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Let's say we have $f_1 : A \to X$ and $f_2 : A \to Y$. What is the most canonical way to denote $f : A \to X \times Y$ that combines $f_1$ and $f_2$ by outputting pair of their respective values for same argument? Sort of like this haskell operation. For sure mathematicians must have some shortcut for such common operation.

Doktor Diagoras
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    Personally choice: $[f_1,f_2]$. Always mentioning that it is the unique function $A\to X\times Y$ determined by $p_i\circ[f_1,f_2]=f_i$ where the $p_i$ denote the projections. – drhab Jul 27 '17 at 11:14
  • Yes, I had intuitive feeling that square brackets would look good in this role. Thank's for confirmation! – Doktor Diagoras Jul 27 '17 at 11:34

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$f(a):=(f_1(a),f_2(a))$ for $a \in A$.

Fred
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  • Well, I hoped for shortcut without "for $a \in A$", point-free style. But if no-one shows up with something more compact, I'll accept it. – Doktor Diagoras Jul 27 '17 at 11:31
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My Personally choice is: $[f_1,f_2]$.

And by introducing I would always mention that it is the unique function $A\to X\times Y$ determined by $p_i\circ[f_1,f_2]=f_i$ for $i=1,2$ where the $p_i$ denote the projections.

Moreover it cannot harm to mention that the function is prescribed by: $$a\mapsto\langle f_1(a),f_2(a)\rangle$$

In my view curled brackets are "overcharged" in mathematics, and I try to withold that a bit :-).

drhab
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As an algebraist/topologist/category theorist, it's always been $f_1 \times f_2$ to me.

Randall
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  • Hm... Are you sure you are not talking about combining $f_1 : A \to X$ with $f_2 : B \to Y$ to get $f_1 \times f_2 : A \times B \to X \times Y$? That's not what question is about. – Doktor Diagoras Jul 28 '17 at 20:38
  • This is how I was "raised." I'm sure different notation abounds. – Randall Jul 28 '17 at 20:59
  • As an aside, the notation I've always seen for the map $A \times B \rightarrow X \times Y$ you reference in your comment is $(f_1, f_2)$, which is attractive given its action.

    Anyway, I suppose all this shows why we have to explicit (or let the context speak clearly) when we use such constructions.

     – Randall Jul 28 '17 at 21:24
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    Also as category theorist? Well, I go along with Doktor Diagoras. Btw, in categories you could also write $\langle f_1,f_2\rangle^{\sharp}:A\to X\times Y$. It is actually the right adjoint in category $\mathbf{Set}$ of arrow $\langle f_1,f_2\rangle:\langle A,A\rangle\to\langle X,Y\rangle$ in category $\mathbf{Set}\times\mathbf{Set}$. – drhab Jul 29 '17 at 08:05