Showing that $X$ has a unique topological group structure. 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!
 A: 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.
A: Let $p^n_i\colon G^n\to G$ be the projection functions given by $p^n_i(g_1,g_2,\dots,g_n)=g_i$. For any pair of functions $f_i\colon Y\to G$, let $(f_1,f_2,\dots,f_n)\colon Y\to G^n$ be the unique function such that $p^n_i\circ(f_1,f_2,\dots,f_n)=f_i$. Note that $p^1_1\colon G\to G$ is the identity function. Finally, fix a singleton $\{*\}$, identify $G^0$ with it, and define $p^n_0\colon G^n\to G^0=\{*\}$ to be the unique function from $G$ to that singleton.
A group structure on $G$ then consists of a unit function $e\colon\{*\}=G^0\to G$, a multiplication function $\mu\colon G^2\to G$, and an inversion function $\chi\colon G\to G$ satisfying the group axioms

*

*associativity $\mu\circ(p^3_1,\mu\circ(p^3_2,p^3_3))=\mu\circ(\mu\circ(p^3_1,p^3_2),p^3_3)\colon G^3\to G^2\to G$

*unit $\mu\circ(p^1_1,e\circ p^1_0)=p^1_1=\mu\circ(e\circ p^1_0,p^1_1)\colon G\to G\times G\to G$

*inverses: $\mu\circ(p^1_1,\chi)=e\circ p^1_0\colon G\to G\times G\to G=G\to G^0\to G$
Analogously, let $p'^n_i\colon X^n\to X$ be the projection functions given by $p'^n_i(x_1,\dots,x_n)=x_i$ and $p'^n_0\colon X^n\to\{*\}=X_0$ the unique function to the singleton. A group structure on $X$ is then a triple of functions for which $e'\colon X^0=\{*\}\to X$, $\mu'\colon X^2\to X$ and $\chi'\colon X\to X$ for which the group axioms, obtained by replacing $p^n_i,e,\mu,\chi$ with $p'^n_i,e',\mu',\chi'$ above, hold.
Consider now a function $p\colon X\to G$ (note that $p^n_i\colon G^n\to G$ are unrelated to $p\colon X\to G$). Let $p^n\colon X^n\to G^n$ be given by $p^n=(p\circ p'^n_1,\dots,p\circ p'^n_n)$ so that $p^n_i\circ p^n=p\circ p'^n_i$. $p\colon X\to G$ is a group homomorphism if and only if $p\circ e'=e=e\circ p^0$, $p\circ\mu'=\mu\circ p^2$, and $p\circ\chi'=\chi\circ p$.
We want to show that if $p\colon X\to G$ is a covering space (with $X$ connected and locally path connected), then for each $e'\colon\{*\}\to X$ satisfying $p\circ e'=e$, there exist unique continuous functions $\mu'\colon X^2\to X$ and $\chi'\colon X\to X$ such that $(e',\mu',\chi')$ is a group structure on $X$ for which $p\colon X\to G$ is a group homomorphism.

