Let $X$ be a topological space and $x_0 \in X.$ Consider the based path space $$\mathcal P_{x_0} (X) : = \left \{\gamma : [0,1] \xrightarrow{\text {continuous}} X\ \bigg |\ \gamma (0) = x_0 \right \}$$ of $X$ at $x_0$ equipped with the compact open topology. Then we get a deformation retract $H : \mathcal P_{x_0} (X) \times [0,1] \longrightarrow \mathcal P_{x_0} (X)$ between $P_{x_0} (X)$ and the constant path $c_{x_0}$ at $x_0$ given by $$H (\gamma, s) = \gamma ((1 - s) \cdot) : t \mapsto \gamma ((1 - s) t).$$ But I am having hard time to show that this map is continuous. In order to show that $H$ is continuous it is enough to show that the inverse image of subbasic open sets in $\mathcal P_{x_0} (X)$ with respect to compact open topology under the map $H$ is open. Any such subbasic open set is of the form $U_{K, V} : =\left \{\gamma \in \mathcal P_{x_0} (X)\ \big |\ \gamma (K) \subseteq V \right \},$ where $K \subseteq [0,1]$ is compact and $V \subseteq X$ is open. Now what will be $H^{-1} (U_{K, V})\ $? It contains all those $(\gamma, s) \in \mathcal P_{x_0} (X) \times [0,1]$ such that $\gamma ((1 - s) K) \subseteq V$ i.e. $\gamma \in U_{(1 - s)K, V}$ and this implies $$H^{-1} (U_{K, V}) = \bigcup\limits_{s \in [0,1]} U_{(1 - s) K, V} \times \{s \}.$$ But why is it open in the product topology?
Any suggestion in this regard would be greatly appreciated. Thanks for investing your valuable time on my question.
Fun Fact $1$ $:$ Let $X$ and $Y$ be topological spaces with $Y$ locally compact. Then the evaluation map $ev : Y^X \times X \longrightarrow Y$ is continuous, where $Y^X$ is the set of all continuous maps from $X$ to $Y.$
Fun Fact $2$ $:$ Let $X, Y$ and $Z$ be topological spaces. Then any continuous map $f : X \times Y \longrightarrow Z$ induces a continuous map $F : X \longrightarrow Z^Y.$ The converse is true if $X$ is locally compact which follows from the above theorem just by observing that given $F : X \longrightarrow Z^Y,$ the induced map $f : X \times Y \xrightarrow{} Z$ is obtained via the following composition $:$
$$X \times Y \xrightarrow{F \times \text{id}} Z^Y \times Y \xrightarrow{ev} Z$$
Why does the required result follow from the above two facts? $:$
In our problem replace $X$ by $\mathcal P_{x_0} (X) \times I,$ $Y$ by $I$ and $Z$ by $X.$ Since $I$ is locally compact it follows from the above facts that the continuity of the map $H : \mathcal P_{x_0} (X) \times I \longrightarrow \mathcal P_{x_0} (X) \subseteq X^I$ defined by $$(\gamma, s) \longrightarrow \gamma ((1 - s) \cdot) : t \mapsto \gamma ((1 - s)t)$$ is equivalent to the continuity of the map $\widetilde H : \mathcal P_{x_0} (X) \times I \times I \longrightarrow X$ defined by $(\gamma, s, t) \mapsto \gamma ((1 - s) t).$ The later map is continuous as it is the composition of the following two continuous maps $:$
$$\mathcal P_{x_0} (X) \times I \times I \xrightarrow{\text {id} \times \delta} \mathcal P_{x_0} (X) \times I \xrightarrow{ev\ \big \rvert_{\mathcal P_{x_0} (X) \times I}} X$$ where $\delta : I \times I \longrightarrow I$ is the continuous map defined by $(s,t) \mapsto (1 - s) t.$
Hence the result follows. $\blacksquare$