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Let $f,g$ be functions in $C^A$ and $C^B$ respectively.

Let $\boxtimes:C^A \times C^B \to (C\times C)^{A \times B}$ s.t.

$f\boxtimes g(a,b)=(f(a),g(b))$

It seems not the tensor product, nor Cartesian product. Then can we call it direct product? But it seems the term 'direct product' often used on operator between structures.

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up vote 7 down vote accepted

For the viewpoint of category theory, your map is just $f\times g$ -- it is the image of $f$ and $g$ under the product bifunctor $(-)\times(-)$.

More verbosely, if you compose $f$ and $g$ with the projection maps from $A\times B$, then you get maps $f\circ \pi_1: A\times B \to C$ and $g\circ \pi_2: A\times B \to C$, which factor through $C\times C$ by the universal property of the latter. The mediating morphism is excatly $f\times g$.

If you consider $\mathbf{Set}$ to be a monoidal category by declaring $\times$ to be $\otimes$, then $f\times g$ is indeed the tensor product $f\otimes g$.

On the other hand, in ordinary set theory, we usually identify a function with its graph: $$f=\{\langle x,f(x)\rangle\mid f(x)\text{ is defined}\},$$ and in that sense your $f\boxtimes g$ is of course not the cartesian product of $f$ and $g$. It is closely related though: $$ f\boxtimes g = \{ \langle\langle a,b\rangle,\langle c,d\rangle\rangle \mid \langle\langle a,c\rangle,\langle b,d\rangle\rangle \in f\times g\}$$ which could be seen as a more vivid justification for Hurkyl's "transpose" terminology.

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I understand. Thank you very much. – Popopo Jun 14 '12 at 5:34

Your map is some transpose of the Cartesian product.

Given a function $f : A \to B$ and another function $g:C \to D$, their Cartesian product is

$$ f \times g : A \times C \to B \times D : (a,c) \mapsto (f(a), g(c)) $$

The notion of "transpose" enters the picture in the sense that you are switching from thinking about a function $A \to C$ to thinking of an element of $C^A$.

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So does the Cartesian product on functions is a bit different from it on relations though all functions are also binary relations? – Popopo Jun 9 '12 at 14:42
And thank you for your answer. – Popopo Jun 14 '12 at 15:05
@Popopo: Sorry, I forgot about your question. When I talk about the Cartesian product of functions, I think in the categorical sense: the product (bi)functor on the category of sets and functions. I don't know what you have in mind for the Cartesian product of relations, but when I hear the phrase, I also think of the (bi)functor on the category of sets and binary relations. Alas, I don't think in terms of this category often, so I don't know what the product actually is on it to be able to compare them. – Hurkyl Jun 14 '12 at 16:37

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