The end-point evaluation map in free path space is homotopy equivalence

Question

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?

Answer 1

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.