I am sure that you will have gathered that I interpreted the "powers" in Mike Barr's question in a different way from the others that replied, and indeed from the way in which he intended it. His "powers" were set-indexed products of the object with itself, whilst mine were in the sense of a cartesian closed category. Nevertheless, it is interesting that we have two conflicting answers. Those who are more familiar with the algebraic examples that Eduardo Dubuc, George Janelidze and Stev Lack have discussed may like to consider where the differences lie, notwithstanding what I am about to say.. Unfortunately, as Mike pointed out to me privately, there was first carelessness, and then an actual error, in what I wrote:

Now if Y and Z are sober and X >---> Y ====> Z is an equaliser, we can form a 3x3 square of objects, whose rows and columns (with one a priori exception) are equalisers, and then check that the last is an equaliser too, ie X is sober.

The carelessness was that I only intended the top row and two columns to be equalisers, not five rows/columns. The error was that $$X--->$$Y apparently needs to be mono. The proof, in all its gory ASCII detail, follows. I apologise for the error. I was reciting results from ASD without checking that all of the relevant hypotheses were present. The correct ASD results are that this is true when either i:X>-->Y is a Sigma-split inclusion, so $i is split epi and $$i is split mono or we have the more general structure that I intend to talk about in Calais. Mike also asked me for a reference for the relationship between my abstract notion of sobriety and the traditional one that every irreducible closed subset is the closure of a unique point. This is discussed (as you might have guessed) in "Sober Spaces and Continuations" TAC 2002 www.PaulTaylor.EU/ASD However, the relationship is conceptual and not extensional. The point is that a sober "space" is one whose points agree with the "primes" of the corresponding algebra, under some Stone-type duality. The paper makes some attempt to set up an extensional equivalence, but such a thing is really not meaningful. Gamma ....................... | : | __________ h _________________________ | | : f | | | i v ----------------> v | X >-------------> Y g Z | | v ----------------> v | | | | | |eta U |eta Y |eta Z | | | | | | | $$ f | | v $$ i v -------------> v |----> $$ X -----------> $$ Y $$ g $$ Z | | | | -------------> | | eta $$ X | | eta$$Y| | eta$$Z| | | |$$etaX | |$$etaY | |$$etaZ | | | | | | | | | | $$$$ f | | v v $$$$ i v v -------------> v v $$$$ X ---------> $$$$ Y $$$$ g $$$$ Z -------------> Suppose that the top row and Y and Z columns are equalisers, i;f = h = i;g and Gamma-->$$X has equal composites to $$$$ X, and therefore continuing to $$$$ Y. (We'll consider $$$$ Z later.) Since eta$$ and $$eta are natural (wrt i), the two squares from $$X to $$$$Y commute. Therefore Gamma-->$$X-->$$Y has equal composites as far as $$$$Y. In other words, it tests the equaliser Y>-->$$Y===>$$$$Y, so there is a unique fill-in (dotted) Gamma-->Y that makes a commutative kite at $$Y. Now, ignore the fact that we have parallel maps f and g etc, and just consider their composite h:Z-->Z with i. By exactly the same argument as we have just used for Y and i, the map Gamma-->$$X-->$$Z has equal composites at $$$$Z, ie it tests Y as the equaliser, and has a uniqe fill-in Gamma-->Z that makes a commutative kite at $$Z. Now, there are two other candidates for the fill-in Gamma-->Z, namely the two composites Gamma--->Y===>Z, so these are all equal. This tests the equaliser X>--->Y====Z, so there is a unique fill-in Gamma-->X. Now here was my error: why should the composite Gamma-->X-->$$X be equal the original map? Well, it is, so long as $$X-->$$Y is mono. Paul Taylor