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The operator '$\circ$', denoting function composition, takes two functions, $f$, and $g$, satisfying $\text{rng}\ f \subseteq \text{dom}\ g$, and returns a function $g\circ f$, satisfying that $\text{dom}\ g\circ f = \text{dom}\ f$, and that $(g\circ f)(x) = g\big(f(x)\big)$, for every $x \in \text{dom}\ f$.

Consider the following, related operator, which we shall denote by '$\langle\rangle$'. This operator takes $n$ functions, $f_1, \dots, f_n$, for any $n \in \{1,2,\dots\}$, and returns a function, $\langle f_1, \dots, f_n\rangle$, satisfying that $\text{dom}\ \langle f_1, \dots, f_n\rangle = (\text{dom}\ f_1)\times \cdots \times(\text{dom}\ f_n)$, and that, for every $x_i \in \text{dom}\ f_i$, $\langle f_1, \dots, f_n\rangle(x_1, \dots, x_n) = \big(f_1(x_1), \dots, f_n(x_n)\big)$.

For instance, if $f_1:\mathbb{N}^2\rightarrow\mathbb{N}$ is the function $f_1(a,b) = a + b$, and if $f_2:\mathbb{N}\rightarrow\mathbb{N}$ is the function $f_2(c) = c^2$, then $\langle f_1, f_2\rangle$ is the function with domain $\mathbb{N}^2\times\mathbb{N}$, which returns $\langle f_1, f_2\rangle\big((a,b),c\big) = (a+b, c^2)$ for every $a, b, c \in \mathbb{N}$.

Does this operator have a standard name and notation, the way the composition operator has a standard name (namely 'composition') and notation (namely '$\circ$')?

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Personally I denote the function as: $$f_1\times\cdots\times f_n$$ If $f_i:A_i\to B_i$ for $i=1,\dots,n$ then: $$f_1\times\cdots\times f_n:A_1\times\cdots\times A_n\to B_1\times\cdots\times B_n$$

It is characterized by the following property:

If $p_i:A_1\times\cdots\times A_n\to A_i$ and $q_i:B_1\times\cdots\times B_n\to B_i$ are the projections for $i=1,\dots,n$ then: $$q_i\circ(f_1\times\cdots\times f_n)=f_i\circ p_i$$for every $i\in\{1,\dots,n\}$.

If $F:\mathbf{Set}^n\to\mathbf{Set}$ is a functor sending objects $(A_1,\dots,A_n)$ to $A_1\times\cdots\times A_n$ then it sends morphisms $(f_1,\dots,f_n)$ to $f_1\times\cdots\times f_n$.

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  • $\begingroup$ Thanks. In what foundations system does this operator exist? ZFC set theory? Second order ZFC set theory? NBG set theory? Category theory? $\endgroup$
    – Evan Aad
    Commented Jun 25, 2019 at 11:52
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    $\begingroup$ I met the operator mainly in category theory. In set-theory it will exist too of course, but no so manifestly I think. $\endgroup$
    – drhab
    Commented Jun 25, 2019 at 12:30
  • $\begingroup$ Thanks. At any rate, this notation is ambiguous, since Category Theory, if I'm not mistaken, is couched in the language of NBG set theory, and in this language a function is a set (or, more generally, a class), and therefore $f_1\times f_2$ can be interpreted as the Cartesian product of the sets (or classes) represented by the functions $f_1$ and $f_2$, which is not the same as the operator I described in my question, and which you described in your answer. And, by the way, does this operator have a standard name in English? Or merely a symbol? $\endgroup$
    – Evan Aad
    Commented Jun 25, 2019 at 12:41
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    $\begingroup$ @EvanAad Category theory isn't "couched" in the language of any particular set theory. You can define the notion of category in pretty much any reasonable foundations. The "default" is usually (nominally) ZFC set theory, but few mathematicians provide definitions directly in these formal languages. Often, most naturally doing category theory requires going beyond ZFC (and thus NBG), e.g. by the addition of inaccessible cardinals. Tarski-Grothendieck set theory is a common choice. Or you could use type theory which is a totally different foundations. This fits category theory a bit better too. $\endgroup$
    – Derek Elkins left SE
    Commented Jun 25, 2019 at 21:08
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    $\begingroup$ @EvanAad As for the notation, drhab is correct that this is a very common notation. What's happening is that $f\times g$ is the functorial action on the arrows $f$ and $g$, which in $\mathbf{Set}$ are functions. There's also the action of the objects which (like for most functors) is also written $X\times Y$ where $X$ and $Y$ are objects. It's usually clear from notation and context which is intended. A structural set theory would allow these distinctions to be drawn more clearly for sets. $\endgroup$
    – Derek Elkins left SE
    Commented Jun 25, 2019 at 21:14

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