessity part of (2), supposet that $\phi: X/H_1 \rightarrow X/H_2$ is an isomorphism of $X$-covering spaces, I can prove that for any $x \in X$, there exists $g_x \in G$, such that $\phi(x) = g_x(x)$. But can I find a universal $g$ such that $\phi(x)=g(x)$ for all $x \in X$? Similar problem occurs in the proof of (3). For the isomorphism part of (3), I can construct a map $\psi: G \rightarrow \text{Aut}(X/H \rightarrow X/G)$, which sends $g \in G$ to an automorphism of the covering space $X/H$ over $X/G$ sending $x \mapsto g(x)$. The kernel of this map is $H$, but I don't know how to show it is surjective.

Thank you very much.

  • $\begingroup$ If you are still interested in this, you can indeed show that $\phi$ is given by multiplication by a universal $g$. To see this, show that you can lift $\phi$ to a deck transformation. $\endgroup$ Commented Jan 6, 2016 at 4:25
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    $\begingroup$ @ShinyaSakai Why is the kernel of $\psi$ $H$? $\endgroup$
    – user319128
    Commented Aug 25, 2016 at 2:51

2 Answers 2


This is really long and I apologize (not to mention years after the fact), but I think it takes care of all of the questions you've posed.

(1) Consider the composition of covering spaces $X \xrightarrow{p_1}Y \xrightarrow{q_1} X/G$, and suppose $p = q_1p_1: X \to X/G$. The deck transformations of $p : X \to X/G$ are precisely the elements of $G$, given by $x \mapsto gx$. Let $H$ be the group of deck transformations of the covering space $p_1 : X \to Y$, and suppose $h \in H$. Then $x \mapsto hx$ is a homeomorphism such that $p_1 h = p_1$. But then $q_1 p_1 h = q_1 p_1$, or $ph = p$, so $h \in G$. Therefore $H \subset G$. Let $p_2, q_2$ be the covering maps in the composition of covering spaces $X \xrightarrow{p_2} X/H \xrightarrow{q_2} X/G$, so $p = q_2 p_2$. So we have the following commutative diagram: $\require{AMScd}$ \begin{CD} X @>p_1>> Y\\ @V p_2 V V @VV q_1 V\\ X/H @>>q_2> X/G \end{CD} Define the map $\phi: X/H \to Y$ by $\phi(Hx) = p_1(x)$. To see that $\phi$ is well-defined, if $Hx_1 = Hx_2$, then $x_1 = hx_2$ for some $h \in H$. So $p_1(x_1) = p_1(hx_2) = p_1(x_2)$ since $p_1 h = p_1$ for every $h \in H$.

We claim that $\phi: X/H \to Y$ is a covering space isomorphism. Note for $Hx \in X/H$, we have $q_2(Hx) = Gx$, and $q_1 \phi(Hx) = q_1 p_1(x) = p(x) = Gx$, so $q_2 = q_1 \phi$. So we need to show $\phi : X/H \to Y$ is a homeomorphism. If $\phi(Hx_1) = \phi(Hx_2)$, then let $y = p_1(x_1) = p_1(x_2)$. Since $X \to Y$ is normal (easy to check since $X \to X/G$ is normal), there is a deck transformation $h \in H$ taking $hx_2 = x_1$, so $Hx_1 = Hx_2$. Obviously $\phi$ is surjective, so $\phi$ is a bijection.

To prove $\phi$ is continuous, suppose $Hx \in X/H$ for some $x \in X$, and let $U \subset Y$ be a neighborhood of $p_1(x) = \phi(Hx)$. Assume $U$ is evenly covered. Let $V \subset p_1^{-1}(U)$ be a neighborhood of $x$ that maps homeomorphically onto both $p_1(V) \subset Y$ and $p_2(V) \subset X/H$. Then $p_2(V)$ is open. If $y \in \phi (p_2(V)),$ then by injectivity, $\phi^{-1}(y) \in p_2(V)$, which lifts to a point in $V \subset p_1^{-1}(U)$. Therefore we have a sequence of bijections $\phi|_{p_2(V)}$, $p_2|_V$, and $p_1|_V$, and the composition $p_1|_V \circ p_2|_V^{-1} \circ \phi|_{p_2(V)}^{-1}$ is a bijection $p_1(V) \to p_1(V)$. Since $p_1 p_2|_V^{-1} \phi^{-1}(y) \in p_1(V)$, we have $y \in p_1(V)$, so $y \in U$. so $\phi(p_2(V)) \subset U$, so $\phi$ is continuous. A similar argument shows $\phi^{-1}$ is continuous. QED.

