How to show that there is an unique multiplication on a group topology in the following problem? Let $(X,+,e)$ be a topological group with identity $e$ and multiplication $+$. Suppose that $X$ is connected, locally path connected, and semilocally simply connected.
Given a subgroup $G$ of $\pi_1(X,e)$, define $P(X,e)$ as the set of all paths in $X$ beginning at $e$. Define an equivalence relation $\sim$ on $P(X,e)$ with $f\sim g$ if and only if $f(1)=g(1)$ and $[f \star g^{-1} ] \in G$. We write the equivalence class of $f$ by $<f>$.
Let $\tilde{X}_G=\{<f> \mid f \in P(X,e)\}$. We define $p:\tilde{X}_G \longrightarrow X$ with $p([f])=f(1)$. I have shown that $p$ is continuous and a covering map.
Now construct a multiplication $\cdot$ on $\tilde{X}_G$ with $<f> \cdot <g>=<f \oplus g>$ where $(f \oplus g )(t)=f(t)+g(t)$ for all $t \in [0,1]$. I also have shown that $(\tilde{X}_G, \cdot, <\varepsilon>)$ is a topological group where $\varepsilon(t)=e$ and $p$ is a group homomorphism.
In book "Topology 2nd " (James Munkres ), it is claimed that there is a unique multiplication on $\tilde{X}_G$ such that $\tilde{X}_G$ becomes a topological group with identity $<\varepsilon>$ and $p$ is a homomorphism. I try to show that if we have another multiplication $\tilde{\cdot}$ on $\tilde{X}_G$ such that $\tilde{X}_G$ becomes a topological group with identity $<\varepsilon>$ and $p$ is a homomorphism, then $<f> \cdot <g>=<f> \tilde{\cdot} <g>$ for all $<f>, <g> \in \tilde{X}_G$ by this way :
Since $p$ is a homomorphism, $$p( <f> \tilde{\cdot} <g>)=p(<f>)+p(<g>)=p(<f> \cdot <g>)=f(1)+g(1).$$ Let $x=f(1)+g(1)$. Because $p$ is a covering map, there is an open neighborhood $U$ of $x$ such that $p^{-1}(U)=\bigcup_{j \in J} V_j$ where $V_j$ is open in $\tilde{X}_G$, $V_i \cap V_j$ is empty and $p|_{V_j}:V_j \longrightarrow U$ is homeomorphism for all $j \in J$. Then there is an unique $V_1$ such that $<f> \tilde{\cdot} <g> \in V_1$, Similary, there is an unique $V_2$ such that $<f> \cdot <g> \in V_2$.
But, I don't have any idea after this step. Thanks.
 A: You said in the comments that you've already shown that $\tilde{X}_G$ is connected, so let's take advantage of that.  Also, I'm gonna use the notation $\langle f\rangle$ instead of $<f>$, just because I think it looks nicer.
Now fix $\langle f\rangle \in \tilde{X}_G$.  Let $$Z = \{\langle g\rangle\in\tilde{X}_G: \langle f\rangle \cdot \langle g\rangle=  \langle f\rangle\tilde{\cdot}\langle g\rangle\}.$$
Note first that by hypothesis, $\langle\varepsilon\rangle\in Z$, so $Z\neq \emptyset$.  Hence, if we can show that $Z$ is both open and closed, then it follows form connectedness of $\tilde{X}_G$ that $Z = \tilde{X}_G$.  Because $\langle f\rangle$ is arbitrary, this will establish the fact that the two multiplications agree.
