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From: "Prof. Peter Johnstone"
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Subject: Re: Subobject classifier algorithm
Date: Wed, 23 Oct 2013 10:52:27 +0100 (BST)
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To: Venkata Rayudu Posina
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Dear Posina,
What you're doing is essentially just using the Yoneda lemma.
If your category is a functor category [C,Set], then Yoneda
tells you that, for each object c of C, \Omega(c) must
correspond bijectively to the set of maps C(c,-) --> \Omega,
and hence to the set of subobjects of C(c,-). So you might as
well *define* it to be the latter set. [Note that I
said `set' rather than `number' as you did, since these sets
may very well be infinite.]
More generally, if your category is a Grothendieck topos E,
then it has a representation as the topos of sheaves on a
site (C,J) whose underlying category C can be taken to be
the full subcategory of E on a generating set of objects.
Then you can similarly conclude that \Omega is (isomorphic to)
the sheaf whose value at a member g of the generating set
is the set of subobjects of g in E. So the answer to your
question `how many objects do we have to check?' is `all the
objects in some generating set'.
Regards,
Peter Johnstone
On Sun, 20 Oct 2013, Venkata Rayudu Posina wrote:
> Dear All,
>
> In continuation of the discussion we had sometime ago regarding
> algorithms for finding truth value objects, I am wondering if the
> following constitutes an algorithm for calculating subobject
> classifiers.
>
> The basic idea is to use the correspondence between parts of an object
> and maps to truth value object from the object to find the truth value
> object.
>
> In general we start with an object (of "simplest" shape such as
> initial and gradually going to less simple ones), enumerate its parts,
> and then look for objects to which the number of maps from the object
> is equal to the number of parts of the object.
>
> In the case of the category of sets, we start with the initial object,
> which has one part. Since there is exactly one function from empty
> set to every set, this doesn't help in identifying the truth value
> set. So we move to [the next] sigleton set, which has two parts. The
> set to which there are exactly two maps from the singleton set is a
> two-element set, which we take as [candidate] truth value set.
> Finally we verify that the two-element set is indeed the truth value
> set by way of checking
>
> parts of an object = maps to truth value object from the object
>
> in the case of [the set after sigleton set] two-element set. (For now
> I'm ignoring the question of how many more objects do we have to
> check.)
>
> The above method does give the correct truth value object in the
> categories of maps, graphs, and dynamical systems in addition to the
> aforementioned case of the category of sets.
>
> In the category of [set] maps, we only have to look at two objects
> before we get to the terminal object, which lets us identify the truth
> value object
>
> w: D --> C
>
> where D = {false, u, true} and C = {false, true}
> with w(false) = false, w(u) = true, w(true) = true (see Sets for
> Mathematics, pp. 114 - 9).
>
> To give one more illustration, in the case of graphs, we have to go
> little beyond terminal object to the generic arrow, whose five parts
> correspond to the five graph maps from the generic arrow to the truth
> value object of graphs (please see bottom-left corner of the cover of
> Conceptual Mathematics).
>
> In all these case we begin with [the simplest] initial object and go
> to next [less simple] object, and at each stage we use
>
> number of parts of an object = number of maps to truth value object
> from the object
>
> to identify (and then verify) the truth value object. All the more
> important is that we have to examine the above correspondence at a few
> simple shapes only (beginning with the initial object) to find the
> truth value object.
>
> Would you be kind enough to let me know if there's something wrong in
> using the above method to find the truth value object (when there's
> one) of a category in general.
>
> Thanking you,
> Yours sincerely,
> posina
>
> http://conceptualmathematics.wordpress.com/
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