For the "only if" direction of (2) and the isomorphism part of (3), we need the following

Lemma: If we have a composition of covering spaces $X \xrightarrow{p_1} Y_1 \xrightarrow{q_1} Z$ and $X \xrightarrow{p_2} Y_2 \xrightarrow{q_2} Z$, and a covering space isomorphism $\phi : Y_1 \to Y_2$, then the map $\phi p_1 : X \to Y_2$ lifts to a deck transformation of $X \to Z$.

Proof: First of all, $(\phi p_1)_* (\pi_1(X)) = \phi_* p_{1*} (\pi_1(X)) \subset p_{2*} \pi_1(X)$, so by Prop. 1.33 in Hatcher's Algebraic Topology, there is a lift $\widetilde{\phi p_1} : X \to X$ of $\phi p_1$. Thus $p_2 \widetilde{\phi p_1} = \phi p_1$. For any such lift, $p\widetilde{\phi p_1} = q_2 p_2 \widetilde{\phi p_1} = q_2 \phi p_1 = q_1 p_1 = p$. So we need to show that some lift $\widetilde{\phi p_1}$ is a homeomorphism $X \to X$. Fix $x_0 \in X$, and let $y_0 = p_1(x_0) \in Y_1$ and $y'_0 = \phi(y_0) \in Y_2$, and let $x_0' \in X$ be a lift of $y'_0$. Then $y'_0 = \phi p_1(x_0)$, so suppose $\widetilde{\phi p_1} : X \to X$ is the unique lift of $\phi p_1$ taking $x_0 \mapsto x_0'$. By similar arguments, there is also a unique lift of the map $\phi^{-1} p_2 : X \to Y_1$ -- denote it $\widetilde{\phi^{-1} p_2}$ -- sending $x_0' \mapsto x_0$. When you compose these lifts, we have: \begin{equation} p_2\left(\widetilde{\phi p_1}\right) \left(\widetilde{\phi^{-1} p_2}\right) = \phi p_1 \left(\widetilde{\phi^{-1} p_2}\right) = \phi \phi^{-1} p_2 = p_2 \end{equation} Thus the composition $\left(\widetilde{\phi p_1}\right) \left(\widetilde{\phi^{-1} p_2}\right)$ is a lift of $p_2$ sending $x_0' \mapsto x_0'$. By uniqueness of lifts, $\left(\widetilde{\phi p_1}\right) \left(\widetilde{\phi^{-1} p_2}\right) = \mathbb{1}_X$, the identity map on $X$. Similarly, $\left(\widetilde{\phi^{-1} p_2}\right)\left(\widetilde{\phi p_1}\right) = \mathbb{1}_X$. Therefore $\left(\widetilde{\phi p_1}\right)$ is a homeomorphism, and thus a deck transformation. QED.

Corollary: Every lift of a covering map $p : \tilde X \to X$ is a deck transformation.

(2) You have the "if" direction, it looks like (consider the map $H_1 x \mapsto H_2 x$ from $X/H_1 \to X/H_2$). For "only if", let $\phi : X/H_1 \to X/H_2$ be the covering space isomorphism, and let $p_1, p_2, q_1, q_2$ satisfy the commutative diagram $\require{AMScd}$ \begin{CD} X @>p_1>> X/H_1\\ @V p_2 V V @VV q_1 V\\ X/H_2 @>>q_2> X/G \end{CD} so that $p = q_i p_i, i = 1,2$ is the covering map $X \to X/G$. Consider $\phi p_1 : X \to X/H_2$. Then by the lemma, $\phi p_1$ lifts to a deck transformation $g \in G$ of $X \to X/G$. So $p_2 g = \phi p_1$, and in particular, $p_2 = \phi p_1 g^{-1}$.

