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* Isbell & MacLane on the insufficiency on skeletal categories
@ 2013-05-24 22:34 Staffan Angere
  2013-05-25 15:47 ` Colin McLarty
  0 siblings, 1 reply; 6+ messages in thread
From: Staffan Angere @ 2013-05-24 22:34 UTC (permalink / raw)
  To: categories

Dear Category theorists,

in Categories for the Working Mathematician, page 164, MacLane relates an argument, "due to Isbell", why one cannot identify all isomorphic objects. I  have not, however, been able to find any publication of Isbell that contains the argument. Does anyone here know if he published it?

I also have a question about the argument itself: why is it made the way MacLane does it, rather than just though noticing that all functions from countable sets are countable, and thus themselves countable, and so isomorphic  to any other countable set? It seems like it would follow directly from this that any functions on the natural numbers have to be equal, if isomorphic (i.e. equinumerious) sets are identical. Why is MacLane doing all the "detours" though products, epics, etc.?

Thanks in advance,
Staffan

[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


^ permalink raw reply	[flat|nested] 6+ messages in thread
* Re: Isbell & MacLane on the insufficiency on skeletal categories
@ 2013-06-07  3:15 Fred E.J. Linton
  0 siblings, 0 replies; 6+ messages in thread
From: Fred E.J. Linton @ 2013-06-07  3:15 UTC (permalink / raw)
  To: Vaughan Pratt; +Cc: categories

On Thu, 06 Jun 2013 07:45:03 PM EDT, by Vaughan Pratt
<pratt@cs.stanford.edu>:

> On 5/25/2013 8:47 AM, Colin McLarty wrote:
>> Of course is also follows that NxN=N.  But it does not follow, and in
fact
>> it is refutable, that the projection functions are the identity function
>> 1_N. Isbell's argument is on p. 164 of my copy of CfWM (1998).
> 
> Why do you need Isbell's long argument, or even any monoidal structure
> on Set, to obtain a contradiction here?  Just use that NxN is a product
> and observe that the pair (3,4) in NxN (as a map from 1 to NxN) would
> have to be both 3 and 4 (as maps from 1 to N) when the projections are
> the identity.

Even more convincing: The equalizer of those two projections from N x N
must be the diagonal in N x N. But if those projections are equal, their
equalizer is all of N x N. Thus every map to N x N factors through the
diagonal there, i.e., no matter what the object A, for every pair of maps
f, g: A --> N, we must have f = g. It will follow that N is terminal.

[Or was that your argument, Vaughan, that I somehow did not recognize?] 

Cheers, -- Fred




[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


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Thread overview: 6+ messages (download: mbox.gz / follow: Atom feed)
-- links below jump to the message on this page --
2013-05-24 22:34 Isbell & MacLane on the insufficiency on skeletal categories Staffan Angere
2013-05-25 15:47 ` Colin McLarty
2013-05-26 20:26   ` Bas Spitters
2013-06-06 22:00   ` Vaughan Pratt
2013-06-07  5:00     ` Colin McLarty
2013-06-07  3:15 Fred E.J. Linton

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