Here is the question I am trying to solve:
Let $G$ be a topological group and $p: X \to G$ be a covering. Let $e \in G$ be the identity element of $G$ and choose $f \in p^{-1}(e).$
$(a)$ Show that $X$ has a unique topological group structure with identity element $f,$ such that $p$ is a (continuous) group homomorphism.
$(b)$ Show that $X$ is abelian if $G$ is.
Some thoughts and hints :
1- I know that I should produce a multiplication map $\mu^{'}$ (Look at the diagram below) and this mutiplication map must be associative and unital (**I do not know why I should show that if later on I also should prove that there exists an inverse, could someone clarify this to me please? **)
I also do not know how to produce this map but I am guessing that we should use the lifting theorem but still I do not know how, could anyone help me in this please?
2- I also know that we want to produce an inverse map $\chi$ on $X$ according to the following diagram
More precisely, since the inverse on a group means the following diagram:
Where we wanna prove that the left arrow of the diagram factors through $e$ i.e., we have $G \to e \to G$ i.e., we have the following diagram:
more precisely, we have the following diagram:
i.e., the equation that should be satisfied is $e = g^{-1} g$ which means that there is a group homomorphism that sends all of $G$ to the trivial group and then includes the trivial group into $G.$
Still I do not know how to show the existence of this inverse.
Note That:
In general, I know I will be using the lifting theorem, in the first case, I wanna show that $X \times X \to G \times G \to G$ lifts over the covering map i.e., we wanna show that the hypotheses of the lifting theorem are satisfied and similarly for the second case.
For (b) I do not know what exactly should I do but I will post it in a separate question
Edit:
1-Is the map $X \times X \to G \times G$ is just $p \times p$?
2- In the link provided by @OsamaGhani (this one Covering space of a topological group is itself a topological group) in the comments, why the author tended to take the identity as the second component in the multiplication defined on the covering space?
3- Also, in the path lifting property, we usually are lifting a path, where is the path we are lifting here? are we using the path lifting property or the homotopy lifting property? I am confused (my gut feeling we will be using homotopy lifting property). No,again, I think there is no given homotopy here to lift
4- I think also there is a typo in the link mentioned above. In this term $\mu(\widetilde{e},x)=x$, I think $\mu$ can not take the first argument an element of $\widetilde{G}.$ Am I correct?
5- I am not accustomed in general to defining a lifting may in two components, how I should choose these two components in this case?
6- How we will keep the endpoint of the first path, the beginning of the second path in our case here?
7- will this link If a covering space of a topological space X has a topological group structure, when we transfer this structure on X? help in defining the multiplication?
EDIT 2
I think if someone can show me why the multiplication map $X \times X \to X$ exists that will be great, I believe we wanna show that $f_* (\pi_1 (Y, y_0)) \subset p_* (\pi_1 (\tilde X, \tilde x_0))$ but I do not know how. Any help will be greatly appreciated!
Best Answer
This excercise is quite long an involved, and I think you'll learn a lot from going through the details yourself. So, this answer will thus be purposely incomplete.
Let's once and for all fix an element $\tilde{e}\in X$ with the property that $p(\tilde{e}) = e\in G$, where $e$ denotes the identity in $G$.
Our first goal is to create a map $\tilde{\mu}:X\times X\rightarrow X$. As hinted in the comments, we already have a map $\mu\circ (p\times p):X\times X\rightarrow G$, so if we can lift that map, we have our desired $\tilde{\mu}$. Such a lift exists if and only if $$\mu_\ast((p\times p)_\ast(\pi_1(X\times X, (\tilde{e},\tilde{e}))\subseteq p_\ast(\pi_1(X,\tilde{e})),$$ so let's check this condition. We need a lemma.
Lemma 1: The induced map $\mu_\ast: \pi_1(G,e)\times \pi_1(G,e)\rightarrow \pi_1(G,e)$ is $\mu_\ast([\gamma],[\alpha]) = [\gamma][\alpha]$ where juxtaposition on the right denotes the group operation in $\pi_1(G,e)$.
Proof: Consider the composition $G\xrightarrow{i_1} G\times G\xrightarrow{\mu} G$ where $i_1(g) = (g,e)$. This composition is the identity on $G$. Using $1$ to denote the trivial loop, it follows that $\mu_\ast([\gamma], 1) = [\gamma]$ for all $[\gamma]\in\pi_1(G,e)$.