We can reformulate the problem by making two observations.
First, the equations $p\circ\mu'=\mu\circ p^2$, $p\circ\chi'=\chi\circ p$, $p^n_i\circ p^n=p\circ p'^n_i$ assert that $\mu'\colon X\to X\to X$, $\chi'\colon X\to X$, and $p'^n_i\colon X^n\to X$ are lifts of $\mu\circ p^2$, $\chi\circ p$, and $p^n_i\circ p^n$ relative to $p\colon X\to G$.
Moreover, if $f'\colon X^n\to X$ and $h'_k\colon X^{m_k}\to X$ are lifts of $f\colon G^n\to G$ and $h_k\colon G^{m_k}\to G$ relative to $p\colon X\to G$, then $f'\circ(h'_1,\dots,h'_n)\colon X^{m_1+\dots+m_k}\to X$ is a lift of $f\circ(h_1,\dots,h_n)\circ p^{m_1+\dots+m_n}\colon X^{m_1+\dots+m_n}\to G$. Indeed, $p\circ f'\circ(h'_1,\dots,h'_n)=f\circ p^n\circ(h'_1,\dots,h'_n)=f\circ(p\circ h'_1,\dots,p\circ h'_n)=f\circ(h_1\circ p^{m_1},\dots,h_n\circ p^{m_n})=f\circ(h_1,\dots,h_n)\circ p^{m_1+\dots+m_n}$.
It follows that not only are the group structure functions $(e',\mu',\chi')$ lifts relative to $p\colon X\to G$ of the group structure functions $(e,\mu,\chi)$ following $(p^0,p^2,p^1)$ respectively, but also each side of each group axiom for the structure functions $(e',\mu',\chi')$ is a lift of each side of the corresponding group axiom for $(e,\mu,\chi)$ following an appropriate $p^n$. In particular, an appropriate uniqueness of lifting would imply that the group axioms hold for the lifts $(e',\mu',\chi')$ if they do for $(e,\mu,\chi)$.
Second, the group axioms imply $\mu'\circ(e',e')=e'=\chi'\circ e'$, i.e. that $\mu'\colon(X\times X,e'\times e')\to(X,e')$ and $\chi'\colon(X,e')\to(X,e')$ are functions between pointed spaces. Likewise, $p'^n_i\circ(e',\dots,e')=e'$, $p^n_i\circ(e,\dots,e)=e$, and $p^n\circ(e',\dots,e')=(e,\dots,e)$ show that $p'^n_i\colon(X,e')^n\to(X,e')$, $p^n_i\colon(G,e)^n\to(G,e)$, and $p^n\colon(X,e')^n\to(G,e)^n$ are also functions between pointed spaces.
Thus it will suffice to show that relative to a covering map $p\colon (X,e')\to(G,e)$ (with $X$ connected and locally path-connected), there are unique continuous lifts of $\mu\circ p^2\colon(X,e')^2\to(G,e)$ and $\chi\circ p\colon(X,e')\to(G,e)$, and that for each group axiom, the lift of the two functions whose equality the axiom asserts, following the appropriate $p^n$, is unique.

We now see the relevance of the unique lifting criterion, which is a combination of the following propositions from Hatcher's Algebraic Topology book.
Proposition 1.33. (lifting criterion) Suppose given a covering space $p\colon(\tilde X,\tilde x_0)\to(X,x_0)$ and a map $f\colon(Y,y_0)\to(X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\tilde f\colon(Y,y_0)\to(\tilde X,\tilde x_0)$ of $f$ exists iff $f_*(\pi_1(Y,y_0))\subseteq p_*(\pi_1(\tilde X,\tilde x_0))$.
Proposition 1.34. (unique lifting property) Given a covering space $p\colon\tilde X\to X$ and a map $f\colon Y\to X$, if two lifts $\tilde f_1,\tilde f_2\colon Y\to\tilde X$ of $f$ agree at one point and $Y$ is connected, then $\tilde f_1$ and $\tilde f_2$ agree on all of $Y$.
In our situation we want to find, for certain functions $f\colon(X,e')^n\to(G,e)$, unique lifts $f'\colon(X,e')^n\to(X,e')$ of $f\circ p^n\colon(X,e')^n\to(G,e)^n\to(G,e)$ relative to $p\colon(X,e')\to(G,e)$. To use both propositions we have to require that $X^n$ be connected, path-connected, and locally path-connected. But it is enough to require that $X$ be connected and locally path-connected, since then $X$ is path-connected, and also $X^n$ are connected and locally path-connected for every $n$ (the latter is an exercise in basic topology).

