%This is a LaTeX file which outlines a possible counterexample %to the conjecture which was on the mailing board at the %beginning of the week. I apologise for forgetting who %originally asked the question, and for the somewhat eccentric %style of the answer! \documentstyle{article} \title{A Counter Example} \date{23 January 1991} \begin{document} \maketitle The question was asked, on the category theory mailing board, as to whether this conjecture is true. \newtheorem{Kan}{Conjecture} \begin{Kan} Let $B$ be a simplicial complex. If homotopy is an equivalence relation on ${\em SS}(A,B)$ for all $A \in ob({\em SS})$, then $B$ is a Kan complex. \end{Kan} I think I have a counter example, and would appreciate it if anyone could spot the flaw(s) in the argument. \rule{0mm}{20mm} Let B be the simplicial set which is the complete directed graph on three vertices. Pictorially, this is the following:- \begin{picture}(200,150) \put(50,110){\circle*{5}} \put(150,110){\circle*{5}} \put(100,20){\circle*{5}} \put(55,112){\vector(1,0){90}} \put(145,108){\vector(-1,0){90}} \put(55,105){\vector(1,-2){40}} \put(90,25){\vector(-1,2){40}} \put(145,105){\vector(-1,-2){40}} \put(110,25){\vector(1,2){40}} \end{picture} This simplicial set is clearly not Kan. \rule{0mm}{10mm} There are nine simplicial set morphisms $I \longrightarrow B$ which we shall call $f_{jk}$ where this maps the non-degenerate element of $I_1$ to the edge from $j$ to $k$ (for $j = k$ this is just the degenerate edge at $j$). We shall denote the non-degenerate element of $I_1$ by $\iota$. \rule{0mm}{10mm} Now, consider any $f:A \longrightarrow B$ for any simplicial set $A$. Since all $n$-simplices in $B$ are degenerate for $n \geq 2$, we have that $f_n(a) = sf_{jk}$ where $s$ is some combination of degeneracy maps; note that if $j = k$, then $s = s_0^n$. Since simplicial morphisms commute with the face and degeneracy maps, this means that $f$ is completely determined by its effect on the $0$-simplices and $1$-simplices. \rule{0mm}{10mm} We require to show that the relation of homotopy is an equivalence relation on the hom-set $\underline{\em SSets}(A,B)$ for any $A$. Clearly it is reflexive, as there is a simplicial set morphism from $(A \times I)$ to $B$ which is dummy in the second variable. \rule{0mm}{10mm} If we now consider any two simplicial morphims $f,g:A \longrightarrow B$ and a homotopy $F:(A \times I) \longrightarrow B$ from $f$ to $g$, we construct a homotopy $G$ from $g$ to $f$ as follows:- \[G_0(a,0) = F_0(a,1) \;\;\; \mbox{and} \;\;\; G_0(a,1) = F_0(a,0) \;\;\; \mbox{for} \;\;\; a \in A_0\] \[ \mbox{If} \;\;\; F_1(a,\iota) = f_{jk}(\iota) \;\;\; \mbox{then} \;\; \; G_1(a,\iota) = f_{kj}(\iota) \;\;\; \mbox{for} \;\;\; a \in A_1 .\] Since these are all the elements of $B$, this determines $G$. Clearly \[ G(a,s_0^n1) = F(a,s_0^n0) = g(a) \;\;\; \forall a \in A_n \] \[ G(a,s_0^n0) = F(a,s_0^n1) = f(a) \;\;\; \forall a \in A_n \] \rule{0mm}{10mm} Lastly, assume we have two homotopies, $F$ from $f$ to $g$ and $G$ from $g$ to $h$. Then, we construct a homotopy $H$ from $f$ to $h$ as follows:- \[ H_0(a,0) = G_0(a,0) = h_0(a) \;\;\; \mbox{and} \;\;\; H_0(a,1) = F_0(a,1) = f_0(a) \;\;\; \mbox{for} \;\;\; a \in A_0 .\] If $F_1(a,\iota) = f_{jk}(\iota)$ and If $G_1(a,\iota) = f_{kl}(\iota)$ then $H_1(a,\iota) = f_{jl}(\iota)$ for $a \in A_1$. This is transitivity, and so the example is complete. \rule{0mm}{20mm} Comments, please! \rule{100mm}{0mm} ${\em Phil}$ \end{document} ====================================== Subj: Kan complexes (twice)