Show that $\phi$ is a fibration.

Question

Show that if

is a pullback square and $p$ is a fibration, then $\phi$ is a fibration. I know the definition of a pullback square and the definition of a fibration, but still, I do not know how to proceed, could anyone help me, please?

Answer 1

There are a number of ways to do this, but this is the most efficient way I'm aware of: Since your square is a pullback square, we can make it more concrete by writing it

$$\require{AMScd}\begin{CD} E\times_BX @>{\pi}>>E \\ @V{\phi}VV @VVV{p} \\ X @>{f}>> B \end{CD} $$

with $E\times_B X = \{(e,x) \in E \times X : p(e)=f(x)\}, \phi(e,x) = x,$ and $\pi(e,x) = e$.

Now suppose we have a homotopy $h: Z \times I \to X$ and a lift $Z \times 0 \cong Z \xrightarrow{\tilde{h}_0} E \times_B X$, with $\phi \circ \tilde{h}_0 (z) = h(z,0)$. Then enlarge the square above to get

$$\require{AMScd}\begin{CD} Z @>{\tilde{h}_0}>>E\times_BX @>{\pi}>> E \\ @VVV @V{\phi}VV @VV{p}V \\ Z\times I @>{h}>> X @>{f}>> B \end{CD} $$

If we define $g = f \circ h: Z \times I \to B$ and $\tilde{g}_0 = \pi \circ \tilde{h}_0: Z \to E$, then it's quick to check that $p \circ \tilde{g}_0 (z) = g(z,0)$ for all $z \in Z$. Then because $p$ is a fibration, there is a homotopy $\tilde{g}: Z\times I \to E$ lifting $g$ against $p$.

From $\tilde{g}$ we should be able to get a homotopy $\tilde{h}: Z \times I \to E\times_BX$ lifting $h$. To define $\tilde{h}(z,t)$ we need a pair $(e,x) \in E \times X$ such that $p(e) = f(x)$. The only course of action that should jump out to us is to get the coordinate in $E$ using $\tilde{g}$ and the coordinate in $X$ using $h$, so let's try $\tilde{h}(z,t) = (\tilde{g}(z,t), h(z,t))$. This is continuous because $\tilde{g}$ and $h$ are.

We need to verify three things: this actually takes values in $E \times_B X$, it lifts $h$ against $\phi$, and lastly that it agrees with $\tilde{h}_0$ on $Z \times 0$.

For the first point, we're supposed to check that $p \circ \tilde{g}(z,t) = f \circ h(z,t)$. But $g = f \circ h$ by definition, and $\tilde{g}$ is lift of $g$ against $p$, so that completes the first verification.

For the second point, we check that $\phi \circ \tilde{h}(z,t) = h(z,t)$. But $\phi$ is projection onto the second factor, so this is immediate.

Finally, for the third point, $\tilde{h}(z,0) = (\tilde{g}(z,0), h(z,0)) = (\pi \circ \tilde{h}_0(z), \phi \circ \tilde{h}_0(z))$ by the definition of $\tilde{g}_0$, the fact that $\tilde{g}$ is a homotopy lifting $g$ against $p$ and extending $\tilde{g}_0,$ and that $\tilde{h}_0$ is a lift of $h$ against $\phi$ on $Z \times 0$. But since $\tilde{h}_0(z) \in E \times_B X$, this equality tells us that $\tilde{h}(z,0)$ and $\tilde{h}_0(z)$ share the same coordinates in $E \times_B X$, so we are done.