Given $Y^I$ the free path space and a map $p:Y^I\to Y, \omega \mapsto \omega (1)$ it is to show that $p$ is a homotopy equivalence. I suppose this requires to show that the map $p$ is in fact a fibration?
The end-point evaluation map in free path space is homotopy equivalence
algebraic-topologyfibrationhomotopy-theory
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This is a very general fact about model categories and homotopy pullbacks, as evidenced by Zhen Lin's comment. It's also proven as a special case of Proposition 4.65 in Hatcher's book. Let me nevertheless spell out the argument precisely for topological spaces.
Define $E_p = \{ (y, \gamma) \in E \times B^{[0,1]} \mid p(y) = \gamma(0) \}$. There's a map (in fact a fibration) $q : E_p \to B$, $(y,\gamma) \mapsto \gamma(1)$, and the homotopy fiber is the fiber $F_p = q^{-1}(b_0)$. The inclusion $$i : F = p^{-1}(b_0) \to F_p$$ is given by $i(y) = (y, \mathrm{cst}_{b_0})$.
Define a homotopy $g_t : E_p \to B$, $(y,\gamma) \mapsto \gamma(t)$. Then $g_0(y, \gamma) = \gamma(0) = p(y)$, so $g_0$ lifts through $p$ by $\bar{g}_0 : E_p \to E$, $\bar{g}_0(y,\gamma) = y$ (i.e. $p \circ \bar{g}_0 = g_0$). Because $E \to B$ is a fibration, by the homotopy lifting property, there is a full lift $\bar{g}_t : E_p \to E$ of $g_t$ through $p$. In other words, $\bar{g}_t$ satisfies the following equation: $$p(\bar{g}_t(y,\gamma)) = \gamma(t).$$
Now restrict everything to the fibers: let $h_t : F_p \to F_p$ be given by $h_t(y,\gamma) = \bigl(\bar{g}_t(y,\gamma), \gamma_{\mid [t,1]} \bigr)$ (because of the previous equation, this is in $F_p$). Then $h_0$ is the identity, whereas $h_1(y,\gamma) = (\bar{g}_1(y,\gamma), \mathrm{cst}_{b_0})$ is in the image of $i : F \to F_p$. And now that $h_t$ is a homotopy between $ih_1$ and the identity, while the restriction of $h_t$ is a homotopy between $h_1i$ and the identity. Thus $F$ and $F_p$ are homotopy equivalent.
Final remark: it's much simpler to prove that $F$ and $F_p$ are weakly homotopy equivalent, because the map $i$ induces easily an isomorphism on all homotopy groups. The square $$\require{AMScd} \begin{CD} E @>>> E_p \\ @VVV @VVV \\ B @>>> B \end{CD}$$ induces a map between the long exact sequences of the respective fibrations: $$\begin{CD} \dots @>>> \pi_n(F) @>>> \pi_n(E) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots \\ @. @VVV @VVV @VVV @VVV @. \\ \dots @>>> \pi_n(F_p) @>>> \pi_n(E_p) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots \end{CD}$$ Since $E$ and $E_p$ are homotopy weakly homotopy equivalent ($PB$ is contractible), and so by the five lemma and induction, the maps $\pi_n(F) \to \pi_n(F_p)$ are isomorphisms.
I'm going to expand on my comments and given a more complete answer.
So far we have that $\newcommand\hofib{\operatorname{hofib}}\hofib(f)\to X$ is a Hurewicz fibration, because it's the pullback of a Hurewicz fibration. Moreover its fiber $F$ is the same as the fiber of $PY\to Y$. This is a general fact that follows from pullback pasting.
Consider the following diagram where $x\in X$ is a point, and both squares are pullbacks. $$ \require{AMScd} \begin{CD} F @>>> f^*P @>>> P\\ @VVV @VVV @VVV\\ x @>>> X @>>> Y \end{CD} $$ In this case $F$ is the fiber of $f^*P$ over $x$. However the outer square is also a pullback, by pullback pasting, so $F$ is also the fiber of $P$ over $f(x)$.
So the fiber of $\hofib(f)\to X$ is $\Omega Y$. All we need to do now is show that the fiber of a Hurewicz fibration is weakly equivalent to its homotopy fiber, where the homotopy fiber is defined by pulling back the path space of the codomain to the domain.
Let's assume then that $f:X\to Y$ is some general Hurewicz fibration (of pointed spaces). Let $F=f^{-1}(*)$ be the point-set fiber of $f$. Let $G = \hofib(f) = f^* PY$. We want to produce a homotopy equivalence between $F$ and $G$. For $PY$ I'm going to assume that we take paths that end at $*$ to fix a convention. Concretely then as a set, $G$ is the set of pairs $(x, \alpha)$ where $x\in X$ and $\alpha : I\to Y$ satisfies $\alpha(0)=f(x)$ and $\alpha(1)=*$.
