Severe Strict Monoidal Category Naivete

Severe Strict Monoidal Category Naivete

[Ellis D. Cooper]

(1) Is strict monoidal category the same as monoid in category of categories?
(2) Is it not true that in a strict monoidal category if
$X\xrightarrow{f}Y\xrightarrow{g}Z$ then $f\square g= g\circ f$?
(3) Is the pentagon axiom automatically satisfied in a strict
monoidal category?

Many thanks for your patience and pointers.



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Re: Severe Strict Monoidal Category Naivete

[Steve Lack]

Dear Ellis,

On 03/12/2010, at 1:55 AM, Ellis D. Cooper wrote:

> (1) Is strict monoidal category the same as monoid in category of categories?

Yes. 

> (2) Is it not true that in a strict monoidal category if
> $X\xrightarrow{f}Y\xrightarrow{g}Z$ then $f\square g= g\circ f$?
 
If I understand correctly, you have arrows f:X->Y and g:Y->Z and you 
are comparing the tensor products f@g:X@Y->Y@Z and g@f:Y@X->Z@Y.
They have different domain and codomain, so cannot be equal.

If you considered a commutative monoid in the category of categories, then these arrows would be equal. But such commutative monoids are very rare. 

> (3) Is the pentagon axiom automatically satisfied in a strict
> monoidal category?
> 

Yes. In that case it asserts that two identity arrows with the same domain and codomain
are equal. 


Steve Lack.

> Many thanks for your patience and pointers.
> 
> 


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Severe Strict Monoidal Category Naivete

[Ellis D. Cooper]

Dear Steve,

At 09:54 PM 12/2/2010, you wrote:
If I understand correctly, you have arrows f:X->Y and g:Y->Z and you
are comparing the tensor products f@g:X@Y->Y@Z and g@f:Y@X->Z@Y. They
have different domain and codomain, so cannot be equal.

I was thinking more about f:X->Y and g:Y->Z and tensoring  f with the
identity morphism of Y to get f@1_Y:X@Y->Y@Y, and also tensoring 1_Y
with g to get 1_Y@g:Y@Y->Y@Z. So I get the composition f@1_Y followed
by 1_Y@g is a morphism from
X@Y->Y@Z, and you made me realize that f followed by g as a morphism
X->Y cannot possibly equal f@1_Y followed by 1_Y@g.

Then again, if the ambient strict monoidal category is symmetric, so
that the latter composition is a morphism X@Y->Z@Y, then to my mind
somehow this is pretty much the same as the composition X->Z of f
followed by g, basically because only the identity morphism of Y is involved.

The context of my inquiry is chemical reaction, as suggested by John
Baez a while ago. That is, if f and g are chemical reactions that
transform X to Y and Y to Z, respectively, then the net effect is
just transformation of X to Y, since the Y produced by f is
completely consumed by g. Bottom line: I would like a correct way to
say that tensoring f with an identity morphism is somehow no different from f.

Ellis



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Severe Strict Monoidal Category Naivete

[Ellis D. Cooper]

Hi David,

At 11:50 PM 12/3/2010, you wrote:
>to push the chemical reaction analogy further, tensoring with an
>identity morphism 1_Y is like having a chemical present that doesn't
>take part in the reaction: it's there are the beginning and end, but
>doesn't change. But you can't take it away (and no, it's not like a
>catalyst, in that your original arrow f was there to begin with).

I agree completely. All I am saying is that since there is no effect
on the reaction by the presence of a neutral chemical, it might just
as well not be mentioned. Perhaps what I am getting at is a quotient
category in which X@Y->Z@Y is identified with X->Z in this specific
situation. I do believe this is how chemists think of their
applications of Hess' Law, which is what my inquiry is really all about.

Ellis



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Re: Severe Strict Monoidal Category Naivete

[David Roberts]

  Hi Ellis,

  to push the chemical reaction analogy further, tensoring with an identity
  morphism 1_Y is like having a chemical present that doesn't take part in the
  reaction: it's there are the beginning and end, but doesn't change. But you
  can't take it away (and no, it's not like a catalyst, in that your original
  arrow f was there to begin with), unless perhaps Y has some sort of dual or
  (weak) tensor inverse, and by now the analogy is stretched beyond breaking
  point.

  David


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