In Hatcher's Algebraic Topology on page $409$ in the third paragraph it is written
Given a fibration $p : E \to B$ with fiber $F = p^{−1} (b_0 )$, we know that the
inclusion of $F$ into the homotopy fiber $F_p$ is a homotopy equivalence. Recall that $F_p$
consists of pairs $(e, \gamma)$ with $e \in E$ and $\gamma$ a path in $B$ from $p(e)$ to $b_0$. The inclusion $F\hookrightarrow E$ extends to a map $i : F_p \to E$, $i(e, \gamma) = e$, and this map is obviously a fibration. In fact it is the pullback via $p$ of the path fibration $PB \to B$.
Why is the fibration $i$ the pullback of the path fibration $\pi: PB \to B$ via $p$?
It is $PB = \{\gamma \in C(I,B): \gamma(0) = b_0\}$ and $\pi(\gamma) = \gamma(1)$ and therefore
$$p^* PB = \{(e,\gamma) \in E \times C(I,B): p(e) = \gamma(1), \ \gamma(0) = b_0\}.$$
But it is $F_p = \{(e,\gamma) \in E \times C(I,B): p(e) = \gamma(0), \ \gamma(1) = b_0\}$, i.e. with $R: C(I,B) \to C(I,B), \gamma \mapsto (t \mapsto \gamma(1-t))$
$$F_p = (\text{id}_E \times R)(p^* PB)$$ and not
$$F_p = p^* PB.$$
What is wrong here?
Best Answer
Whether or not something's wrong here is a bit subtle: Generally speaking, pullbacks are only defined up to (unique) isomorphism, and you've already noted that $p^* PB$ and $F_p$ are naturally isomorphic (via the map "reversing the paths"), so with this view everything's fine. However, since Hatcher explicitly defines the pullback as a certain subset of the product, I tend to agree with you that the statement is not quite correct.