On Lifts in Kan Simplicial Sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:26:53Z http://mathoverflow.net/feeds/question/84485 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84485/on-lifts-in-kan-simplicial-sets On Lifts in Kan Simplicial Sets Michael Placke 2011-12-29T00:33:03Z 2011-12-29T03:42:42Z <p>In a Kan simplicial set $X_\bullet$ we have the lifting property, that is for any $n$-tupel $\left(x_0,\ldots,\hat{x_j},\ldots,x_n \right)$ of $(n-1)$-simplices $x_k \in X_{n-1}$ with $d_i(x_k)=d_{k-1}(x_i)$ for $i &lt; k$ there is a $n$-simplex $y \in X_n$ with $d_k(y)=x_k$ for $k \neq j$. Moreover $x_j:=d_j(y)$ fills in the missing $(n-1)$ simplex in the tupel above.</p> <p>Now when we keep $x_j:=d_j(y)$ fixed, is $y$ unique? </p> <p>In other words, is $y$ the only $n$-simplex with $d_k(y)=x_k$ for all $k$ (including $k=j$)?</p> <p>Or still in other words, is a $n$-simplex uniquely determined by its boundary? (Of course it is NOT uniquely determined by the "horn" $\left(x_0,\ldots,\hat{x_j},\ldots,x_n \right)$)</p> http://mathoverflow.net/questions/84485/on-lifts-in-kan-simplicial-sets/84490#84490 Answer by Hiro Lee Tanaka for On Lifts in Kan Simplicial Sets Hiro Lee Tanaka 2011-12-29T03:42:42Z 2011-12-29T03:42:42Z <p>The answer is "No, it is not unique." As a simple example consider the triangulation of a circle with two vertices and two non-degenerate edges. You can write down the associated Kan complex. If you fix the two vertices you find more than one edge that fills them.</p> <p>A more flimsy example: Just take the singular complex of any topological space X that's not a point. If you take a null-homotopic map of S^n into X, there are in general a whole ton of ways to fill in S^n by a disc. It's an easy exercise to translate this argument into the language of filling in a simplex with specified boundary.</p>