We claim that for $h_1 \in H_1$, the deck transformation $gh_1 g^{-1} : X \to X$ is a lift of $\phi p_1 g^{-1} : X \to X/H_2$. Indeed, since $p_1 h_1 = p_1$, we have $p_2 gh_1 g^{-1} = \phi p_1 h_1 g^{-1} = \phi p_1 g^{-1}$. But by the corollary, since $gh_1g^{-1}$ is then a lift of $p_2 : X \to X/H_2$, $gh_1 g^{-1}$ is a deck transformation of $X \to X/H_2$. But these are precisely the elements of $H_2$, so $gh_1 g^{-1} = h_2$ for some $h_2 \in H_2$. QED. $ \newcommand{\Aut}{\mathrm{Aut}} $

(3) Suppose we have the composition of covering spaces $X \xrightarrow{q} X/H \xrightarrow{r} X/G$, and $p = rq : X \to X/G$ is the covering map. To prove $\Aut(X/H \to X/G) \cong G/H$, let's consider your homomorphism $\psi : G \to \Aut(X/H \to X/G)$ given by $\psi(g) = (Hx \mapsto H(gx))$. As you said, its kernel is $H$ (make sure to also prove the map is well-defined). To prove $\psi$ is surjective, suppose $f : X/H \to X/H$ is a deck transformation of $X/H \to X/G$. By the lemma (using $Y_1 = Y_2 = X/H$), the map $fq$ lifts to a deck transformation $g \in G$ of $X \to X/G$, so $qg = fq$. Then, $f(Hx) = fq(x) = qg(x) = H(gx)$ for every $Hx \in X/H$, so $f = (Hx \mapsto H(gx)) = \psi(g)$. So $\psi$ is surjective, and by the first isomorphism theorem, $G/H \cong \Aut(X/H \to X/G)$.

Now, you again seem to have the "if" direction worked out for this one (use the established isomorphism with the definition of a normal covering space). For "only if", the argument is almost entirely algebraic. You can use the fact that if $\phi : G_1 \to G_2$ is a group homomorphism, $H \lhd G$, and $\phi(G_1) \lhd G_2$, and if $\phi' : G_1 / H \to G_2 / \phi(H)$ is the induced map on quotient groups, then $\phi'(G_1/H) \lhd G_2/\phi(H)$. (This is an easy computation.) By Prop. 1.40 in Hatcher, we have $G \cong \pi_1(X/G) / p_*(\pi_1(X))$ and $H \cong \pi_1(X/H) / q_*(\pi_1(X))$,and by Prop. 1.39, $r_*(\pi_1(X/H)) \lhd \pi_1(X/G)$. By our fact from group theory about normal subgroups of quotient groups, we have $r_*'(\pi_1(X/H) / q_*(\pi_1(X))) \lhd \pi_1(X/G) / p_*(\pi_1(X))$ (note $r_*'$ is injective).

Let $\phi : \pi_1(X/G) \to G$ be the homomorphism sending a homotopy class $[\gamma] \in \pi_1(X/G)$ to the deck transformation $g \in G$ sending the basepoint $x_0 \in X$ to $gx_0$ along the lifted path $\tilde \gamma$ in $X$ of $\gamma$, ie. $\tilde \gamma(0) = x_0$ and $\tilde \gamma(1) = gx_0$. Define $\psi : \pi_1(X/H) \to G$ similarly. These are the homomorphisms used in Hatcher's proof of Prop. 1.39 to show that $G \cong \pi_1(X/G) / p_*(\pi_1(X))$ and $H \cong \pi_1(X/H) / q_*(\pi_1(X))$ (in particular, $\psi(\pi_1(X/H)) = H$).

We claim that $\psi = \phi r_*$. For a loop $\gamma$ in $X/H$, if $\psi[\gamma] = h$, then there is a path $\tilde \gamma$ in $X$ with $\tilde \gamma(0) = x_0$ and $\tilde \gamma(1) = hx_0$. Consider the loop $r\gamma$ in $X/G$. The path $\tilde \gamma$ is also the unique lift of $r\gamma$ with $\tilde \gamma(0) = x_0$, and so $\phi[r \gamma] = \phi r_*[\gamma] = h$. Hence $\phi r_* = \psi$, as desired.

