Michael Barr's letter of 19 Jan to the bulletin board about this seems to me to be based on a misunderstanding. He says he shows how to fill in a 1-horn; but all he does is to produce the third side of a triangle when two sides are given - while what the Kan property demands is that one be able to fill in the whole triangle, interior and all. My difficulty was to see how to get any non- -trivial stuff on the interior at all, using "homotopy is an ER". At any rate, I have now received a counter-example from Bill Dwyer at Notre Dame; here it is: -------------------------------------------------- I'm beginning to think that the answer to your question is "no". Here's a possible counterexample and a line of argument. Suppose that E is the category with one object x and one non-identity morphism e: x-->x, where e^2=e. Then (1) If C is a category and F, G: C --> E are functors, then F and G are connected by a natural transformation. In fact, the function which assigns to any morphism of C the map e: x --> x is a natural transformation from F to G. There are only four diagrams to check: 1 1 e e x ---> x x ---> x x ---> x x ---> x e | | e e | | e e | | e e | | e V V V V V V V V x ---> x x ---> x x ---> x x ---> x 1 e 1 e (2) Let N be the nerve functor from categories to simplicial sets and L its left adjoint. L assigns to any simplicial set A the category whose objects are the vertices of A and whose morphisms are generated by the one-simplices of A, with one relation for each two-simplex. Let f, g: A --> N(E) be two simplicial maps. These correspond to functors F, G: L(A) --> E. As in (1) there is a natural transformation between F and G, which passes to a homotopy H : N(L(A)) x \Delta[1] ----> N(E) between N(F) and N(G). Composing H with the map A x \Delta[1] -----> N(L(A)) x \Delta [1] induced by the adjunction map A --> N(L(A)) gives a homotopy between f and g. In other words, for any two maps f, g: A --> N(E) there is a homotopy from f to g, so that in particular homotopy is an equivalence relation on maps A -> N(E). (3) N(E) is not a Kan complex. The nerve of a category is a Kan complex iff the category is a groupoid. For instance, in N(E) the following cofiguration cannot be extended to a two-simplex, because e does not have an inverse. x A e / / x ---> z 1 I think this is ok. Let me know if you find a hole in it. Bill Dwyer ---------------------------------------------------------- Max Kelly, 24 Jan. ======================================