$Z$ is open Let $\langle g\rangle\in Z$.  This means that $\langle f\rangle \cdot\langle g \rangle = \langle f\rangle \tilde{\cdot}\langle g\rangle.$  Set $x = p(\langle f \rangle \cdot \langle g\rangle).$
Because $p$ is a covering, there is an open neighborhood $V$ of $x$ for which $p^{-1}(V) = \coprod V_i$ with each $p|_{V_i}:V_i\rightarrow V$ a homeomorphism.  Since $p(\langle f \rangle \cdot\langle g\rangle)\in V$, there is some $V_i$, say $V_1$, which contains $\langle f\rangle \cdot \langle g\rangle.$
Because $(\tilde{X}_G,\cdot)$ is a topological group, left multiplication by $\langle f\rangle^{-1}$ is a homeomorphism, so $W:=\langle f \rangle^{-1}\cdot V$ is an open set.  Likewise, $\tilde{W} = \langle f \rangle^{-1}\tilde{\cdot} V$ is an open set.  (Careful, the notation $\langle f\rangle^{-1}$ used for $W$ refers to the $\cdot$-inverse, while it refers to the $\tilde{\cdot}$-inverse for $\tilde{W}$.
Note that $\langle g\rangle = \langle f \rangle^{-1}(\langle f\rangle \cdot \langle g\rangle)$, so $\langle g\rangle\in W$.  That is, $W$ is an open neighborhood of $\langle g\rangle$.  Likewise, $\tilde{W}$ is an open neighborhood of $\langle g\rangle$.  Thus, $U:= W\cap \tilde{W}$ is an open neighborhood of $\langle g\rangle$.
We claim that $U\subseteq Z$.  So see this, pick $\langle h\rangle \in U$.  Then $$\langle f\rangle \cdot \langle h\rangle \in \langle f \rangle \cdot U\subseteq \langle f \rangle \cdot W = \langle f\rangle \cdot \langle f\rangle^{-1} \cdot V_1 = V_1,$$ so $\langle f\rangle \cdot \langle h\rangle \in V_1$.  Likewise, $\langle f\rangle \tilde{\cdot} \langle h\rangle \in V_1$.
Further, $p(\langle f\rangle \cdot \langle h \rangle) = f(1) + h(1) = p(\langle f\rangle \tilde{\cdot} \langle h \rangle)$ by hypothesis.  Because $p|_{V_1}$ is injective, it now follows that $\langle f \rangle \cdot \langle h\rangle = \langle f\rangle \tilde{\cdot}\langle h \rangle$.  In otherwords, $h\in Z$.  This is establishes that $U\subseteq Z$, so $Z$ is open.
$Z$ is closed.  We will show the complement is open.  So, suppose $\langle g\rangle \notin Z$.  As in the open case, if we set $x = p(\langle f\rangle \cdot \langle g\rangle)$ (which is equal to $p(\langle f\rangle \tilde{\cdot} \langle g\rangle)$ because both are equal to $f(1) + g(1)$), we have an open set $V$ around $x$ for which $p^{-1}(V) = \coprod V_i$ with $p|_{V_i}:V_i\rightarrow V$ a homeomorphism.
We let $V_1$ be the $V_i$ containing $\langle f\rangle \cdot \langle g\rangle$ and we let $V_2$ be the $V_i$ containing $\langle f\rangle \tilde{\cdot} \langle g \rangle$.  Note that $V_1\neq V_2$ (because other wise, injectivity of $p$ on $V_1$ would force $\langle f\rangle \cdot \langle g\rangle = \langle f\rangle \tilde{\cdot} \langle g\rangle,$), so $V_1\cap V_2 = \emptyset$.
Create the open set $U$ similarly to the open case:  $U = \langle f\rangle^{-1} V_1\cap \langle f\rangle^{-1}\tilde{\cdot} V_2$.  Since $\langle f\rangle \cdot \langle g\rangle \in V_1$ and $\langle f\rangle \tilde{\cdot} \langle g\rangle \in V_2$, we see that $\langle g\rangle \in U$.
We claim that $U$ is a subset of the complement of $Z$.  To that end, let $\langle h\rangle \in U$.  Then $\langle f \rangle \cdot \langle h\rangle \in V_1$ while $\langle f\rangle \tilde{\cdot} \langle h\rangle \in V_2$.  Because $V_1\cap V_2 = \emptyset$, $\langle f\rangle \cdot \langle h\rangle \neq \langle f\rangle \tilde{\cdot}\langle h\rangle$.  That is, $\langle h\rangle \notin Z$.  Thus, $U$ is a subset of the complement of $Z$, so $Z$ is closed.
