Let $\phi: A \rightarrow B$ be a continuous homomorphism between Lie groups (or any appropriate generalization). Is there any relationship between the degree of $\phi$ as a covering map of and the number of elements in its kernel?
The motivation for this question comes from the spin groups: $Pin(s,t)$, $Spin(s,t)$, and $Spin^+(s,t)(s,t)$. If $\lambda$ is the map $$\lambda: Pin(s,t) \rightarrow O(s,t) \\ u \mapsto R_u$$ where $$R_u(x) = \begin{cases} -x \quad x \parallel u \\ x \quad x \perp u \end{cases}$$ then $\lambda$ has kernel $\{1, -1\}$. From this fact it somehow follows that $\lambda$ is a double covering of Lie group. I do not see where this conclusion comes from, and I suspect that there might be a relation between the degree of $\phi$ as a covering map of and the number of elements in its kernel but I am not sure.
Is it important that $\phi$ be open as well? Surjectivity is implied since we assume $\phi$ is also a covering map.