When is the lifting correspondence an isomorphism of groups? Let $p:E\to B$ be a covering map, and let $b_0\in B$. Chose $e_0$ such that $p(e_0)=b_0$. Given an element $[f]\in \pi_{1}(B,b_0)$, let $\tilde{f}$ denote the lifting of $f$ to a path in E which begins at $e_0$. Let $g([f])$ denote the end point of $\tilde{f}(1)$ of $\tilde{f}$. Then $g$ is a well defined set map $$g:\pi_{1}(B,b_0)\to p^{-1}(b_0)$$(This is what I mean by the lifting correspondence.)
I am aware of the following:
$\textbf{Theorem:}$ If $E$ is simply connected, then $g$ is a bijection.
My question is, does this imply that $g$ is a group isomorphism?  If not, when is this the case?
Thanks,
 A: Yes, $g$ is a group isomorphism. But in order for this statement to make any sense, we first need to define the group structure on $p^{-1}(b_0)$!
One sensible way to do this is as follows. Let $H$ be the group of deck transformations for the covering $p : E \to B$. In other words, each element $h \in H$ is a homeomorphism $h : E \to E$ that is compatible with the projection $p : E \to B$, i.e. that satisfies $p \circ h = p$. It is immediate from this definition that every $h \in H$ sends $e_0$ to some point in $ p^{-1}(b_0)$, and with more work (see Hatcher pages 70-71), one can verify that this correspondence between the elements of $H$ and the points in $p^{-1}(b_0)$ is one-to-one and onto. Having thus identified the set $p^{-1}(b_0)$ with the group $H$, it is now meaningful to ask whether or not your map $g : \pi_1(B, b_0) \to p^{-1}(b_0)$ is a group isomorphism, and the answer is yes.
As I mentioned, details of this are provided in Hatcher on pages 70-71, though you may as well prove it yourself. You already know that $g$ is a bijection, so it only remains to check that $g$ is a group homomorphism, and this follows by an argument using uniqueness of path lifts. A few months ago, I wrote an answer about a more general version of this result - see the fourth bullet point in that answer for a hint on how to complete the proof.
