# Streamlining of Hatcher's Algebraic Topology Proposition 1.40 sufficient conditions

In Hatcher's Algebraic Topology Proposition 1.40, it is said that, for a covering action of a group $$G$$ on a space $$Y$$,

(b) $$G$$ is the group of deck transformations of the covering map $$Y\to Y/G$$ if $$Y$$ is path-connected.

It seems to me that $$Y$$ connected is sufficient, since the unique lifting property only required connectedness. Am I right?

(c) $$G$$ is isomorphic to $$\pi_1(Y/G)/p_*(\pi_1(Y))$$ if $$Y$$ is path-connected and locally path-connected, but it seems to me by looking closely at the proof that $$Y$$ path-connected is sufficient.

Indeed the full power of the lifting criteria does not seem needed to build the deck transformation from the action of $$g\in G$$, since only loops are concerned. Am I right?

Edit Following the comments of @FShrike, here is my approach when I only suppose $$Y$$ path-connected.

(i) Let the covering map be $$p: Y\to Y/G$$. Choose $$y_0\in Y$$ and $$b_0=p(y_0)\in Y/G$$. Let $$\gamma:I\to X/G$$ be a loop with $$\gamma(0)=\gamma(1)=b_0$$

(ii) By the path-lifting theorem there exists a unique path $$\widetilde{\gamma}: I\to X$$ with $$\widetilde{\gamma}(0)=y_0$$ and $$p\circ\widetilde{\gamma}=\gamma$$. Since $$p(\widetilde{\gamma}(0))=p(\widetilde{\gamma}(1))$$, then there is and only one because of unique lifting property $$g\in G$$ with $$\widetilde{\gamma}(1)=g.y_0$$ It is easy to check that such $$g$$ does not depend of the choice of $$\gamma$$ inside the homotopy class of $$\gamma$$ in $$\pi_1(Y/G, b_0)$$. Therefore we get a well defined map $$\pi: \pi_1(Y/G, b_0)\to G$$

(iii) It is easy to check that this map is a morphism of groups and that its kernel is the isotropy group of the monodromy action $$p_*(\pi_1(Y, y_0))$$.

(iv) Now to show that this map is surjective is where the path-connected hypothesis on $$Y$$ is used: let $$g\in G$$. There exists a path $$\sigma: I\to Y$$ with $$\sigma(0)=y_0$$ and $$\sigma(1)=g.y_0$$, hence $$g=\pi(p\circ\sigma)$$ Quite straightforward proof if I am not wrong.

Edit 2 In his proof of (c) of Proposition 1.40, Prof. Hatcher uses the proof of (b) of proposition 1.39, which uses the lifting criterion 1.33 whose proof indeed requires local path-connectedness hypothesis.

This is a bit of an overkill in my opinion, since proposition 1.39 is about proving the same thing as proposition 1.40 but in the case where $$G$$ is supposed to be the group of deck transformations $$G(Y)$$.

To show that $$G(Y)$$ is actually a covering action, we really need the hypothesis of local path-connectedness to be able to use the lifting criterion.

But in proposal 1.40 we already begin with a group $$G$$ having a covering action on $$Y$$!, and so its results just come from algebraic view, no need for local path-connectedness.

• Well, one silly point is that if a space is not path-connected, then it's hard to talk about path lifting and/or homotopy... Jul 5, 2023 at 23:33
• Sorry, you're right. Looking at my photos of my notes I can even see where I've crossed out the "path" in "path connected". (b) is correct for just $Y$ connected. However $(c)$ requires the local path connectivity assumption since it relies on propositions $1.33,1.37$ Jul 5, 2023 at 23:42
• @FShrike thank you very much for your effort +1! For (c) I have carefully constructed a proof not using the full power of the propositions you mentioned. It is too late for me to write it now. Will do it tomorrow. Jul 5, 2023 at 23:47
• @brunoh I'd be interested in a proof that doesn't rely on local connectivity. Be warned, there are examples of path connected but not locally path connected spaces where the lifting of proposition $1.33$ fails even if the criterion is met... see exercise $1.3.7$ Jul 5, 2023 at 23:58
• @FShrike yep I am aware that is why I asked the question. Will see if my proof (easy one) works. Thank you again. Jul 6, 2023 at 0:10

