Restriction of the pullback of a covering map

Question

Suppose $p:E\to B$ is a covering map, and let $f:A\to B$ be a continuous map, with $A,B,E$ connected and locally path-connected. Consider the pullback $f^*E$ and let $\pi:f^*E\to A$ be the projection. According to May's Algebraic Topology, if $D$ is a component of $f^*E$, then $\pi|_D:D\to A$ is a covering map. But how do we know that $D\to A$ is surjective?

Answer 1

Recall that $f^*E=\{(a,e) \mid f(a)=p(e)\}$. If $(a,e) \in D$, it suffices to show that for every other $a' \in A$, there exists $e' \in E$ such that $(a',e') \in D$. Consider a path $\gamma$ from $a$ to $a'$, which becomes a path $f \circ \gamma$ from $f(a)$ to $f(a')$. By the lifting property, this path lifts to a path $\tilde{\gamma}$ from $e$ to some $e'$. Then $(a',e') \in f^*E$ and is in the same component as $(a,e)$, so we are done.