Proving continuity of a map from $X \times X$ to $PX$ Let $X$ be contractible space and $F$ be a homotopy between constant map $c_a$ and $1_X$ ( $F(x,0)=a , F(x,1)=x)$. we define path $s(x,y) : I \rightarrow X$ as follows:
$$\begin{align} 
 s(x,y)(t)  =
 \begin{cases}
F(x,1-2t)  & \text{ $0\leq t \leq \frac{1}{2}$ } \\
F(y,2t-1),  & \text{$\frac{1}{2} \leq t \leq 1$}
\end{cases}
 \end{align}$$
It is easy to show that $s(x,y)$ define a path in $X$ from $x$ to $y$.  
Now we define map $f : X \times X \rightarrow PX$ , $f(x,y)=s(x,y)$.
I want to prove that $f$ is continuous.
Here $PX$ is path space of $X$ with compact-open topology.
for proving i couldn't find some open subset in $X$ for every open subset in $PX$ at arbitrary point.
Can some body help me?
Thanks. 
 A: Reminder: The topology on $PX$ is given by the subbase
$$\mathcal{B} = \{ B( K, W ) : K \subseteq I \text{ is compact}, W \subseteq X \text{ is open} \}$$
where $B(K, W) = \{ \gamma \in PX : \gamma[K] \subseteq W \}$. We have to show that for fixed $B = B(K, W)$, the set 
$$f^{-1}[B] = \{ (x, y) \in X \times X : f(x, y) \in B \}$$ 
is open. 

Let $(x_0, y_0) \in f^{-1}[B]$ and $\gamma = f(x_0, y_0) \in B$. 
By continuity of $F$, the set $G = \{ (z, t) \in X \times I : F(z, t) \in W \}$ is open. Let 
$$\begin{align*}
K_1 & = \left\{ 1-2t : t \in K \cap \left[ 0, \textstyle \frac{1}{2} \right] \right\}, \quad \Gamma_1 = \{ x_0 \} \times K_1 \\
K_2 & = \left\{ 2t-1 : t \in K \cap \left[ \textstyle \frac{1}{2}, 1 \right] \right\}, \quad \Gamma_2 = \{ y_0 \} \times K_2
\end{align*}$$
so that $F[ \Gamma_1 \cup \Gamma_2 ] = \gamma[K] \subseteq W$, hence $\Gamma_1 \cup \Gamma_2 \subseteq G$. 

Every $(x_0, t)$ where $t \in K_1$ has an open neighborhood $(x_0, t) \in U_t \times V_t \subseteq G$. The family $\{ U_t \times V_t : t \in K_1 \}$ is an open cover of $\Gamma_1$. Since it is compact, we have open $U_1, \ldots, U_n \subseteq X, \ V_1, \ldots, V_n \subseteq [0, 1]$ such that 
$$\Gamma_1 \subseteq \bigcup_{k=1}^n U_k \times V_k \subseteq G.$$
Finally consider $\displaystyle U = \bigcap_{k=1}^n U_k, \ V = \bigcup_{k=1}^n V_k$ and note that $\Gamma_1 \subseteq U \times V \subseteq G.$
Do the same for the set $\Gamma_2$ to obtain $U' \subseteq X, \ V' \subseteq [0, 1]$ such that $\Gamma_2 \subseteq U' \times V' \subseteq G$.

It suffices to show that if $x \in U, \ y \in U'$ then $f(x, y) \in B$, hence $(x_0, y_0) \in U \times U' \subseteq f^{-1}[B]$. But it is simply a matter of checking.