$$\newcommand{\N}{\mathcal{N}}\newcommand{\G}{\mathcal{G}}\newcommand{\deck}{\operatorname{Deck}}\newcommand{\X}{\widetilde{X}}\newcommand{\x}{\widetilde{x}}\newcommand{\op}{{^\mathsf{op}}}$$Let's assemble the necessary ingredients. What I will do is go through my notes and point out exactly where each hypothesis is used for each "ingredient". I'd like to note that I write the group action differently to Hatcher; for me, if $$f,g$$ are homotopy path classes, $$fg$$ denotes the homotopy path class of the composite path $$a\overset{g}{\to}b\overset{f}{\to}c$$.

A lot of this comes down to how one defines "normal". I've written this partially in response to a request so that this question gets properly answered and I think it's good to have the hypotheses clearly pointed out.

By proposition $$1.34$$, for a connected covering $$p:\X\to X$$ the elements of $$\deck(\X;X)$$ are completely determined by their values at a single point.

Hatcher defines the covering to be a normal covering if for all $$x\in X$$, $$\x_1,\x_2\in p^{-1}(x)$$ there exists $$\tau\in\deck(\X;X)$$ with $$\tau(\x_1)=\x_2$$. If the covering is normal and connected, there is a unique such $$\tau$$.

If the base space $$X$$ is globally and locally path connected, connected coverings $$p:\X\to X$$ are normal if and only if $$H:=p_\ast\pi_1(\X;\x_0)$$ is a normal subgroup of $$G:=\pi_1(X;p(\x_0))$$ for some $$\x_0\in\X$$ if and only if this is true for all $$\x_0\in\X$$. Write $$x_0:=p(\x_0)$$.

• Using global path connectivity of $$X$$, it's not too hard to argue that the cover is normal if and only if it is "normal at $$\x_0$$"; i.e. for all $$\x_1\in p^{-1}(p(\x_0))$$ there is a deck transformation $$\x_0\mapsto\x_1$$. This can be used to get the "iff. [...] is true for some $$\x_0$$" into a statement about all $$\x_0\in\X$$
• We establish that for $$\x_1,\x_2$$ in the same $$p$$-fibre, $$p_\ast(\pi_1(\X;\x_1))=p_\ast(\pi_1(\X;\x_2))$$ if and only if there exists a deck transformation taking $$\x_1\to\x_2$$. This step requires, in the direction (normal subgroup $$\to$$ normal cover)): (a) local path connectivity of $$X$$ to obtain local path connectivity of $$\X$$ and thus to deduce certain lifts exist by proposition $$1.33$$ (b) connectivity of $$\X$$ to use uniqueness of lifts to deduce the lifts are in fact deck transformations. In the converse, normal cover implies normal subgroup, it's a completely trivial statement since $$\x_1,\x_2$$ are in the same $$p$$-fibre iff. there is a deck transformation taking one to the other, in which case the equality of fundamental groups under $$p$$ is immediate
• For $$g\in G$$, there is a unique lift of $$g$$ (no hypotheses needed, see the proof of $$1.7$$) to a homotopy path class $$[\gamma]:\x_0\to\x_1$$ for some $$\x_1\in p^{-1}(x_0)$$. The conjugate $$gHg^{-1}\le G$$ is precisely the image $$p_\ast\{[\gamma\cdot f\cdot\gamma^{-1}]:f\in\pi_1(\X;\x_0)\}=p_\ast\pi_1(\X;\x_1)$$. We get that $$gHg^{-1}=H$$ if and only if $$p_\ast\pi_1(\X;\x_1)=p_\ast(\pi_1(\X;\x_0))$$; we find that $$H$$ is normal in $$G$$ if $$p_\ast(\pi_1(\X;\x_0))=p_\ast(\pi_1(\X;\x_1))$$ for all $$\x_1\in p^{-1}(x_0)$$. The "if" turns into an "if and only if" if $$\X$$ is assumed to be path connected. If the cover is normal, then this equation easily holds - see the first bullet point. In the other direction, we need the local path connectivity assumption on $$X$$ as well as the (path) connectivity of $$\X$$ to find $$H$$ is normal in $$G$$ if and only if the cover is "normal at $$\x_0$$" by the first bullet point, hence normal.

