I need to find an isometry between $\mathbb{B}(X \times Y; M)$ and $\mathbb{B}(X;\mathbb{B}(Y; M))$ where $\mathbb{B}(X; M)$ is the space of bounded functions of X to M. X and Y are any sets and M is a metric space.
My first move was to define $\phi: \mathbb{B}(X \times Y; M) \rightarrow \mathbb{B}(X;\mathbb{B}(Y;M))$ such that $$\phi(f) = \psi_{f} $$ and define $\psi_{f}: X \rightarrow \mathbb{B}(Y;M)$. But here is the problem, I cant see how can I map $X$ into $\mathbb{B}(Y;M)$.
Any help is appreciated!
Notice that to define $\psi_f : X \rightarrow \mathbb{B} \left( Y, M \right)$, you have to define for each $x \in X$, a function $\psi_f \left( x \right) : Y \rightarrow M$, which has to be bounded.
Now, to achieve this, we have to use things which we already know (which are given to us). Even before we start defining $\psi_f$, we have fixed a function $f: X \times Y \rightarrow M$. So, the natural thing to do here is to use this function.
For a fixed $x \in X$, it is easy to see that $f \left( x, y \right)$ is bounded as $y \in Y$ varies. Therefore, the most natural vay to define $\psi_f$ is
$$\psi_f \left( x \right) \left( y \right) = f \left( x, y \right).$$
Now, I suppose the other parts of the result about isometry will follow easily.