# 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?

• Any covering map is a local homoemorphism by definition; this has nothing to do with groups. May 26 '20 at 12:54

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.
• I now understand. Can you give me an example where the covering group is not locally diffeomorphic to the group? The covering map $e=\exp(2\pi it):\mathbb R\to S^1$ is actually a local diffeomorphism. May 26 '20 at 15:12
• Well the covering map is always a local diffeomorphism since $\widetilde G$ inherits a smooth structure from $G$. In general a covering map is always a local homeomorphism. May 26 '20 at 15:16