Kan complexes Max is right; I forgot what I was doing. Nevertheless, I learned long ago that a simplicial set is Kan iff the homotopy relation on maps to that set is an equivalence. I tried to prove it here with no references and I cannot, but I can certainly give a plausibility argument. I'm not sure I ever knew the whole proof. It must be in papers of Kan, in the late 50s. The plausibility argument is that if x^0 and x^1, say, are such that d^0x^0 = d^0x^1 and I want to find a 2-simplex of which x^0 and x^1 are d^0 and d^1, I begin by showing that x^0 ~ s^0d^0x^0 = s^0d^0x^1 ~ x^1. It follows that there is a simplex of which x^0 and x^1 are two of the faces. Then you have to find one of which it is the first two faces and I am not sure how to proceed. To see that x^0 ~ s^0d^0x^0, we calculate that d^0s^1x^0 = s^0d^0x^0, while d^1s^1x^0 = x^0. Dwyer's example seems to be that the nerve of a category isn't Kan. Yes, in fact about 1970, Boardman and Vogt showed that the nerve of a category always satisfies the "middle Kan" condition, meaning that it satisfies the condition for all missing indices except the first and last, but is not in general Kan. But it is not clear what is the point of including Dwyer's example. It is certainly not an example of a complex in which homotopy is an equivalence relation. Michael ++++++++++++++++++++++++++