First of all, we always have a map $i:F\to G$, given by $x\mapsto (x,*)$, where $*\in PY$ denotes the constant path at the base point. To get a map $G\to F$ we need to use the fact that $f$ is a Hurewicz fibration. Consider the square $$ \begin{CD} G @>\pi_1>> X \\ @VG\times \{0\}VV @VVfV \\ G\times I @>>\operatorname{ev}> Y, \end{CD} $$ where the bottom map is $(x,\alpha,t)\mapsto \alpha(t)$. The fact that this square commutes is the fact that if $(x,\alpha)\in G$, then $f(x) = \alpha(0)$.
Thus there exists a lift $H:G\times I \to X$. $H_0=\pi_1$, and $f\circ H_1 = *$, since $\alpha(1)=*$ for all $(x,\alpha)\in G$. Thus $H_1$ actually lands in the fiber, $F$. For convenience let's let $h=H_1$. Let's keep in mind for later use that $H$ is a homotopy between $\pi_1$ and $h$.
We now have $i:F\to G$ and $h:G\to F$. We just need to check that these maps are homotopy inverses of each other. Now $h\circ i$ is homotopic to $\pi_1\circ i$, but $\pi_1\circ i = 1_F$ by definition of $i$. For the other composition $i\circ h$, consider some element $(x,\alpha)\in G$. $h(x,\alpha) = x'$ for some $x'\in F$, and $H(x,\alpha,-)$ is a path from $x$ to $x'$ in $X$ lifting $\alpha$. This gives a path in $G$ from $(x', *)$ to $(x,\alpha)$ by $t\mapsto (H(x,\alpha,t), s\mapsto \alpha(t+s)$ (extend $\alpha$ so that $\alpha(t+s)=*$ when $t+s>1$). But this is a homotopy from $1_G$ to $i\circ h$: $$ (g,t)\mapsto \left(H(g,t), s\mapsto \operatorname{ev}(g,t+s)\right), $$ as desired.
Best Answer
You can directly show that this is a homotopy equivalence. The proposed inverse will be $s: Y \longrightarrow Y^I$ sending $y \mapsto k_y$, the constant path at $y$. One direction is clear: $p(s(y)) = k_y(1) = y$.
For the other direction, we want to show that $s \circ p: Y^I \longrightarrow Y^I$ is homotopic to the identity. Note that $s(p(\gamma)) = k_{\gamma(1)}$. The idea is to use the fact that $[0, 1]$ is homotopy equivalent to $\{1\}$. Now, we define the homotopy $H: Y^I \times I \longrightarrow Y^I$ via $(\gamma, t) \mapsto (s \mapsto \gamma((1 - t)s + t))$. At $t = 0$ this is the identity, and at $t=1$ this is $s \circ p$.
We now need to show that $H$ is continuous. I'll mainly be using the adjoint property of the compact-open topology. See Tammo tom Dieck's Algebraic Topology section 2.4 for details. Anyways, this adjoint property says that showing $H$ is continuous is equivalent to showing $H^\wedge: Y^I \longrightarrow (Y^I)^I$ via $\gamma \mapsto (t \mapsto H(\gamma, t))$ is continuous. Furthermore, there is a homeomorphism $\alpha: Y^{I \times I} \longrightarrow (Y^I)^I$ via $f \mapsto (t \mapsto (s \mapsto f(s, t)))$. The inverse is given by $g \mapsto ((t, s) \mapsto g(t)(s))$. As this is a homeomorphism, $H^\wedge$ is continuous iff its composition with this inverse map is continuous. Tracing out the definition, this composition $Y^I \xrightarrow{H^\wedge} (Y^I)^I \xrightarrow{\alpha^{-1}} Y^{I \times I}$ is as follows:
$$ \begin{align*} \gamma &\mapsto (t \mapsto H(\gamma, t)) \\ &= (t \mapsto (s \mapsto \gamma((1 - t)s + t))) \\ &\mapsto ((t, s) \mapsto \gamma((1 - t)s + t)). \end{align*} $$
In other words, if we let $f: I \times I \longrightarrow I$ be given by $f(t, s) = (1-t)s + t$ (which witnesses the homotopy equivalence $I \simeq \{1\}$), then $(\alpha^{-1} \circ H^\wedge)(\gamma) = \gamma \circ f$. Because $f$ is continuous, the composition map $\gamma \mapsto \gamma \circ f$ is continuous in the compact-open topology. Thus, $H^\wedge \circ \alpha^{-1}$ is continuous, so $H^\wedge$ is continuous, and finally $H$ itself is continuous. We have shown $p \circ s = id$ and $s \circ p \simeq id$, so $p$ is indeed a homotopy equivalence.