# On criterion of normal covering maps

I tried to find the criterion for a covering map to be normal. However, the criterion I came up with and proved seems to be a little bit weaker than what is stated in other textbooks. I don't know where I got wrong. Here is the statement:
$$\newcommand{\cover}{\widetilde} \newcommand{\lift}{\widetilde}$$ Let $$p:\cover{X}\rightarrow X$$ be a covering map, $$X$$ be path-connected, and $$x_0$$ be the base-point of $$X$$.

1. $$\pi_1(X,x_0)$$ acts transitively on $$p^{-1}(x_0)$$ if and only if $$\cover{X}$$ is also path-connected.
2. The stabiliser of $$\lift{x_0}\in p^{-1}(x_0)$$ is the image of $$p_*:\pi_1(\cover{X}, \lift{x_0})\rightarrow \pi_1(X, x_0)$$ which must be a subgroup of $$\pi_1(X,x_0)$$.
3. there is a bijection (in terms of sets): $$\frac{\pi_1(X,x_0)}{\pi_1(\cover{X},\lift{x_0})}\longleftrightarrow p^{-1}(x_0)$$
4. $$\frac{\pi_1(X,x_0)}{\pi_1(\cover{X},\lift{x_0})}$$ is a group, i.e. $$p_*(\pi_1(\cover{X},\lift{x_0}))$$ is a normal subgroup of $$\pi_1(X,x_0)$$, if $$\cover{X}$$ is path-connected.\

The proof is as follows:
Proof: We firstly deal with 1, then 2 and 3 are only simple corollaries of 1. If $$\cover{X}$$ is path-connected, then for any $$\lift{x_0}, \lift{x_0}'\in p^{-1}(x_0)$$, there must a path $$\lift{\gamma}:\lift{x_0}\mapsto \lift{x_0}'$$, and $$p_*(\lift{\gamma})=\gamma$$ is a loop based at $$x_0$$, belonging to $$\pi_1(X, x_0)$$, therefore, $$\lift{x_0}\cdot \gamma=\lift{x_0}'$$. Conversely, if $$\pi_1(X,x_0)$$ acts transitively on $$p^{-1}(x_0)$$, and suppose the contrary that $$\cover{X}$$ is not path-connected and let $$y,z$$ in two different path components. Then, since $$X$$ is path-connected, there is a path between $$p(y)$$ and $$p(z)$$, denoted as $$\gamma$$, then the lift of $$\gamma$$ that starts at $$y$$ ends at $$z':=\lift{\gamma}(1)$$. $$p(z')=(p\circ \lift{\gamma})(1)=\gamma(1)=p(z)$$, so $$z,z'\in p^{-1}(p(z))$$ while they are in different path components. By a change of base point, we have $$\pi_1(X,p(z))$$ acts transitively on $$p^{-1}(p(z))$$, which results a path between $$z$$ and $$z'$$, contradicting with them being in different path components.
