Inclusion of space loop is a Serre fibration Let $i: X \longmapsto Y$ be an inclusion, and define $$\Omega_X Y:= \left\lbrace \gamma : I \longmapsto Y : \gamma(1) \in X \right\rbrace$$
Is it true that the map $$\bar{i} : \Omega_x Y \longmapsto Y$$
$$\gamma \longmapsto \gamma(0)$$
Is always a Serre Fibration?
Any solution would be appreciated, don't know where to start due to scarcity of hypothesis.
 A: We recall that a map $f:A\rightarrow B$ is a Hurewicz fibration if and only if it admits a lifting function $W_f\rightarrow A^I$, where $W_f$ is the mapping path space of $f$ and $A^I$ is the space of maps $I\rightarrow A$ in the compact-open topology. See, for instance, Algebraic Topology, E. Spanier, p. 92.
Below I'll work with the following definitions, letting
$$\Omega_XY=\{\ell:I\rightarrow Y\mid \ell(0)\in X\}$$
and $\overline i:\Omega_XY\rightarrow Y$ be defined by $\overline i(\ell)=\ell(1)$. Apologies for being obtuse.
Define the mapping path space $W_{\overline i}$ to be the pullback of the arrows
$$\Omega_XY\xrightarrow{\overline i} Y\xleftarrow{e_0} Y^I$$
where $e_0(m)=m(0)$. Thus
$$W_{\overline i}=\{(m,\ell)\in Y^I\times \Omega_XY\mid \ell(1)=m(0)\}.$$
By adjunction we identify $(\Omega_XY)^I$ with the subspace of $Y^{I^2}$ given by
$$(\Omega_XY)^I\cong\{f:I^2\rightarrow Y\mid f(0,t)\in X\;\forall \;t\in I\}.$$
Then a lifting function for $\overline i$ is a map $\lambda:W\rightarrow (\Omega_XY)^I$ satisfying
$$\lambda(\ell,m)(s,0)=\ell(s),\qquad \lambda(\ell,m)(1,t)=m(t).$$
To this end define
$$\lambda(\ell,m)(s,t)=\begin{cases}\ell((1+t)s)&s\leq\frac{1}{1+t}\\
m((1+t)s-1)&s\geq\frac{1}{1+t}.\end{cases}$$
This is the required lifting function.
