# General name for a multidimensional function that maps each coordinate independently

This is a question about terminology. I hope it is not too silly but I haven't been able to find a clear answer. Basically, is there a more or less standard name for a function from an n-dimensional space into another n-dimensional space that maps each coordinate independently?

In formal terms, let's suppose we have two sets $A$ and $B$ and we can define functions mapping one set to the other $f \colon A \to B$. If, for an integer $n > 1$ we take the Cartesian products $A^n$ and $B^n$, then we can define functions $g \colon A^n \to B^n$. One simple form of such a function would be $$g(a_1, \dots, a_n) = (f_1(a_1), \dots, f_n(a_n)),$$ where each $f_i$ is a function from $A$ to $B$. But of course there will also be more general $g$ functions that don't map each coordinate independently. I'm looking for the English words that express this distinction as in $g$ is a 'whatever' function vs $g$ is a 'non-whatever' function.

I am aware of certain particular cases, like the diagonal vs non-diagonal linear maps in a vector space. But I'm wondering if there are any standard names for this distinction in the general case where we simply have sets, Cartesian products, and functions.

Let $n=2$ for simplicity. Assume you are given two maps $f_i : A_i \to B_i$. Then the function you are referring to is the cartesian product $f_1 \times f_2$.

Here we are thinking of $f_i$ as a subset of $A_i \times B_i$:

Definition A function $f : A\to B$ is subset (also called $f$) of $A\times B$, so that (a) for all $a\in A$, $\{ b\in B : (a, b)\in f\}$ is nonempty, and (b) if $(a, b), (a, b')\in f$, then $b= b'$.

Of course the above implies that it is unambiguous to write, for all $a\in A$, $f(a)$ so that $f(a) \in \{ b\in B : (a, b)\in f\}$. So $(a, f(a) ) \in f$.

From the above definition of functions as subsets, we see that $f_1\times f_2$ is a subset of $(A_1\times B_1) \times (A_2 \times B_2)$ and this latter set is identified as $(A_1 \times A_2) \times (B_1 \times B_2)$.

Claim: $f_1 \times f_2$, treated as a subset of $(A_1 \times A_2) \times (B_1 \times B_2)$, is a function $A_1\times A_2 \to B_1\times B_2$ and $$\tag{1}(f_1\times f_2)(a_1, a_2) = (f_1(a_1), f_2(a_2))$$ for all $a_i \in A_i$.

To prove the claim that $f_1\times f_2$ is a function $A_1\times A_2 \to B_1\times B_2$, let $(a_1, a_2) \in A_1\times A_2$. Then first of all, $$(a_i, f_i(a_i)) \in f_i \subset A_i \times B_i$$ for $i=1,2$. So $$\tag{2}(a_1, a_2, f_1(a_1), f_2(a_2)) \in f_1\times f_2.$$

Secondly, if $$(a_1, a_2, b_1, b_2), (a_1, a_2, b'_1, b'_2) \in f_1\times f_2,$$ then $$(a_i, b_i)\in f_i,\ \ \ (a_i, b'_i)\in f_i.$$ This implies $b_i = b'_i$ since $f_i$ is a function. Thus $(b_1, b_2) = (b'_1, b'_2)$ and so $f_1\times f_2$ defines a function. Now (1) follows from (2).

• This does not really answer the question, which is to find a generic name for functions that can be written as a cartesian product of functions. – J.-E. Pin Feb 9 '18 at 5:01
• @J.-E.Pin : From the question it seems to me that the questioner did not realize that those functions are just the cartesian product (If they realized I guess they would not ask this question). – user99914 Feb 9 '18 at 5:15
• @John, thanks for your answer. I have decided to accept it even if, as J.-E. Pin pointed out above, I was looking for a couple of opposing terms rather than an explanation that the "simple form" I mentioned is also a Cartesian product. But I really appreciate the explanation and it is true that this means that I can make the distinction I was looking for by referring to "n-dimensional functions that are a Cartesian product" vs. those "that are not a Cartesian product". While this is wordier than I wanted, I suppose there is no common simpler way of putting it. – Ángel José Riesgo Feb 14 '18 at 9:28