Letting $\phi' : \pi_1(X/G) / p_*(\pi_1(X)) \to G$ and $\psi' : \pi_1(X/H) / q_*(\pi_1(X)) \to H$ be the corresponding isomorphisms, this implies that $\phi' r_*' = \psi'$. So, for $h \in H$ and $g \in G$, if we let $[\beta'] = r_*' \psi'^{-1}(h)$ and $[\gamma'] = \phi'^{-1}(g)$, because $r_*'(\pi_1(X/H) / q_*(\pi_1(X))) \lhd \pi_1(X/G) / p_*(\pi_1(X))$, we have $[\gamma'][\beta'][\gamma']^{-1} \in r_*'(\pi_1(X/H) / q_*(\pi_1(X)))$, and so \begin{align} ghg^{-1} &= \left(\phi'[\gamma']\right) \left(\psi' r_*'^{-1} [\beta']\right) \left(\phi'[\gamma']\right)^{-1} = \left(\phi'[\gamma']\right)\left(\phi'[\beta']\right)\left(\phi'\left([\gamma']^{-1}\right)\right) \\ &= \phi'\left([\gamma'][\beta'][\gamma']^{-1}\right) \end{align} Because $[\gamma'][\beta'][\gamma']^{-1} \in r_*'(\pi_1(X/H) / q_*(\pi_1(X)))$, and $\phi' r_*' = \psi'$, this implies \begin{equation} ghg^{-1} = \psi' r_*'^{-1}\left([\gamma'][\beta'][\gamma']^{-1}\right) \in H. \end{equation} Hence $H \lhd G$. QED.

  • $\begingroup$ Are the diagrams showing up for you? They aren't for me, I see "math processing error". It turns out, on my end, that the @> label >> constructions were causing the issue; putting the label in braces, as in @>{p_1}>> caused it to render. But if it's not an issue for anyone else, I won't bother to edit. $\endgroup$
    – pjs36
    Commented Jan 3, 2017 at 19:39
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    $\begingroup$ Really? That's odd. Yeah I see the diagrams. $\endgroup$
    – D Ford
    Commented Jan 3, 2017 at 19:47
  • $\begingroup$ Odd indeed! Oh well, I've been seeing funny things with Mathjax lately that I suspect are local to me. At any rate, this looks like a nice answer that I'd love to upvote, but unfortunately, it's way beyond me! :) $\endgroup$
    – pjs36
    Commented Jan 3, 2017 at 19:51
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    $\begingroup$ If possible, could you please explain that why we have $ϕ_∗p_{1∗}(π_1(X))⊂p_{2∗}π_1(X)$ here? Also may I please ask about the "if" direction of part (b)? $\endgroup$
    – Y.X.
    Commented Aug 23, 2017 at 10:57
  • $\begingroup$ For your first question, since we assume $p = q_1 p_1 = q_2 p_q : X \to Z$ is itself a covering (as it is in parts (b) and (c) applying this lemma, though perhaps I should have made that more explicit), we have $q_2 p_2 = q_2 \phi p_1$. In particular, $q_{2*} p_{2*}(\pi_1(X)) = q_{2*} \phi_* p_{1*}(\pi_1(X))$. Since these are injective homomorphisms, we can actually conclude $p_{2*}(\pi_1(X)) = \phi_* p_{1*}(\pi_1(X))$. $\endgroup$
    – D Ford
    Commented Aug 29, 2017 at 22:54

I have posted the answer for (a) months ago, but I think it was completely wrong, so let me try to rewrite it (but please scold me brutally if the following does not make sense).

We consider $X \overset{p}\longrightarrow C \overset{q}\longrightarrow X/G$. We may assume that $p$ and $q$ are surjective. Consider $\text{Aut}(p) \leqslant \text{Aut}(qp)\simeq G$ (the canonical isomorphism given by the hypothesis on $X$). Then we have $X \twoheadrightarrow X/\text{Aut}(p)$ respects the relation of $X \twoheadrightarrow C$, so as quotients, they are isomorphic, which gives a desired isomorphism with $H = \text{Aut}(p)$.

Comment. I am not so sure if this is enough since I have not found $H$ as a subgroup of $G$ but an isomorphic copy of a subgroup.


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