show that the path space fibration is a fibration I am reading Rational Homotopy Theory (Felix, Halperin & Thomas). They say it follows easily from the exponential law that for some topological k-space $X$
$$\Omega X\hookrightarrow_i PX\to_p X$$ is a fibration where the last map is defined by $\gamma\to\gamma(0)$.
I cannot do it however. Given any k-space $Z$, continuous maps $f:Z\times \{0\}\to PX$ and $g:Z\times I\to X$ such that the $gi=pf$ 
we have to find a map $K:Z\times I\to PX$ such that $Ki=f$ and $pK=g$. Could someone help me define $K$?
 A: First of all the path space fibration map should be defined as $\pi: P(X) \rightarrow X$ $\pi(\alpha) = \alpha(1)$. Switzer books on algebraic topology has a straightforward proof of this fact. My thesis also contains a proof but that has not been published yet. Here is a rough proof, I havent edited it yet, I will let you fill in the details.
Let $(X,x_0)\in{\ptop}$, define the space $$P(X) = (X,x_0)^{(I,0)}.$$ equipped with the compact open topology. This is space of paths in $X$ starting at $x_0$, give this space basepoint $\omega_0$, the constant path at $x_0$. We define the map $\pi: (P(X),\omega_0) \rightarrow (X,x_0)$ by $\pi(\alpha) = \alpha(1)$ for $\alpha\in{P(X)}$, we call this map the \emph{space path fibration}. We will show that $\pi$ is a fibration with fiber $\Omega X$.
\medskip
Let $(Z,z_0)\in{\ptop}$ and $g_t:(Z,z_0) \rightarrow (X,x_0)$ be a homotopy and $h_0: (Z,z_0) \rightarrow (PX,\omega_0)$ be a map such that $g_0 = \pi \circ h_0$. Define $H': (Z \times I \times I, (z_0,0,0) \rightarrow (X,x_0)$ by
\begin{equation*}
H'(z,t,s) =\begin{cases}
        h_0(z)(s(t+1))  \quad &\text{if} \ 0 \leq s \leq \frac{1}{t+1} \\
         g_{s(t+1)-1}(z)  \quad &\text{if} \, \frac{1}{t+1} \leq s \leq 1.\\
     \end{cases}
\end{equation*}
This is continuous and indeed pointed. This defines a map $H: Z \times I \rightarrow X^I$ by $H(z,t)(s) = H'(z,t,s)$. We have $H(z,t)(0) = H(z,t,0) = h_0(z)(0) = x_0$ so $H: Z \times I \rightarrow PX$. Moreover $H(z_0,t) = \omega_0$ so this collapses to a map $h_t:(Z,z_0) \rightarrow (PX,\omega_0)$. Finally $H(z,0)(s) = h_0(z)(s)$ and so $H$ is a homotopy of $h_0$. Lastly $\pi \circ h_t = g_t$. Therefore this is the desired homotopy lift.
\medskip
Lastly $\pi^{-1}(x_0) = \{ \alpha\in{PX} \mid \alpha(1) = x_0\} = \Omega X$.