For 4, if $$\cover{X}$$ is path-connected, then $$\pi_1(X,x_0)$$ acts transitively on $$p^{-1}(x_0)$$. Therefore, for any $$\beta\in \pi_1(X,x_0)$$, it can be lifted up to a path $$\lift{\beta}:\lift{x}\mapsto \lift{x_0}$$ where $$\lift{x}$$ is in the same fibre with $$\lift{x_0}$$, then for any path $$\gamma\in \pi_1(\cover{X}, \lift{x_0})$$, $$\lift{\beta}\cdot \gamma\cdot\lift{\beta}^{-1}$$ is a loop based at $$\lift{x}$$. Hence $$\beta\cdot p_*(\pi_1(\cover{X}, \lift{x_0}))\cdot\beta^{-1}=p_*\pi_1(\cover{X}, \lift{x})$$. But by the criterion of existence of covering transformations, there exists a unique covering transformation $$F:\cover{X}\rightarrow \cover{X}$$ such that $$F(\lift{x_0})=F(\lift{x})$$. Then, since $$p\circ F=p$$, $$p_*\pi_1(\cover{X}, \lift{x_0})=p_*\circ F_*(\pi_1(\cover{X}, \lift{x_0}))=p_*\pi_1(\cover{X}, \lift{x})=\beta\cdot p_*\pi_1(\cover{X}, \lift{x_0})\cdot\beta^{-1}$$, so $$p_*\pi_1(\cover{X}, \lift{x_0})$$ is normal.
The criterion of the existence of covering transformation is: let $$x\in X$$ and $$\lift{x_1}\in p_1^{-1}(x), \lift{x_2}\in p_2^{-1}(x)$$ where $$\cover{X_1}$$ is additionally path-connected and locally path-connected, then there exists a covering transformation $$F:\cover{X_1}\rightarrow \cover{X_2}$$ such that $$F(\lift{x_1})=\lift{x_2}$$ if and only if $$(p_1)_*(\pi_1(\cover{X_1}, \lift{x_1}))\subset (p_2)_*(\pi_1(\cover{X_2}, \lift{x_2}))$$
The proof of this criterion is as follows:
If there exists a covering transformation $$F:\cover{X_1}\rightarrow \cover{X_2}$$ such that $$F(\lift{x_1})=\lift{x_2}$$, then, for any $$w\in\pi_1(\cover{X_1}, \lift{x_1})$$, $$(p_1)_*(w)=(p_2\circ F)_*(w)=(p_2)_*(F\circ w)$$, but $$F\circ w$$ is a loop based at $$\lift{x_2}$$, so $$(p_1)_*(w)\in (p_2)_*(\pi_1(\cover{X_2}, \lift{x_2}))$$.
Conversely, assume that $$(p_1)_*(\pi_1(\cover{X_1}, \lift{x_1}))\subset (p_2)_*(\pi_1(\cover{X_2}, \lift{x_2}))$$, and we want to construct a covering transformation $$F:\cover{X_1}\rightarrow \cover{X_2}$$ such that $$F(\lift{x_1})=\lift{x_2}$$. Firstly, we define $$F(\lift{x_1})=\lift{x_2}$$, and then try to extend $$F$$ from this domain of single point. That requires us to assign a value $$\lift{y_2}$$ for any $$\lift{y_1}\in \cover{X_1}$$. Naturally, let $$\lift{\gamma}:\lift{x_1}\mapsto \lift{y_1}$$, cast it down to $$\gamma:=(p_1)_*(\lift{\gamma})$$, lift it up to $$\lift{\gamma'}$$ started at $$\lift{x_2}$$, and then we can define $$F(\lift{y_1}):=\lift{\gamma'}(1)\in\cover{X_2}$$. Then we need to check that: 1. $$F$$ is well-defined. 2. $$F$$ is continuous. 3. $$F$$ is a covering transformation.

