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Gmane
From: =?iso-8859-1?Q?Joyal=2C_Andr=E9?= <joyal.andre <at> uqam.ca>
Subject: Re: Equality again
Newsgroups: gmane.science.mathematics.categories
Date: Tuesday 1st June 2010 14:38:04 UTC (over 7 years ago)
Dear Marco,

We could use of the dotted-equality symbol only when the 
canonical isomorphism under consideration is part of
a contractible network of isomorphisms. The network does
not need to be explicitly identified if the context is clear enough. 


For example, the dotted equality

(A times B)times C =. A times (B times C)

is refering to the associativity constraint.
The dotted equality

  A times B =. B times A

is refering to the symmetry constraint. But
the dotted equality

  A times A =. A times A

is ambiguous and should be excluded (actually, it
is not ambiguous, since the identity of A times A
is denoted  A times A = A times A ).

I am proposing a rule of thumb, not a new formalism.
Mathematics is as much an art as it is an exact science.

Best,
André


-------- Message d'origine--------
De: [email protected] de la part de Marco Grandis
Date: mar. 01/06/2010 02:36
À: Prof. Peter Johnstone; [email protected]
Objet : categories: Re: Equality again
 

On 27 May 2010, at 13:30, Prof. Peter Johnstone wrote:

>
> TeX provides a command \doteq for an equality sign with a dot over it;
> this is used in other areas of mathematics to mean "is approximately
> equal to", but as far as I know it hasn't yet been used by category-
> theorists. Perhaps we could use it to mean "is canonically
> isomorphic to"?
>
> I'd also like to use it (or something like it) between pairs of
> morphisms, meaning that (they are not equal but) they become equal
> when composed with the appropriate canonical isomorphisms (to which
> I can't be bothered to give names) in order to match up their domains
> and codomains. (Of course, this is simply saying that they are
> canonically isomorphic as objects of the functor category [2,C],
> where C is the category in which they live.)
>
> Peter Johnstone

Dear Peter,

Isn't this very dangerous?

1. First, I think you are referring to some (specified) *coherent*
(contractible) system of isomorphisms,
otherwise you can easily prove that 1 = - 1 (see an example below).

2. Even in that case, we know that coherence can be a delicate thing.
Let us take the cartesian product in Set (or the tensor product in a
symmetric monoidal category).
Would you write XxY =. YxX for the symmetry isomorphism s?
Then by XxX =. XxX do you mean s or the identity?
For XxXxX =. XxXxX we have six permutations of variables, generated
by sxX and Xxs; and so on.
I hope nobody will suggest some complicated trick to account for this;
transpositions and permutations are already there, known to
everybody; but we have to name them.


3. Coming back to point 1, "canonical" isomorphisms need not be
coherent.
There are a lot of such situations; I like to refer to the induced
isomorphisms in homological algebra,
because much of my early work was linked with that.

...

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CD: 4ms