Pullback of maps out of a connected & locally path connected space

Question

I'm reading Peter May's "Concise Course in Algebraic Topology", and I'm having trouble interpreting this yellow-highlighted line.

When I've seen the expression "pullback of $f$ along $g$" previously, $g$ and $f$ have had shared codomains. So I have two questions:

• Diagrammatically, what does the setup look like here? (i.e. what is May saying?)
• Why does $D$ cover $A$ in this situation?

Answer 1

In the chapter "Covering spaces" May makes the assumption

Throughout this chapter, we assume that all given spaces are connected and locally path connected.

You can of course introduce the concept of covering spaces in a more general setting by omitting this assumption. Most authors do so.

In this general setting one can easily prove that the pullback of a covering map $p : E \to B$ along a map $f :A \to B$ is again a covering map. This pullback is of course a map $p' : E' \to A$:

$\require{AMScd}$ \begin{CD} E' @>{f'}>> E \\ @V{p'}VV @V{p}VV \\ A @>{f}>> B\end{CD}

May's statement (which I modified a little bit to adjust it to our general setting) is this:

The pullback of a map $f :A \to B$ along a covering map $E \to B$ is a covering map $p : D \to A$.

I think the pullback of $f$ along $E \to B$ is a map $D \to E$ which is not what we want. But perhaps that is only a question of notational conventions.

May's statement includes the assumption "where $A$ is connected and locally path connected". This is due to his basic assumption on spaces. Moreover, he adds "and $D$ is a component of the pullback". This is due to the fact that our above $E'$ is not necessarily connected. But restricting $p' : E' \to A$ to such $D$ yields again a covering map - which now fits into his framework. An example for this phenomenon is the covering map $p : \mathbb R \to S^1$. Pulliing back along the inclusion of a one-point space $P$ into $S^1$ yields the covering map $\mathbb Z \to P$.