Given a continuous function $f: X \to Y$ a covering $p: E \to Y$, what are there conditions for me to say that there exists a lift $g: X \to E$ s.t. $p \circ g = f$? Under what conditions is this lift unique for a starting point $x_0$?
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$\begingroup$I prefer to answer this question using groupoids and the notion of covering morphism of groupoids: this is a morphism $q: H \to G$ of groupoids such that for each $e \in Ob(H)$ and $g \in G$ starting at $q(e)$ there is unique $h \in H$ starting at $e$ such that $q(h)=g$. If $p:E \to Y$ is a covering map, then $q=\pi_1(p): \pi_1(E) \to \pi_1(Y)$ is a covering morphism of groupoids. A full exposition of this approach is in Chapter 10 of Topology and Groupoids (T&G), as it was in the 1968, 1988 (differently titled) editions of this book. Thus covering morphisms of groupoids form a good algebraic model of covering maps of spaces.
Now given a map $f: X \to Y$ from a path connected space $X$ and points $x \in X, e \in E$, $f$ lifts to a map $g : X \to E$ such that $pg=f$ and $f(x)=e$ if and only if the corresponding morphism of fundamental groupoids $\pi_1(f): \pi_1(X) \to \pi_1(Y)$ lifts to a morphism of groupoids $\phi: \pi_1(X) \to \pi_1(E)$ such that $\phi(x)=e$ and (of course) $q \phi= \pi_1(f)$. Further, such a lift is unique. This is 10.5.3 of T&G.
A bit of study of the properties of covering morphisms also relates this existence condition to the usual condition that $\pi_1(f) $ maps the fundamental group $\pi_1(X,x)$ into the image under $q$ of $\pi_1(E,e)$.
(Some similar results are in Peter May's "Concise ...." book, which refers to the 1988 edition of T&G as "idiosyncratic"; in fact T&G is still "out of line" in many ways! For example, this approach allows for an answer to the question: what local conditions on a space $Y$ are sufficient for a covering morphism $q: H \to \pi_1(Y)$ to be realised by a covering map $p: E \to Y$. Also May does not use the fundamental groupoid on a set of base points, and so fails to use groupoids to determine the fundamental group of the circle!)
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