Universal covering group is locally homemorphic to the group In Knapp's book on Lie group he claims that for any separable metrizable topological group $G$ which is path-connected, locally connected and locally simply connected, the universal covering space admits a unique topological group structure such that the covering map $e:\widetilde{G}\to G$ is a continuous group homomorphism where $\widetilde{G}$ is the universal covering space.
For this the multiplication on $\widetilde{G}$ is defined to be the unique lift of the map $m\circ (e\times e):\widetilde{G}\times\widetilde{G}\to G.$ The identity he chooses any $\widetilde{1}\in e^{-1}(1).$ Now here comes my doubt. Clearly, multiplicative identity is unique. Hence we must have that the cardinality of the fiber of $e^{-1}(1)$ is one. This means $e$ must be a local homeomorphism. Since $G$ is connected we must have that cardinality of $e^{-1}(g)$ is one for all $g\in G.$ Thus $e$ is a local homeomorphism (even a local diffeomorphism) which is onto. So locally $\widetilde{G}$ is just $G$!!! Is my argument alright. Can one back up my argument with an explicit example?
 A: This is incorrect. You choose the identity $\widetilde 1 \in e^{-1}(1)$ before you choose the lift in order to make it well defined. This is because if $p:(C,c_0) \rightarrow (X,x_0)$ is a covering space and $f: (Y,y_0) \rightarrow (X,x_0)$ is a map so that $f_* (\pi_1(Y)) \subset p_*(\pi_1(C))$ then there doesnt exist a unique lift $g:Y \rightarrow C$ unless you require that $g(y_0) = c_0$, only then is the lift unique. 
In your situation $C = \widetilde G$, $X = G$ and in order to define the multiplication uniquely you have to choose a basepoint $\widetilde 1 \in e^{-1}(1)$ before taking the lift of $\widetilde G \times \widetilde G \rightarrow G$.
Look at the case $G = S^1, \widetilde G = \mathbb R$ and $e^{2 \pi it}:\mathbb R \rightarrow S^1$. There are many lifts of $\mathbb R \times \mathbb R \rightarrow S^1$, if $f:\mathbb R \times \mathbb R \rightarrow \mathbb R$ is any one of them you can construct many more by simply defining $g(x) = f(x) + n$ where $n$ is any integer. However if you require that your lift has to take $0 \in \mathbb R$ to $1 \in S^1$ there is only one lift.
