Bijection between continuous maps $C(X\times A,Y)$ and continuous maps $C(X,Y^A)$ 
If $A$ is a discrete space, and $X,Y$ are any spaces, I want to show there is a bijection between the sets of continuous maps $C(X\times A,Y)$ and $C(X,Y^A)$. Here, $Y^A$ is the set of all continuous maps from $A$ to $Y$ (which is all maps from $A$ to $Y$ since $A$ is discrete), and we topologize $Y^A$ with the product topology.

Here is what I've tried.
Define the map $\Psi:C(X\times A,Y)\to C(X,Y^A)$ via $f\mapsto\Psi(f)$ where for each $x\in X$ the map $\Psi(f)(x)$ is defined by $(\Psi(f)(x))(a)=f(x,a)$. It was easy to show that $\Psi$ is injective. I'm stuck on surjective. My intuition was to pick $h\in C(X,Y^A)$ and define the map $\tilde{h}:X\times A\to Y$ via $(x,a)\mapsto h(x)(a)$. Clearly $\Psi(\tilde{h})=h$, so I'm pretty sure this is what I want to show surjectivity, but I cannot figure out why $\tilde{h}$ is continuous. I tried picking an open $U$ in $Y$ and showing $\tilde{h}^{-1}(U)$ is open in $X\times A$, but all I can say about this is that $U^A$ is open in $Y^A$, so $h^{-1}(U^A)$ is open in $X$. I'm not sure where to go from here.
Maybe I can use the fact that since $A$ is discrete, we have a homeomorphism between the cartesian product and the coproduct $X\times A\cong\coprod_AX$? Any suggestions? By the way, I want to show this directly, i.e. by exhibiting an actual bijection, rather than hitting it with any sledgehammers.
 A: You give $Y^A$ the product topology, i.e. you take $Y^A = \prod_{a \in A} Y_a$ with $Y_a = Y$. As a set, the product $\prod_{a \in A} Y_a$ is defined as the set of all functions $\varphi : A \to Y$ between the sets $A, Y$. But since $A$ is discrete, the set of all these functions is nothing else than the sets all continuous maps $f : A \to Y$ between the spaces $A, Y$.
Let $p_a : Y^A \to Y$ denote the projection on the factor $Y = Y_a$. It is given by $p_a(f) = f(a)$.
The universal property of the product says that a function $h : X \to Y^A$ is continuous iff all $p_a \circ h : X \to Y$ are continuous. Note that $(p_a \circ h)(x) = p_a(h(x)) = h(x)(a)$.
So let $h : X \to Y^A$ be continuous and $U \subset Y$ be open. Then all $(p_a \circ h)^{-1}(U)$ are open in $X$ and, because $A$ is discrete, all $(p_a \circ h)^{-1}(U) \times \{a\}$ are open in $X \times A$. Thus
$$(\overline h)^{-1}(U) = \{(x,a) \in X \times A \mid \overline h(x,a) = h(x)(a)  = (p_a \circ h)(x) \in U \} \\ = \{(x,a) \in X \times A \mid x \in (p_a \circ h)^{-1}(U) \}  \\ = \bigcup_{a \in A} (p_a \circ h)^{-1}(U) \times \{a\}$$
which is open in $X \times A$.
To get a more general picture, I recommend to learn about the compact-open topology on arbitrary function spaces $Y^Z$ = set of all continous maps $f : Z \to Y$ . You will see that if $A$ is discrete, then the compact-open topology on $Y^A$ agrees with the product topology. Moreover, if $Z$ is locally compact (note that discrete spaces have this property), then you have a canonical bijection
$$e : C(X \times Z,Y) \to C(X,Y^Z), e(\phi)(x)(z) = \phi(x,z) .$$
This is known as the exponential map. In this forum you will find a lot of questions about it.
