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