Let $\pi: P \to B$ denote a principal $G$-bundle over base $B$, and let $f: B' \to B$ be a continuous map from another space $B'$ to $B$.
I've been reading Stephen Mitchell's Notes on Principal Bundles and Classifying Spaces. On page 3, he says that sections of the pullback bundle $f^{*}P \to B'$ are in bijective correspondence with lifts in the diagram
Now, I understand that this fits into a bigger commutative diagram, namely
where $\text{pr}_2$ is the projection onto the second factor (thinking, as usual, of the pullback bundle as $f^{*}P = \{ (b', p) \in B' \times P ~|~ f(b') = \pi(p) \}$).
This suggests that given a section $s' : B' \to f^{*}P$, one can construct the lift by defining $\widetilde{f} = \text{pr}_2 \circ s'$.
Question: What about the converse? That is, given a lift $\widetilde{f}: B' \to P$ that fits into the above commutative diagram, how does one construct a (local) section of $\pi' : f^{*}P \to B'$?
In the textbook, “Lecture Notes on Algebraic Topology,'' by Davis and Kirk [cf. pg. 168], it is stated that the cross-section problem for pullback bundles is equivalent to the so-called relative lifting problem. On the one hand, the existence of a commutative square is equivalent to the object in the top left corner being isomorphic to the pullback bundle (this is also mentioned in Mitchell's notes referred to above). However, without resorting to a local trivialization, is there a way to understand the claimed bijective correspondence?
Best Answer
Consider the map $s': B' \to B'\times P$ given by $s'(b') = (b', \tilde{f}(b'))$.
As $\tilde{f}$ is a lift of $f$ through $\pi$, we have $\pi\circ\tilde{f} = f$, so $\pi(\tilde{f}(b')) = (\pi\circ\tilde{f})(b') = f(b')$, i.e. $s'(b') \in f^*P$. Therefore, we can regard $s'$ as a map $s' : B' \to f^*P$.
Now note that $\pi'(s'(b')) = \pi'(b', \tilde{f}(b')) = b'$, so $s'$ is a section of $\pi' : f^*P \to B'$.
Here's another approach. Consider the maps $\operatorname{id}_{B'} : B' \to B'$ and $\tilde{f} : B' \to P$. By the universal property of the pullback, there is a unique map $s' : B' \to f^*P$ such that $\pi'\circ s' = \operatorname{id}_{B'}$ (i.e. $s'$ is a section of $\pi' : f^*P \to B'$) and $\operatorname{pr}_2\circ s' = \tilde{f}$. Moreover, these two equalities imply that $s'(b') = (b', \tilde{f}(b'))$ as above.