Prove that the universal covering space of a Lie group is unique

Solution 1:

The problem is that the map $\Psi$ is not unique, but you can force it to have $\Psi(\tilde e)=\hat e$ in a unique way. The universal covering space of a space is indeed unique up to isomorphism, but not up to a unique isomorphism. For instance, the covering map $\mathbf R \to S^1$ has many automorphisms (which are in bijection with $\pi_1(S^1) = \mathbf Z$). You obtain unicity of the isomorphism only in the category of pointed spaces. Thus, in fact, given the pointed space $(G,e)$ and your two covering spaces $(\tilde G, \tilde e)$ and $(\hat G, \hat e)$, the universal property of the covering space implies that there is a unique map of pointed spaces $\Psi: (\tilde G, \tilde e) \to (\hat G, \hat e)$.