Take the 2-minute tour ×  If we have a Serre fibration $p: E
\rightarrow B$ with fiber of homotopy type $S^{k-1}$, then we
can create a fibration with contractible fiber by first taking
the mapping cylinder $M_p$ of $p$ to get a map $M_p \rightarrow
B$ (not necessarily a fibration) with contractible "fiber" and
then applying the "space of paths" construction to get a
fibration $M_p \times_B B^I \rightarrow B$. My question is, is
this last part of the construction necessary, or is the mapping
cylinder $M_p$ already a Serre fibration?  I tried lifting a
homotopy $f_t: X \times I \rightarrow B$ with starting point
$\tilde f_0: X \rightarrow M_p$ by cutting $X$ into the closed
preimage $C$ of $B \subset M_p$ and the open preimage $U$ of $E
\times [0,1) \subset M_p$. On $C \times I$ we set $\tilde f_t(x)
= f_t(x) \in B \subset M_p$. On $U \times I$ we lift $f_t|U: U
\times I \rightarrow B$ to $g_t: U \times I \rightarrow E$ and
then set $\tilde f_t(x) = (g_t(x),(1-t)(t$ coordinate of $\tilde
f_0(x)) + t(1))$. This