Talk:Closed monoidal category

more common name

Maybe Monoidal closed category is the more common term? Geometry guy 21:01, 14 May 2007 (UTC)

Notation

Does anyone have a reference for the notation used in this article? I'm suspicious of the notation $A\Leftarrow -$ for the right adjoint to $-\otimes A$. In the case of Set this would make $A\Leftarrow B$ the set of functions from A to B!? It seems like $-\Leftarrow A$ would be better.

In any case, I would prefer to switch notation to that used in the reference by Kelly. He uses $[A,-]$ for the right adjoint to $-\otimes A$ and $[\![A,-]\!]$ for the right adjoint to $A\otimes -$. This notational appears to be slightly more common. It's also consistent with the notation used at closed category. Any thoughts? -- Fropuff (talk) 00:25, 15 February 2008 (UTC)

Your first question has an easy answer: I made a typo when writing that section, which I've fixed now. It's vastly more common to use lollipop-shaped symbols than to use $\Leftarrow$ or $\Rightarrow$, especially in linear logic, but I'm not sure how get those symbols here.

I like $[A,B]$ for the internal hom in symmetric or braided monoidal categories, and that notation is indeed very common. But I don't think that using $[A,B]$ versus $[\![A,B]\!]$ makes the right/left distinction clear, in the cases where that distinction really matters, and I don't think it's caught on. John Baez (talk) 16:46, 7 August 2009 (UTC)

Here: $\multimap$ its \multimap there is no \leftspoon or \rightspoon.

Well, I think that Fropuff is saying, in a roundabout manner, that one wants to write currying as

${\text{Hom}}(X,Y\Rightarrow Z)\simeq {\text{Hom}}(X\otimes Y,Z)$

so that X,Y and Z are in sequential order on both sides, whereas the current left-right definition in this article mixes these up in a non-intuitive way. I'm trying to figure out if this is just definitional, or whether some knee-jerk sloppiness from symmetric or braided ideas has crept in. Note that currently, both internal+hom at the nLab and closed+monoidal+category at the nLab writes these in the 'natural' sequential order. linas (talk) 13:01, 23 August 2012 (UTC)

OK, so the very first formula in the "definition" section is this:

$B\mapsto A\otimes B$

If instead, we write

$B\mapsto B\otimes A$

then for the third formula, we would get:

$\mathbf {C} (B\otimes A,C)\cong \mathbf {C} (B,A\Rightarrow C)$

right? This seems to me to be a better approach. To use a different notation, this mean the left-adjoint functor is

$-\otimes A:{\mathcal {C}}\to {\mathcal {C}}$

and the right adjoint functor is just as before:

$[A,-]:{\mathcal {C}}\to {\mathcal {C}}$

which is both what ncatlab says, and is what Fropuff says Kelley says, and I like that. So, yeah, I'm sort of gearing up to change the article to say this. Right? I'm not being stupid, am I? linas (talk) 03:30, 24 August 2012 (UTC)

Dammit. So, if I make the changes above, the first section looks pretty, and is in harmony with Kelley, with nlab, and with other WP articles. But then, the next section turns ugly: the current formula

$\mathrm {eval} _{A,B}:A\otimes (A\Rightarrow B)\to B$

which is pleasent, would have to be written as

$\mathrm {eval} _{A,B}:(A\Rightarrow B)\otimes A\to B$

which seems more awkward, somehow .... or does it? Hmmm... But I think harmonizing with everyone else is probably the most important thing to do. Its late, I'm going to bed.linas (talk) 03:56, 24 August 2012 (UTC)

OK I flipped everything around, correctly, I believe. linas (talk) 00:54, 25 August 2012 (UTC)

citation tag

I added the citation needed thing on the equivalent condition a while back. I think that in order for the second condition to imply the first, we have to assume that evaluation is natural in the second argument. -Sam W 67.170.52.2 (talk) 10:42, 27 March 2022 (UTC)