Cancelling coproducts?
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ? So far I have neither a proof nor a counterexample. If this cancellation principle exists it'll be slightly subtle, since 0+N ~= 1+N, and so we'd only be able to cancel adding a restricted class of (finite?) objects. I'd be grateful for either a proof or a counterexample! Michael Abbott 13-Nov-2002 13:41:53 -0400,662;000000000000-00000000
Yes, it is true. Let me use = to mean isomorphic. Begin by showing that if U + V = U + W = 1, then V = W. This is because U + V = 1 implies that U x V = 0. But then V = V x U + V x W = V x W and hence V is contained in W and vice versa. Now suppose that X + 1 = Y + 1. Then the 1 on the left can be written as 1 = U_0 + U_1, being the pullback with Y and 1. Similarly, X = V_0 + V_1. Moreover, Y = U_0 + V_0 and 1 = U_1 + V_1. But then U_0 = V_1 and then X = V_0 + V_1 = V_0 + U_0 = Y. On Wed, 13 Nov 2002, Michael Abbott wrote:
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ?
So far I have neither a proof nor a counterexample. If this cancellation principle exists it'll be slightly subtle, since 0+N ~= 1+N, and so we'd only be able to cancel adding a restricted class of (finite?) objects.
I'd be grateful for either a proof or a counterexample!
Michael Abbott
13-Nov-2002 19:35:17 -0400,1945;000000000000-00000000
On Wed, 13 Nov 2002, Michael Abbott wrote:
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ?
This is true in any topos -- and, as far as I can see, the argument works in arbitrary extensive categories. Unpacking the isomorphism and its inverse, one gets isomorphisms X \cong X_1 + X_2, 1 \cong U + V and Y \cong Y_1 + Y_2 where X_1\cong V \cong Y_1 and X_2\cong Y_2.
So far I have neither a proof nor a counterexample. If this cancellation principle exists it'll be slightly subtle, since 0+N ~= 1+N, and so we'd only be able to cancel adding a restricted class of (finite?) objects.
It works for a bit more than the class of (Kuratowski-)finite objects in a topos, since the argument above works if 1 is replaced by any subobject of 1. My guess is that, in a topos, it works for all \tilde{K}-finite objects in the sense of "Sketcehs of an Elephant", section D5.4. Whether it works for all R-finite objects in the sense of section D5.5, I'm not sure. Peter Johnstone 14-Nov-2002 22:43:21 -0400,1921;000000000000-00000000
O course the property is well known - see the article of S. Schanuel in Como '90 proceedings, Springer LNM 1488 (pg. 381, property 4) of section 5). Carboni. At 09.41 13/11/2002 +0000, you wrote:
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ?
So far I have neither a proof nor a counterexample. If this cancellation principle exists it'll be slightly subtle, since 0+N ~= 1+N, and so we'd only be able to cancel adding a restricted class of (finite?) objects.
I'd be grateful for either a proof or a counterexample!
Michael Abbott
15-Nov-2002 12:19:35 -0400,1345;000000000000-00000000
Can I just add to the positive replies ... and also add a rather obvious remark (after Michael Barr's proof): a necessary and sufficient condition for A to be cancellable in an extensive category, that is X + A ~= Y + A => X ~= Y for every X and Y is that A itself has the property of being "absolutely cancellable" i.e. Z + B ~= A `~= Z' + B => Z ~= Z'. If A is not absolutely cancellable then the failure of the above implication means that A + Z' ~= A + Z yet Z ~/= Z' so that it is certainly not cancellable. Clearly all subobjects of 1 are absolutely cancellable (and so certainly cancellable) and sums of cancellable things are cancellable ... -robin On 13 Nov, Michael Abbott wrote:
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ?
So far I have neither a proof nor a counterexample. If this cancellation principle exists it'll be slightly subtle, since 0+N ~= 1+N, and so we'd only be able to cancel adding a restricted class of (finite?) objects.
I'd be grateful for either a proof or a counterexample!
Michael Abbott
14-Nov-2002 22:43:22 -0400,1923;000000000000-00000000
Michael Abbott writes:
Am I being dim here? In an extensive category (coproducts are preserved by pullbacks and are disjoint), can I cancel thus: X+1 ~= Y+1 ?=>? X ~= Y ?
Several people have already pointed out that `1 is cancellable' in this sense. There is also a more obvious, but nonetheless usuful, kind of cancellation property which holds in extensive categories: for arbitrary Z, if X+Z is isomorphic to Y+Z via an isomorphism commuting with the injections of Z, then X is (canonically) isomorphic to Y. Steve Lack. 19-Nov-2002 21:14:18 -0400,6099;000000000001-00000000
participants (6)
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Aurelio Carboni -
Dr. P.T. Johnstone -
Michael Abbott -
Michael Barr -
Robin Cockett -
Steve Lack