The boundary map of Kan simplicial sets

2011-12-18T16:20:18Z

I don't know whether or not the following is a research level question, but since it is concerned with simplicial sets and since they are very popular these days, I think this is the right place to ask:

Suppose $S_\bullet$ is a simplicial set,

$\partial(S_n) = \lbrace (x_0,...,x_n)| d_i(x_j)=d_{j-1}(x_i) \text{ for } i < j \rbrace \subset \times^{n+1}S_{n-1}$

is the simplicial kernel (simplicial boundery) of $S_\bullet$ in dimension $n$ and

$\Lambda^n_k = \lbrace \left(x_0,\ldots,\hat{x}_k,\ldots,x_n\right)| d_i(x_j)=d_{j-1}(x_i) \mbox{ for } i < j \mbox{ and } i,j \neq k \rbrace \subset \times^{n}S_{n-1}$

is the $k$-horn in dimension $n$.

Then if the simplicial set is Kan (has the Kan property) the map $$ \begin{array}{cccl} Kan(n,k): & S_n &\rightarrow& \Lambda^n_k \ ; & x & \mapsto & \left(d_1(x),\ldots,\widehat{d_k(x)},\ldots,d_n(x) \right) \end{array} $$ is surjective.

Now my question is:

If $S_\bullet$ is Kan, is the 'boundary map' $$ \begin{array}{cccl} \partial_n: & S_n &\rightarrow& \partial(S_n)\ ; & x & \mapsto & \left(d_1(x),\ldots,d_n(x) \right) \end{array} $$ surjective, too?

Answer by Karol Szumiło for The boundary map of Kan simplicial sets

2011-12-18T18:33:30Z

The surjectivity of your "Kan maps" is equivalent to the lifting property against all horn inclusions. Similarly the surjectivity of "boundary maps" is equivalent to the lifting property against all boundary inclusions $\partial \Delta^n \to \Delta^n$. The horn inclusions generate acyclic cofibartions and similarly boundary inclusions generate cofibrations. This means that "boundary maps" are surjective if and only if $S$ is a contractible Kan complex.

The boundary map of Kan simplicial sets - MathOverflow

The boundary map of Kan simplicial sets

Answer by Karol Szumiło for The boundary map of Kan simplicial sets