TITLE: In a fibration, can a deformation retraction of the base be lifted to the total space?
QUESTION [10 upvotes]: Given a fibration $p:E \rightarrow B$ and if $A$ is a deformation retract of $B$. Is it true that $p^{-1}(A)$ is a deformation retract of $E$?. If this is not true, can some conditions be imposed on $p$, $E$ or $B$ to make that statement true?. If so, the theorem is still valid for a strong deformation retract?

REPLY [4 votes]: There is another situation when we can lift a strong deformation retraction
(SDR). If your fibration $p:E\to B$ is a discrete covering ($\pi^{-1}(b)$
has the discrete topology for every $b\in B$), then no additional
assumptions are needed.
Indeed, by definition, the fibration $p$ has the homotopy lifting
property from arbitrary spaces. To be concrete, in the diagram, we
can lift the the bottom map $H$ to the dashed arrow $\tilde{H}$,
so that $H\circ(id_{I}\times p)=p\circ\tilde{H}$:
$$
\begin{eqnarray*}
I\times E & \stackrel{\tilde{H}}{\dashrightarrow} & E\\
id_{I}\times p\downarrow\quad &  & \downarrow p\\
I\times B & \stackrel{H}{\to} & B
\end{eqnarray*}
$$
In general, $\tilde{H}$ is only a weak retraction. However, when
the fibers of $p$ are discrete, and $H$ is an SDR to $A\subset B$,
then $\tilde{H}$ also an SDR from $p^{-1}(B)=E$ to $p^{-1}(A)$.
To prove this, using $H(t,a)=a$ for
all $t$, and all $a\in A\subset B$, we get:
$$
\tilde{H}(t,c)\in p^{-1}(H\circ(id_{I}\times p)(t,c))=p^{-1}(H(t,p(c)))=p^{-1}(p(c)).
$$
Hence, $\tilde{H}(t,c)=c$, for all $c\in p^{-1}(a)$, as wanted, by discreteness
and continuity.