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 defines a continuous lift on $C$ and on $U$ separately. If the continuous lift on $U$ extends to the closure of $U$ then we're done. The map $U \rightarrow E$ could be nasty though near the boundary of $U$. Perhaps a better approach is to first construct a map from $X \times I$ that is only "close to" a lift, then use obstruction theory (I'm not an expert on this) to show that it is homotopic to some lift.