I'm reading May's book, 'A Concise Course in Algebraic Topology' and I'm confused about what he means by the induced functor from a covering space. First, here are some helpful/relevant definitions.
(Definition 1) A covering space is a triple $(p, E, B)$ where $p: E \to B$ is a surjective continuous map from the total space to the base space such that for any point $b \in B$ there exists a neighborhood of $b$ whose inverse under $p$ is a disjoint union of open sets in $E$, each of which is homeomorphic to the neighborhood of $b$ via the restriction of $p$.
(Definition 2) A star in a small groupoid $C$ is the collection of objects $C(x, Z)$, for arbitrary objects $Z \in |C|$, and morphisms constructed through composition; that is, for a morphism between objects $C(Z, Z')$, $StC(x)(A,B) = C(Z,Z') \circ C(x, Z)$
(Definition 3) A covering of small groupoids is a triple $(p, E, B)$ such that $p$ is a surjective functor of objects and restricts to a bijection between the stars of objects in $E$; that is, $p: StB(e) \to StE(p(e))$ is bijective for each object.
My question is, how do we get the functor in the following proposition:
(Prop) For a covering of spaces $(p, E, B)$, the induced functor $\Pi(p): \Pi(E) \to \Pi(B)$ is a covering of groupoids.
On objects it is simply the underlying function $p:E \to B$. On arrows, i.e. on paths, it is given by composition: given $[f]:x \to Y$ homotopy class of a path in $\Pi(E)$, you define $\Pi(p)([f]):=[p \circ f]$.
Finally, this functor is a covering of groupoid thanks to the well-known fact that a covering admits a lift for any path in the base space, after having chosen a basepoint in the fiber of its starting point.