$f:X\times Y\rightarrow Z$ continuous then $f_y:X \rightarrow Z$ continuous $X,Y,Z$ are topological spaces, and given any $y \in Y$, $f_y(x)=f(x,y)$
I tried taking an open set $U_Z \subset Z$. Since $f$ continuous, $f^{-1}(U_Z) \subset X \times Y$ is open. Since the projection $p_X:X \times Y\rightarrow X$ is open, $p_X(f^{-1}(U_Z)) \subset X$ is open. I want to show that $f_y^{-1}(U_Z) \subset X$ is open. I know that $f_y^{-1}(U_Z)\subset p_X(f^{-1}(U_Z))$. If $f_y^{-1}(U_Z)$ is open as a subspace of $p_X(f^{-1}(U_Z))$, then this should do it. But, given any $x \in f_y^{-1}(U_Z)$, I'm not being able to find an open set $x \in V \subset p_X(f^{-1}(U_Z))$ such that $V\subset f_y^{-1}(U_Z)$
 A: For fixed $y\in Y,\ f_y:X\to Z: x\mapsto f(x,y).$ Fix $x_0\in X$ and suppose $f_y(x_0)\in V\subseteq Z$ is open in $Z$. Using continuity of $f$, we can find an open basic set $(x_0,y)\in U_X\times U_Y$ in $X\times Y$ such that $f(U_X\times U_Y)\subseteq V.$ But then, $f_y(U_X)\subseteq V,$ so $f_y$ is continuous at $x_0.$
A: Let $y\in Y$ be fixed. Let $i_{y}:X\rightarrow X\times Y$ be defined
by $i_{y}(x)=(x,y)$. We to go show that $i_{y}$ is continuous. Recall a fact: Let $Z_{1},Z_{2}$ be topological spaces, $\theta:Z_{1}\rightarrow Z_{2}$
be a map. If there exists subbase $\mathcal{SB}$ for the topology
of $Z_{2}$ such that $\theta^{-1}(U)$ is open for each $U\in\mathcal{SB}$,
then $\theta$ is continuous.
For our case, $\mathcal{SB}=\{U\times Y\mid U\mbox{ is an open subset of }X\}\cup\{X\times V\mid V\mbox{ is an open subset of }Y\}$
is a subbase for the product topology on $X\times Y$. By direct
verification, it is clear that $i_{y}^{-1}(A)$ is open for each $A\in\mathcal{SB}$.
Therefore $i_{y}$ is continuous.
Finally, $f_{y}=f\circ i_{y}$, so $f_{y}:X\rightarrow Z$ is continuous.
