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• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeOct 8th 2009

pairing — pretty simple, but not to be confused with the product

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeOct 9th 2009

copairing, with examples at interval object.

domain, as disambiguation

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeOct 9th 2009
This comment is invalid XHTML+MathML+SVG; displaying source. <div> <blockquote> <a href="http://ncatlab.org/nlab/show/copairing">copairing</a> </blockquote> <p>Thanks. Reminds me that I don't have the font installed that you use for the coproduct...</p> </div>
• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeOct 9th 2009
This comment is invalid XHTML+MathML+SVG; displaying source. <div> <blockquote> <a href="http://ncatlab.org/nlab/show/pairing">pairing</a> — pretty simple, but not to be confused with the product </blockquote> <p>Was that a hind in my direction, yb the way? I think I may have written "product" for "pairing" here and there. Not that I can't tell one from the other, but maybe I was not using good terminology.</p> </div>
• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeOct 9th 2009

It wasn't meant to be a hint, but I wrote it to link it from interval object, where you had not only written ‘product’ for ‘pairing’ but had also denoted copairing as if it were a coproduct.

But I fixed all that.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeOct 9th 2009

Do you mean writing $f \coprod g : x \coprod y \to z$ ? Isn't that standard notation?

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeOct 10th 2009

It's not the standard that I'm aware of.

If $f\colon x \to z$ and $g\colon y \to z'$, then $f \amalg g\colon x \amalg y \to z \amalg z'$.

But if $z'$ is the same as $z$, then we also have $[f,g]\colon x \amalg y \to z$.

The relation between these (or rather, their duals) is discussed in detail at pairing.

There are lots of notations for the copairing $[f,g]$ and I suppose that somebody might write it as $f \amalg g$, but that conflicts with the idea that $\amalg\colon C^2 \to C$ is a functor, which is reflected in my $f \amalg g$. Of course, if you keep track of the targets, then there's no actual conflict, but it doesn't seem right to me.

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeOct 10th 2009
• (edited Oct 10th 2009)

I'm not sure that there is a strong consensus on notation here; my own has been to use $\langle f, g\rangle$ for pairing and $(f, g)$ for copairing. Like Toby, I would reserve $\bigsqcup$ (or more often just $+$) for the bifunctor.

Wait: how do I activate LaTeX formulas here again? [Note: I've edited after figuring this out]

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeOct 10th 2009

Okay, testing...

$x^2 + y^2$
• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeOct 10th 2009

Okay.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeOct 10th 2009

I agree that there's no consensus on notation, but I also agree that $\sqcup$ or $+$ should be reserved for the bifunctor.