Why do we ask that condition in Kan complex?

Question

Let $\{X_n\}_{n=0}^\infty$ be simplicial set with faces $d_i:X_n\to X_{n-1} $, a simplicial set is called a Kan Complex if for any $x_0,...,x_{k-1},x_{k+1},...,x_{n+1}$ if $d_i(x_j)=d_{j-1}(x_i)$ for $i<j$, then there is $x\in X_{n+1}$ such that $d_i(x)=x_i$ for $i\neq k$. My question is why do we ask to leap at $k$? What happens if instead we ask that for any $x_0,...,x_n\in X_n$ with $d_i(x_j)=d_{j-1}(x_i)$ there is $x\in X_{n+1}$ with $d_i(x)=x_i$?

Answer 1

The usual condition says that a map $\Lambda^k[n+1] \to X$ can be extended to $\Delta[n+1] \to X$. Your other condition says that a map $\partial\Delta[n+1] \to X$ can be extended to $\Delta[n+1] \to X$. The latter condition implies the first condition, but even more so, it implies that $X$ is contractible. One can see this by considering geometric realizations, or doing some homotopy theory in simplicial sets.