Question

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 $i < k$ 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.

Now when we keep $x_j:=d_j(y)$ fixed, is $y$ unique?

In other words, is $y$ the only $n$-simplex with $d_k(y)=x_k$ for all $k$ (including $k=j$)?

Or still in other words, is a $n$-simplex uniquely determined by its boundary? (Of course it is NOT uniquely determined by the "horn" $\left(x_0,\ldots,\hat{x_j},\ldots,x_n \right)$)

Answer 1

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.

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.