Actually, that $$p_2\circ F=p_1$$ is already obvious from the definition, because for any $$\lift{y_1}\in \cover{X_1}$$, $$p_2\circ F(\lift{y_1})=p_2\circ \lift{\gamma'}(1)=\gamma(1)=(p_1\circ \lift{\gamma})(1)=p_1(\lift{y_1})$$.
Then, to check that $$F$$ is well-defined, let $$\lift{\gamma_1}$$ and $$\lift{\gamma_2}$$ be two paths from $$\lift{x_1}$$ to $$\lift{y_1}$$, notice that $$\lift{\gamma_1}\cdot\lift{\gamma_2^{-1}}\in \pi_1(\cover{X_1}, \lift{x_1})$$, so $$(p_1)_*(\lift{\gamma_1}\cdot\lift{\gamma_2^{-1}})=\gamma_1\cdot\gamma_2^{-1}\in (p_2)_*(\pi_1(\cover{X_2}, \lift{x_2}))$$. Therefore, we can lift $$\gamma:=\gamma_1\cdot\gamma_2^{-1}$$ to $$\lift{\gamma'}\in p_*(\pi_1(\cover{X_2}, \lift{x_2}))$$. Notice that $$\lift{\gamma'}|_{[0,1/2]}$$ is a lift of $$\gamma_1=p_*(\lift{\gamma_1})$$, so by Uniqueness of lifts lemma, $$\lift{\gamma'}|_{[0,1/2]}$$ is just $$\lift{\gamma_1'}$$, and similarly $$\lift{\gamma'}_{[1/2,1]}$$ is simply $$\lift{\gamma_2'}^{-1}$$, which implies that $$\lift{\gamma_1'}$$ and $$\lift{\gamma_2'}$$ end at the same point. So $$F$$ is well-defined.
In addition, we check that $$F$$ is continuous. Define a basis of $$\cover{X_2}$$ compromising all elementary sheets over all elementary neighbourhoods in $$X$$. Then, it suffices to show that for any elementary sheet $$U\subset \lift{X_2}$$, we have $$F^{-1}(U)\subset \lift{X_1}$$ is also open. Equivalently, it is to show that for any $$v\in F^{-1}(U)$$ there exists an open neighbourhood $$v\in V\subset F^{-1}(U)$$ containing $$v$$. Let $$w:=p_1(v)\in X$$, by the definition of covering spaces, there exists an open neighbourhood $$w\in W\subset X$$ such that $$(p_2)^{-1}(W)=\amalg_{k\in I}{C_k}$$ where $$C_k$$ are open and $$p_2|_{C_k}$$ is a homeomorphism for any $$k$$. Apparently, $$U$$ is one of the $$C_k$$, say $$C_t$$. Since $$W$$ is open, $$(p_1)^{-1}(W)$$ must be open. $$\cover{X_1}$$ is locally path-connected, so we can find a path-connected open neighbourhood of $$v$$ such that $$v\in V\subset (p_1)^{-1}(W)$$. We claim that $$F(V)\subset U$$. For any point $$v'\in V$$, take a path $$\gamma: v\mapsto v'$$ in $$V$$, then follow the definition of $$F$$, we have $$F(v')=\lift{(p_1)_*\gamma}(1)$$. Notice that $$\gamma$$ lies entirely in $$V\subset (p_1)^{-1}(W)$$, so $$(p_1)_*\gamma$$ lies entirely in $$W$$ which is an elementary neighbourhood, and thus $$\lift{(p_1)_*\gamma}$$ lies entirely in $$\amalg_{k\in I}{C_k}$$. However, that means $$\lift{(p_1)_*\gamma}$$ must be in only one of the $$C_k$$, which must be $$U$$, because $$\gamma(0)=v\in V\subset F^{-1}(U)$$. Therefore, $$F(v')\subset U$$, and this holds for any $$v'\in V$$, so $$F(V)\subset U$$.
Furthermore, if $$\cover{X}:=\cover{X_1}=\cover{X_2}$$, we can use the criterion of the existence of covering transformations twice, to obtain two covering transformation $$F_1:\cover{X_1}\rightarrow\cover{X_2}$$ and $$F_2:\cover{X_2}\rightarrow \cover{X_1}$$, which are essentially the same. Thus, for any $$\lift{y_1}\in \cover{X_1}$$, we can obtain $$F_2\circ F_1(\lift{y_1})$$ in this manner: take a path $$\gamma: \lift{x_1}\mapsto \lift{y_1}$$, cast it down to $$X$$ and lift up again to $$\cover{X_2}$$ such that the lifting $$\lift{(p_1)_*\gamma}$$ begins at $$\lift{x_2}$$, then $$F_1(\lift{y_1})=\lift{(p_1)_*\gamma}(1)$$, but obviously when we apply the same to $$\lift{(p_1)_*\gamma}$$, we would get $$\gamma$$, so $$F_2\circ F_1(\lift{y_1})=\gamma(1)=\lift{y_1}$$. Similarly, we have $$F_1\circ F_2(\lift{y_2})=\lift{y_2}$$. Hence we know that if $$\cover{X_1}=\cover{X_2}$$, the covering transformation we obtained is a deck transformation.

