Why is the Mapping Cylinder a fibering? If $p:E \rightarrow B$ is a fibration with fiber $F$, in other words we have the fiber space $(E,B,F,p)$. Let $M_p$ be the mapping cylinder of $p$, $M_p = ((E\times[0,1]) \sqcup B)/\sim  $, in which $(x,0)\sim p(x)$. Then we can define the map $$\overline{p}:M_p \rightarrow B$$ given by $\overline{p}([x,t])=p(x)$ for $(x,t)\in E\times [0,1]$, and $\overline{p}(b)=b$ for $b\in B$.
My question is , how to show that $\overline{p}$ is a fibration with fiber $C(F)$ (the cone of $F$, i.e., $C(F)=F\times[0,1]/\sim$ with $(x,0)\sim(y,0)$ for all $x,y\in F$).
I tried to show that $\overline{p}$ satisfies the homotopy lifting property, but had no luck in defining the lifting homotopy in a general case. I think that in some point I have to use the fact that $p$ itself is a fibration, but I don't see how to use this fact. Also I have tried to use the fact that $M_f$ strong deformation retracts onto $B$, but again had no luck in showing that $\overline{p}$ is a fibration.
Edit: I think I understand why the fiber would be $C(F)$, if we simply annalyse $\overline{p}^{-1}(b)$ for $b\in B$, we get that it must be equal to $((p^{-1}(B)\times[0,1])\sqcup \{b\})/\sim$ with $(x,0)\sim b = p(x)$ for all $x$, thus the fiber is $C(p^{-1}(B)) = C(F)$. It remains to show that $\overline{p}$ is in fact a fibration.
 A: Here is a partial answer which makes use of some additional assumptions. We'll assume


*

*$p:E\rightarrow B$ is a locally trivial map with typical fibre $F$.

*$B$ is locally compact Hausdorff.


Under these assumptions we'll prove that the projection $q:M_p\rightarrow B$ is locally trivial.
The key point to notice is that if $X,Y$ are spaces and $X$ is locally compact, then
\begin{array}{ccc}
X\times Y &\xrightarrow{in_0} & X\times Y\times I \\
pr_X\downarrow & &\ \downarrow \\
X &\rightarrow& X\times CY
\end{array}
is a pushout diagram, where $CY=(Y\times I)/(Y\times\{0\})$. This is easy to see because the square is obtained by applying the functor $X\times(-)$ to the pushout diagram defining $CY$. The functor $X\times(-)$, as a left adjoint, preserves all colimits, and in particular preserves the pushout.
Now take the map $p$ and form the mapping cylinder $M_p$ as the pushout in the square
\begin{array}{ccc}
E &\xrightarrow{in_0} & E\times I \\
p\downarrow & &\ \downarrow \\
B &\rightarrow& M_p.
\end{array}
Our assumptions that $p$ is locally trivial and that $B$ is locally compact means that any $x\in B$ has a compact neighbourhood $K\subseteq B$ over which $p$ is trivial. Because $B$ is Hausdorff, $K$ is compact Hausdorff, and hence locally compact. Thus we have a pushout
\begin{array}{ccc}
K\times F &\xrightarrow{in_0} & K\times F\times I \\
pr_K\downarrow & &\ \downarrow \\
K &\rightarrow& K\times CF.
\end{array}
which induces a map $\varphi:K\times CF\rightarrow M_p$ factoring the projection $K\times CF\rightarrow K\subseteq B$.
Let $\overline p:M_p\rightarrow B$ be the canonical map. The inverse image $\overline p^{-1}(K)$ is the pushout of $p^{-1}(K)\cap B\cong K$ against $p^{-1}(K)\times I\cong K\times F\times I$. This is exactly $K\times CF$, and we conclude that $\varphi$ is a homeomorphism onto its image in $M_p$.
Thus we have produced a chart for $\overline p$ in a neighbourhood of any given point $x\in B$. Our conclusion follows: the map $\overline p:M_p\rightarrow B$ is locally trivial.
Discussion: For our statements $p$ need not be a fibation. Even if $p$ is a fibration, then we cannot conclude that $\overline p$ is a fibration. If we have the additional assumption that $B$ is paracompact, then both $p$ and $\overline p$ are necessarily fibrations. This will be the case, for instance, if $B$ is a manifold or CW complex.
I do not know if the result you have asked for holds in the generality that $p$ is a fibration. What is true is that if $f:X\rightarrow Y$ is any map, then the projection $M_f\rightarrow Y$ is a Dold fibration. That is, $f$ is homotopy equivalent to a fibration in the category $Top_Y$ of spaces over $Y$. This means that $f$ will satisfy the weak homomotopy lifting property. The map I suggest in the comments is an example of a Dold fibration. It is not always possible to lift homotopies exactly, but for any $\epsilon>0$ a solution can be found for the cylinder $[-\epsilon,1]$ which lifts the required homotopy over $[0,1]$.
Thus $\overline p$ will aways be a Dold fibration, but there has been nothing gained by the assumption that $p$ is a fibration. If you are interested in producing a counterexample to see that $\overline p$ may fail to be an actual fibration, then here is a strategy. The map $\overline p$ is a homotopy equivalence. Thus for it to be a fibration it is necessary and sufficient that a diagonal filler $X\rightarrow M_p$ can be found for any strictly commuting diagram
\begin{array}{ccc}
A &\xrightarrow{f} & M_p \\
\downarrow & &\ \downarrow \overline p \\
X &\xrightarrow{g}& B
\end{array}
in which $A\subseteq X$ is a closed cofibration. I would suggest to take $B=I$ and $A=\{1\}$, with $f$ mapping to a point in $E\times \{1\}\subseteq M_p$, and $g$ being a path in $B$. The idea is that any diagonal map would need to 'jump' over the whole interval in $M_p$ and would be discontinuous. I'm afraid I don't have time right now to try and make this rigourous. I'll remark again that I don't know the full answer to your query.
