Let $\tilde{G},G,H$ be Lie groups with $H$ connected, $f:H\to G$ a Lie group homomorphism, and $p:\tilde{G}\to G$ be a covering map that is also a Lie group homomorphism. Suppose $\tilde{f}:H\to \tilde{G}$ is a (not necessarily continuous) lift of $f$ that is a group homomorphism. Then is it necessarily true that $\tilde{f}$ is continuous, and hence a Lie group homomorphism?
Actually I want this result to show that the map $\rho:U(V)\to \text{Spin}^c(V)$ defined here An embedding of $U(n)\to \text{Spin}^c(n)$ is continuous. This map $\rho$ is a lift of the map $U(V)\to SO(V)\times S^1$ given by $A\mapsto (i(A),\det(A))$ where $i:U(V)\to SO(V)$ is inclusion. (Note that there is a double covering $\text{Spin}^c(V)\to SO(V)\times S^1$.)
Best Answer
$\newcommand{\wt}{\widetilde}\newcommand{\by}{\times}\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}\newcommand\Q{\mathbb Q}$
The $\wt f$ you describe will not always be continuous. As a simple case, say $G=H=\R$ and let $\wt G=\R\by\Q$ with covering map given by projection onto $\R$. Let $f:\R\to\R$ be the identity, so choosing a lift $\wt f:\R\to\R\by\Q$ amounts to choosing a group homomorphism $\R\to\Q$. There are plenty of these. For example, observe that $\R$ is a $\Q$-vector space and fix an (uncountable) basis $\{e_i\}_{i\in I}$. Let $\lambda_i:\R\to\Q$ be projection onto the $i$th factor, i.e. given $r\in\R$, one has $$r=\sum_{i\in I}\lambda_i(r)e_i.$$ Then, $\wt f(r):=(r,\lambda_i(r))$ is a lift of $f$ and a group homomorphism for all $i\in I$, but these maps are typically not continuous.