Therefore, since the criterion of the existence of covering transformation demands the local path-connectivity of $$\cover{X}$$, we must add one more condition. However, isn't it a bit stronger than what is stated in http://www.homepages.ucl.ac.uk/~ucahjde/tg/html/gal-03.html?
By the way, the criterion of the existence of covering transformation stated in http://www.homepages.ucl.ac.uk/~ucahjde/tg/html/gal-02.html ultimately based on the lifting criterion stated here:http://www.homepages.ucl.ac.uk/~ucahjde/tg/html/gal-01.html. But it seems to be that the author misused the lifting criterion because he did not require the covering spaces to be locally path-connected while that is required in the lifting criterion. PS: from the link given below, it seems that we couldn't guarantee that the fundamental group acts transitively on the fiber, but what would be the sufficient condition then? By the way, the proof of 1 is originally given in: https://dec41.user.srcf.net/notes/II_M/algebraic_topology.pdf, Lemma 3.1.15.(i).

• See youtube.com/watch?v=-ZDWxlEpHj8&t=854s for an example de covering that is not normal but connected Commented Apr 6, 2023 at 19:08
• I examined youtube.com/watch?v=-ZDWxlEpHj8&t=854s and the proof in it, but I don't see how the condition of transitive action is used. Commented Apr 7, 2023 at 8:35
• Does the fundamental group of the base space really act on the fiber in this case? Commented Apr 7, 2023 at 16:24
• Yes, because we can lift any homotopy class of loops in the fundamental group such that the lifting starts at some specified point in the fiber, so that we can define the result of the action of the loop on this specified point as the end point of the lifting. Commented Apr 8, 2023 at 14:36

Well done for deriving these results by yourself! That's very impressive. Statements 1, 2 and 3 are true, and you've got the right ideas for the proofs of these.

However, statement 4 is not true as written. Just from reading your post, I can't tell if this is an actual misunderstanding or if this is just because you wrote your post in a hurry. But I'll point it out regardless.

Please take a close look at this passage in your proof of statement 4:

[We have proved that] $$\beta\cdot p_*(\pi_1(\widetilde{X}, \widetilde{x_0}))\cdot\beta^{-1}=p_*(\pi_1(\widetilde{X}, \widetilde{x})),$$ [where $$\beta$$ is an element of $$\pi_1(X, x_0)$$ that lifts to a path from $$\widetilde{x}$$ to $$\widetilde{x_0}$$].

But by the criterion of existence of covering transformations, there exists a unique covering transformation $$F:\widetilde{X}\rightarrow \widetilde{X}$$ such that $$F(\widetilde{x_0})=\widetilde{x}$$.

Then, since $$p\circ F=p$$, $$p_*(\pi_1(\widetilde{X}, \widetilde{x_0}))=p_*(F_*(\pi_1(\widetilde{X}, \widetilde{x_0})))=p_*(\pi_1(\widetilde{X}, \widetilde{x}))=\beta\cdot p_*(\pi_1(\widetilde{X}, \widetilde{x_0}))\cdot\beta^{-1},$$ so $$p_*\pi_1(\widetilde{X}, \widetilde{x_0})$$ is normal.

This is circular logic. A correct version of this would say:

We have proved that $$\beta\cdot p_*(\pi_1(\widetilde{X}, \widetilde{x_0}))\cdot\beta^{-1}=p_*(\pi_1(\widetilde{X}, \widetilde{x})),$$ where $$\beta$$ is an element of $$\pi_1(X, x_0)$$ that lifts to a path from $$\widetilde{x}$$ to $$\widetilde{x_0}$$.

Therefore, the criterion of existence of covering transformations tells us that the following two conditions are equivalent:

• $$p_*(\pi_1(\widetilde{X}, \widetilde{x_0})) = \beta\cdot p_*(\pi_1(\widetilde{X}, \widetilde{x_0}))\cdot\beta^{-1}$$.
• There exists a covering transformation $$F: \widetilde{X} \to \widetilde{X}$$ such that $$F(\widetilde{x_0}) = \widetilde{x}$$.

So statement 4 should have said something like this:

$$p_*(\pi_1(\widetilde{X},\widetilde{x_0}))$$ is a normal subgroup of $$\pi_1(X,x_0)$$ if and only if the group of covering transformations acts transitively on the fibre $$p^{-1}(x_0)$$.

If the above condition holds, then $$\frac{\pi_1(X,x_0)}{p_\star(\pi_1(\widetilde{X},\widetilde{x_0}))}$$ is a group. This group is isomorphic to the group of covering transformations.

[Actually, my final claim about the isomorphism requires additional work to prove. Do give it some thought.]

$$\widetilde{X}$$ does need to be locally path-connected for our argument to be valid. As you pointed out, we need $$\widetilde{X}$$ to be locally path-connected in order to invoke the criterion for the existence of covering maps.

If you want to refer to an authoritative source, then look at Hatcher. You'll see that locally path-connected condition is one of the assumptions in the theorem we just proved. This is Hatcher's Proposition 1.39.

[A small word of caution. One thing that tripped me when I first read Hatcher's Proposition 1.39 is that Hatcher only explicitly say that the base space $$X$$ is locally path-connected. However, the criterion for the existence of covering maps requires the covering space $$\widetilde{X}$$ to be locally path-connected, hence my confusion. I eventually figured out that the base space being locally path connected implies that the covering space is locally path connected. The proof is easy.]