Using Proposition 1.33. we can show that lifts of $\mu\circ p^2\colon(X,e')^2\to(G,e)^2\to(G,e)$ and $\chi\circ p\colon(X,e')\to(G,e)\to(G,e)$ relative to $p\colon (X,e')\to(G,e)$ exist. To that end, we need to determine when $f\colon (G,e)^n\to(G,e)$ has $(f\circ p^n)_*\pi_1((X,e')^n)\subseteq p_*\pi_1(X,e')$. For simplicitly, let $p_*\pi_1(X,e')=H$.
Recall Hatcher's Proposition 1.12. $\pi_1(X\times Y)$ is isomorphic to $\pi_1(X)\times\pi_1(Y)$ where $X$ and $Y$ are path-connected.
We actually have more: given continuous functions $g\colon X_1\to X_2$ and $h\colon Y_1\to Y_2$, then $(g\circ p^2_1,h\circ p^2_2)_*\colon\pi_1(X_1\times Y_1)\to\pi_1(X_2\times Y_2)$ corresponds to $(g_*\circ p_2^1, h_*\circ p^2_2)\colon\pi_1(X_1)\times\pi_1(Y_1)\to\pi_1(X_2)\times\pi_1(Y_2)$.
Thus $(f\circ p^n)_*\pi_1((X,e')^n)=f_*\circ p^n_*(\pi_1(X,e')^n)=f_*(H^n)$ where $H^n=H\times\cdots\times H\subseteq\pi_1(G,e)\times\cdots\times\pi_1(G,e)=\pi_1(G,e)^n\cong\pi_1(G,e)^n$.
Now a trick. Given $n$ loops $\gamma_1,\dots,\gamma_n$ in $G$ with basepoint $e$, on the one hand, the product loop $(\gamma_1,\dots,\gamma_n)\in\pi_1(G,e)^n$ satisfies $(\gamma_1,\dots,\gamma_n)=(\gamma_1,c_e,\dots,c_e)(c_e,\gamma_2,c_e,\dots,c_e)\cdots(c_e,\dots,c_e,\gamma_n)$ where $c_e$ is the constant loop at $e$. On the other hand, $f_*(c_e,\dots,c_e,\gamma_i,c_e,\dots,c_e)=f^i_*(\gamma_i)$ where $f^i_*\colon\pi_1(G,e)\to\pi_1(G,e)$ is induced by $f^i\colon(G,e)\to(G,e)$ is given by $f^i(g)=f(\underbrace{e,e,\dots,e,g}_i,e,\dots,e)$.
Thus we have $f_*(\gamma_1,\dots,\gamma_n)=f_*((\gamma_1,c_e,\dots,c_e)(c_e,\gamma_2,c_e,\dots,c_e)\cdots(c_e,\dots,c_e,\gamma_n))=f_*(\gamma_1,c_e,\dots,c_e)f_*(c_e,\gamma_2,c_e,\dots,c_e)\cdots f_*(c_e,\dots,c_e,\gamma_n)=f^1_*(\gamma_1)\cdots f^n_*(\gamma_n)$.
In particular, $f_*(H^n)=f^1_*(H)f^2_*(H)\cdots f^n_*(H)$, the right-hand side being a product of subsets in $\pi_1(G,e)$. But $H=p_*\pi_1(X,e')$ is the image of a group under a homomorphism, so contains the identity $c_e$. It follows that $f^i_*(H)=c_e\cdots c_eHc_e\cdots c_e=f^1_*(c_e)\cdots f^i_*(H)\cdots f^n_*(c_e)\subseteq f^1_*(H)f^2_*(H)\cdots f^n_*(H)$.
Thus the condition $f_*(H^n)\subseteq H$ requires $f^i_*(H)\subseteq H$ for each $i$. But conversely, because $H$ is the image of a group under a homomorphism and so closed under the group operation, $f^i_*(H)\subseteq H$ for each $i$ implies $f_*(H^n)=f^1_*(H)\cdots f^n_*(H)\subseteq H\cdots H=H$.
To summarize: $f\colon(G,e)^n\to(G,e)$ is such that $f\circ p^n\colon(X,e')^n$ has a lift $f'\colon(X,e')^n\to(X,e')$ if and only if $f^i_*(H)\subseteq H$ for each $f^i\colon(G,e)\to (G,e)$ given by $f^i(g)=(\underbrace{e,\cdots,e,g}_i,e,\cdots,e)$ and $H=p_*\pi_1(X,e')\subseteq\pi_1(G,e)$.

Using the above, it follows that any $\mu\colon (G,e)^2\to (G,e)$ satisfying $\mu(x,e)=x=\mu(e,x)$ has a lift since those equations assert that $\mu^1=\mathrm{id}_G=\mu^2$, and then $\mu^i_*(H)=\mathrm{id}_{G,*}(H)=H$.
Seeing that $\chi_*(H)\subseteq H$ is more subtler: one can do it using the inverse axiom and functoriality of the fundamental group. Path-connectedness of $G$ (implied by path-connected of the covering space $X\to G$) implies that $(e_*,\mu_*,\chi_*)$ also satisfy the group axioms, so define a group structure on $\pi_1(G,e)$. But since $\mu_*\colon\pi_1(G,e)\times\pi_1(G,e)\to\pi_1(G,e)$ is a group homomorphism and $e_*=c_e$ is the identity of $\pi_1(G,e)$, the Eckmann-Hilton argument implies that $\mu_*$ is the group operation of $\pi_1(G,e)$. Then $\chi_*$ is the inverse function of $\pi_1(G,e)$, whence $\chi_*(H)=H$ since $H$ being a subgroup is closed under inverting.

