```
X ---h--> X'
\ /
pi pi'
\ /
v v
S
```

The product of these objects is given by the fiber product or pullback:
$X \times_S Y \equiv \{ (x, y) \in X \times Y: \pi_X(x) = \pi_Y(y) \}$

along with the map $\pi_{X \times Y}: X \times_Y \rightarrow S; \pi((x, y)) \equiv \pi_X(x)$.
See that for consistenty, we could also have defined this as $\pi((x, y)) \equiv \pi_Y(y)$.
Since our condition is that $\pi_X(x) = \pi_Y(y)$, it all works out.
Said differently, we consider the product of fibers over the same base-point.
```
E
pi|
v
B'-f-> B
```

Given this, we would like to pullback the bundle $E$ along $f$ to get a new
bundle over $B'$.This is defined by:
$E'_f \equiv \{ (b', e) : f(b') = \pi(e) \} \subseteq B' \times E$

This is equipped with the subspace topology. We have the projection map
$pi': E'_f \rightarrow B$, $\pi'((b', e)) \equiv b'$. The projection into
the second factor gives a map $h: E'_f \rightarrow E$, $h((b', e)) \equiv e$.
This makes the obvious diagram commute:
```
E' -h-> E
|pi' |pi
B' -f-> B
```

Any section $\sigma: B \rightarrow E$ of $E$ induces a section of $E'$
$\sigma': B' \rightarrow E'$, by producing the function (given as a relation):
$\begin{aligned}
\sigma': B' \rightarrow E' \\
\sigma(b') \equiv (b', \sigma(f(b')) \in_? E' \simeq B' \times E
\end{aligned}$

This has codomain $E'$. To check, if $(b', \sigma(f(b'))$ is in $E'$,
we need $f(b') = \pi(\sigma(f(b'))$. But this is true since $\sigma$
is a section, and thus $\pi(\sigma(f(b')) = f(b')$.
Moreover, we need to check that $\sigma'$ is indeed a section of $B'$. For this,
we need to check that $\pi'(\sigma'(b')) = b'$. Chasing definitions, we find
that this is:
$\begin{aligned}
&\pi'(\sigma'(b'))
&= \pi'(b', \sigma(f(b')))
&= b'
\end{aligned}$

Hence we are done, we have indeed produced a legitimate section.