Does $\mathbb{R}^n\times\mathbb{R}^m$ imply $\mathbb{R}^{n+m}$ mapping?

Question

This might be a very basic question, but does the function $$ f:\mathbb{R}^n\times\mathbb{R}^m\to \mathbb{R} $$ necessarily imply the existence of a function $$ g:\mathbb{R}^{n+m}\to\mathbb{R}? $$ If so, is the reverse implication true, in general?

For example, take $n=m=2$, and $g(x_1,x_2,y_1,y_2)=x_1x_2y_1y_2$. What is $f$ in this case? Take $\mathbf{x}=(x_1,x_2)^T$ and $\mathbf{y}=(y_1,y_2)^T$.

If $f$ is not always constructible, is there a name for the class of functions that are?

Edit: If $g(x_1,x_2,y_1,y_2)=x_1y_1+x_2y_2$, for example, we have some type of closed-formula for $f$, given by $f(\mathbf{x},\mathbf{y})=\mathbf{x}\cdot \mathbf{y}$. In general, however, some representations in terms of vectors might not be possible, and that is my main question. Understandably, none of this should really matter once we see it as a function of all vector components. I am, however, particularly curious about the constructability of such functions when "partitioned" to vector-based components, such as $\mathbf{x}$ and $\mathbf{y}$ in the example.

Answer 1

I will approach this question in two ways. $\newcommand{\R}{\mathbb{R}}$

First, let's consider the simple case of $\R^2 \times \R^2$ versus $\R^4$. You want to know whether a function $f : \R^2 \times \R^2 \rightarrow \R$ always corresponds to a function $g : \R^4 \rightarrow \R$, and vice-versa.

• For the first direction of this correspondence, suppose we have a function $f(\textbf{x}, \textbf{y})$, where $\textbf{x}, \textbf{y} \in \R^2$. How do we get a corresponding function $g(z_1, z_2, z_3, z_4)$? Simply note that $(z_1,z_2)$ and $(z_3,z_4)$ are elements of $\R^2$, and let $g(z_1, z_2, z_3, z_4) = f((z_1,z_2), (z_3,z_4))$.

• It seems that you are having more trouble with the reverse direction of the correspondence, where we start with a function $g(z_1,z_2,z_3,z_4)$. How do we define the corresponding function $f(\textbf{x},\textbf{y})$? The key is to realize that $\textbf{x},\textbf{y}$, being elements of $\R^2$, each have two components within themselves. That is, there exist unique numbers $x_1,x_2 \in \R$ such that $\textbf{x} = (x_1,x_2)$, and similarly $\textbf{y} = (y_1,y_2)$. Then, you can define $f$ in terms of these components, rather than directly in terms of $\textbf{x},\textbf{y}$: $$ f((x_1,x_2),(y_1,y_2)) = g(x_1,x_2,y_1,y_2). $$ This is what people do all the time in these situations, and it's mathematically sound. However, you might be unsatisfied by this answer, because I think you really want your formula for $f$ in terms of $\textbf{x},\textbf{y}$ instead of the components. In that case, what you can do is appeal to the component projections maps. These are special maps that just pick out one or the other of the components in $\R^2$ (or $\R^n$): $$ \pi_1 : \R^2 \rightarrow \R, \text{ given by } \pi_1(x_1,x_2) = x_1 \\ \pi_2 : \R^2 \rightarrow \R, \text{ given by } \pi_2(x_1,x_2) = x_2 $$ Then, given your function $g$, you can define $f$ by $$ f(\textbf{x}, \textbf{y}) = g(\pi_1 \textbf{x}, \pi_2 \textbf{x}, \pi_1 \textbf{y}, \pi_2 \textbf{y}). $$ This is really the same thing as what I presented first, but it makes it more clear that this formula really is in terms of $\textbf{x}, \textbf{y}$.

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Now, the more general approach. You want to have a correspondence between functions $f : \R^n \times \R^m \rightarrow \R$ and functions $g : \R^{n+m} \rightarrow \R$. What you need to do is notice that there is an isomorphism between $\R^n \times \R^m$ and $\R^{n+m}$, namely the map $$ h : \R^{n+m} \rightarrow \R^n \times \R^m \text{ given by } \\ h(z_1, z_2, ..., z_n, z_{n+1}, ..., z_{n+m}) = ((z_1, ..., z_n), (z_{n+1}, ..., z_{n+m})). $$ The fact that this map is an isomorphism means that it has an inverse $h^{-1} : \R^n \times \R^m \rightarrow \R^{n+m}$ (I'll let you think about what this inverse is). Then, the correspondence you want is quite simple.

• Given $f : \R^n \times \R^m \rightarrow \R$, it corresponds to $f \circ h : \R^{n+m} \rightarrow \R$;
• Given $g : \R^{n+m} \rightarrow \R$, it corresponds to $g \circ h^{-1} : \R^n \times \R^m \rightarrow \R$.

This is in fact the same correspondence I described above, put more concisely and generally. $\newcommand{\Hom}{\operatorname{Hom}}$

(This can be generalized to the setting of category theory. Indeed, aside from the fact that $X^n \times X^m \cong X^{n+m}$ (which is in fact also true in the setting of abstract categorical products), the above argument can be generalized as follows. For any two isomorphic objects $A,B$, the functions $A \rightarrow C$ are in correspondence with the arrows $B \rightarrow C$. That is, the sets $\Hom(A,C)$ and $\Hom(B,C)$ are isomorphic. This is a result of the fact that $\Hom(-,C)$ is a (contravariant) functor, and functors preserve isomorphism.)