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?