In conclusion, under no other hypotheses at all, a normal cover $$p:\X\to X$$ implies that the subgroups $$H$$ are always normal in $$G$$, but the local path connectivity of $$X$$ and global path connectivity of $$\X$$ are crucial in obtaining the converse implication.

We also would like to prove $$\N(H)/H\cong\deck(\X;X)\op$$ holds - but when? We may drop the contravariance in "$$\mathsf{op}$$" because every group is isomorphic to its dual, but my preferred conventions make the contravariance the natural choice.

We put $$\Phi:\N(H)\to\deck(\X;X)\op$$ by assigning to every $$g$$ its unique lifted homotopy path class $$[\gamma]:\x_0\to\x_1$$, and $$g\in\N(H)$$ means $$\x_1$$ must be such that $$p_\ast(\pi_1(\X;\x_0))=p_\ast(\pi_1(\X;\x_1))$$, so under the full hypotheses of local path connectivity of $$X$$ and connectivity of $$\X$$ there is a unique deck transformation taking $$\x_0$$ to $$\x_1$$; we define $$\Phi(g)$$ to be this unique deck transformation. It's then not too hard, assuming path connectivity of $$\X$$, to check $$\Phi$$ is a surjective homomorphism. The kernel of $$\Phi$$ is always $$H$$ (if $$\Phi$$ is well-defined, that is) so the first isomorphism theorem kicks in.

If $$\N(H)=G$$ i.e. $$H$$ is normal in $$G$$ but we don't already know the cover is normal, we again need the local path connectivity hypothesis to define $$\Phi$$. But if the cover is given to be normal, then $$\N(H)=G$$ is true without any other hypothesis and $$\Phi$$ is well-defined without any need for local path connectivity of $$X$$, so $$G/H$$ embeds as a subgroup of $$\deck(\X;X)\op$$; if $$\X$$ is also assumed to be path connected, we get the full isomorphism $$G/H\cong\deck(\X;X)\op$$.

In conclusion, $$G/H\cong\deck(\X;X)\op$$ is always true if the cover is path connected and normal in the sense of Hatcher. If the cover is only normal in the sense that $$H$$ is normal in $$G$$, then we need $$X$$ to be locally path connected to make this work.

In the case of (left) group covering actions of $$\G$$ on a space $$Y$$, it's always true that $$q:Y\twoheadrightarrow Y/\G$$ is a normal covering. If $$Y$$ is connected, you were quite right in saying that $$\G\cong\deck(Y;Y/\G)$$ follows (if $$Y$$ is not connected we can at least say $$\G$$ embeds as a subgroup of $$\deck(Y;Y/\G)$$).

Now for the main question - when can we say $$\G\op\cong\pi_1(Y/\G;q(y_0))/q_\ast(\pi_1(Y;y_0))$$? ($$=G/H$$) We know the cover is normal. Therefore, we need only have $$Y$$ path connected to make everything work. If we don't even assume $$Y$$ is path connected, we can just conclude that $$\pi_1(Y/\G;q(y_0))/q_\ast(\pi_1(Y;y_0))$$ embeds as a subgroup of $$\G\op$$.

The local path connectivity of $$X$$ is only needed in the instance that $$\N(H)\neq G$$ or that $$\N(H)=G$$ but we haven't, a priori, been told that the cover is a normal cover.