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What is the name for the operation from mappings $f:X\to Y_1$ and $g:X\to Y_2$ with the same domain to mapping $h:X\to Y_1 \times Y_2$ defined as $h(x)=[f(x), g(x)]$? I named it pairing, but it is unlikely others will call it the same way.

More generally given $f_i:X\to Y_i, i \in I$, $h:X\to \prod_i Y_i$ is defined as $h(x):=(f_i(x))_{i \in I}$.


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You "map them to the product space." So you could call it "product mapping." it is known to be continuous if the components are. – gnometorule Feb 1 '13 at 18:06

It appear clear that $h(x) = [f(x),g(x)]$ should probably be written: $h(x) = \left(f(x), g(x)\right)$ - and called an ordered pair, with range being the "ordered" Cartesian product of the ranges/images of the two functions involved.

In the general case, we'd have an ordered "$n$-tuple", where $n$ denotes the number of "arguments" (in this case, the number of functions whose "ranges"/"images" are factors in cross product to which $h$ is mapping) or in your notation, $n = |I|$.

ADDED to address comment below:

There is no operation from f, g to h, rather, there is an operation h defined in terms of f and g:

$h(x)$ is the Cartesian Product $f(x) \times g(x)$, for $x \in X$. More generally, $h(x)$ is the Cartesian Product of $f_i(x)$, $i\in I$, for $x \in X$.

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ordered n-tuple is just a value $h$ can take. I meant the operation from $f$ and $g$ to $h$. – Tim Feb 1 '13 at 18:14
Thanks for the reply an edit! – Tim Feb 1 '13 at 18:33
Does that help matters? – amWhy Feb 1 '13 at 18:33
I think I understand the operation on mappings is defined in terms of the operation on sets. I know the name of the operation on sets. Just wonder if there is a name for the operation on mappings. – Tim Feb 1 '13 at 18:35
@Tim is defining an operation on functions, specifically, a map $$\prod_{i\in I}{}^XY_i\to{}^X\prod_{i\in I}Y_i:\langle f_i:i\in I\rangle\mapsto h\;.$$ – Brian M. Scott Feb 1 '13 at 19:27

Unless I'm misunderstanding the question, I think the term you may be looking for is Cartesian product.

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I meant the operation that maps mappings to a mapping, not maps sets to a set. – Tim Feb 1 '13 at 18:22

I write ${}^AB$ for the set of functions from $A$ to $B$. You’re interested in the map

$$\Phi:\prod_{i\in I}{}^XY_i\to{}^X\prod_{i\in I}Y_i:\langle f_i:i\in I\rangle\mapsto\Big(x\mapsto\langle f_i(x):i\in I\rangle\Big)\;.$$

More generally, you could have a family $\{X_i:i\in I\}$ instead of a single $X$, with functions $f_i:X_i\to Y_i$ for $i\in I$, and look at the map

$$\Phi:\prod_{i\in I}{}^{X_i}Y_i\to{}{^\left(\prod_{i\in I}X_i\right)}\prod_{i\in I}Y_i:\langle f_i:i\in I\rangle\mapsto\Big(\langle x_i:i\in I\rangle\mapsto\langle f_i(x_i):i\in I\rangle\Big)\;.$$

The map $h=\Phi\big(\langle f_i:i\in I\rangle\big)$ is called the Cartesian product of the maps $f_i$ and written $\prod_{i\in I}f_i$.

Technically it’s actually isomorphic to that Cartesian product. A typical element of $h$ is an ordered pair

$$\Big\langle\langle x_i:i\in I\rangle,\langle f_i(x_i):i\in I\rangle\Big\rangle\;,$$

and a typical element of $\prod_{i\in I}f_i$ is of the form

$$\Big\langle\langle x_i,f_i(x_i):i\in I\Big\rangle\;,$$

but the correspondence between the two is obvious.

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