Repeating this using $i_2(g) = (e,g)$, we find that $\mu_\ast(1,[\gamma]) = [\gamma]$ for all $[\gamma]\in \pi_1(G,e)$.
Finally, any element in $\pi_1(G,e)\times \pi_1(G,e)$ has the form $([\gamma],[\alpha]) = ([\gamma],1)(1,[\alpha])$ for some $[\gamma],[\alpha]\in pi_1(G,e)$. Since $\mu_\ast$ is a homomorphism, we therefore have $$\mu_\ast([\gamma],[\alpha]) = \mu_\ast([\gamma],1)\mu_\ast(1,[\alpha]) = [\gamma][\alpha],$$ as claimed. $\square$
With the lemma in hand, we can now prove:
Proposition 2: We have $\mu_\ast( (p\times p)_\ast(\pi_1(X\times ,(\tilde{e}, \tilde{e}))))\subseteq p_\ast(\pi_1(X,\tilde{e})).$
Proof: For ease of writing, I'm going to write $H:=p_\ast(\pi_1(X,\tilde{e})).$ Then our goal is to show that $\mu_\ast(H\times H)\subseteq H$. But, from Lemma 1, $\mu_\ast(H\times H) = HH\subseteq H$ since $H$ is a subgroup. $\square$
So, we have a lift $\tilde{\mu}:X\times X\rightarrow X$. Notice that $\tilde{\mu}(\tilde{e},\tilde{e})$ projects to $e$. Thus, we may pick $\tilde{\mu}$ so that $\tilde{\mu}(\tilde{e},\tilde{e}) =\tilde{e}$. Once we have made this choice, $\tilde{\mu}$ is unique. Here is the first important fact regarding $\tilde{\mu}$.
Proposition 3: The projection map $p$ resepcts $\tilde{\mu}$ in the sense that $p\circ \tilde{\mu}(x_1,x_2) = \mu(p(x_1), p(x_2))$. In other words, $p$ is a homomorphism (except we don't yet know that $\tilde{\mu}$ gives a group structure.)
Proof: This is exactly what is meant by saying that $\tilde{\mu}$ is a lift of $\mu\circ (p\times p). $\square$
But how can we verify that $\tilde{\mu}$ is a group multiplication? There are three things we need to verify: associativity, the existence of an identity, and the existence of inverses.
Here's the argument for inverses (which I believe is the hardest case). Something similar works for the other two properties which need verification.
First, we need an inverse map. To that end, consider the composition $X\xrightarrow{p}G\xrightarrow{inv} G$ where $inv(g) = g^{-1}$ is the inverse map on $G$. I will leave it to you to verify that this composition has a unique lift $\tilde{inv}:X\rightarrow X$ with $\widetilde{inv}(\tilde{e}) = \tilde{e}$.
But why is $\widetilde{inv}$ an inverse map? We need to verify that $\tilde{\mu}(x,\widetilde{inv}(x)) = \tilde{e}$ for all $x\in X$. Said another way, we need to verify that the composition $X\xrightarrow{(i_1, \widetilde{inv})}X\times X\xrightarrow{\tilde{\mu}} X$ the constant map with image $\tilde{e}$.
To that end, consider the composition $X\xrightarrow{p}G\xrightarrow{c_{e}} G$ where $c_e$ denotes the map which is constantly $e$: $c_e(g) = e$ for all $g\in G$. I claim that the map $\tilde{\mu}\circ(i_1,\widetilde{inv})$ lifts this map. Indeed, we have $$p(\tilde{\mu}((i_1,\widetilde{inv})(x)) = \mu(p\times p)(x, \widetilde{inv}(x)) = \mu(p(x), p(\widetilde{inv}) = \mu(p(x), inv(p(x)) = \mu(p(x), p(x)^{-1}) = e.$$
So, the map $\tilde{\mu}\circ(i_1, \widetilde{inv})$ and the map $c_{\tilde{e}}:X\rightarrow X$ are both lifts of the same map agreeing at one point, so they agree everywhere. That is, $\widetilde{inv}$ really is the inverse map.