Fibrant Simplicial Set

Let $X$ be a simplicial set. Then $X$ is fibrant if and only if every simplicial map $f:\Lambda^i[n]\to X$ has an extension for each $i$ .

Assume that $X$ is fibrant. Let $f:\Lambda^i[n]\to X$ . The elements $f(d_0\sigma_n),f(d_1\sigma_n),\dots,f(d_{i-1}\sigma_n),f(d_{i+1}\sigma_n),\dots,f(d_n\sigma_n)$ are matching faces with respect to $i$ . This is because for $j\geq k$ and $k,j+1\neq i$ ,

$\begin{aligned} d_jf(d_k\sigma_n)&=f(d_jd_k\sigma_n)\\ &=f(d_kd_{j+1}\sigma_n)\\ &=d_kf(d_{j+1}\sigma_n) \end{aligned}$
Thus, since $X$ is fibrant, there exists an element $w\in X_n$ such that $d_jw=f(d_j\sigma_n)$ for $j\neq i$ . Then, the representing map $g=f_w:\Delta[n]\to X$ , $f_w(\sigma_n)=w$ , is an extension of $f$ .

Conversely let $x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1}$ be any elements that are matching faces with respect to $i$ . Then the representing maps $f_{x_j}:\Delta[n-1]\to X$ for $j\neq i$ defines a simplicial map $f:\Lambda^i [n]\to X$ such that the diagram

Screen Shot 2016-01-26 at 11.21.11 PM
commutes for each $j$ .

By the assumption, there exists an extension $g:\Delta[n]\to X$ such that $g|_{\Lambda^i[n]}=f$ . Let $w=g(\sigma_n)$ . Then $d_jw=d_jg(\sigma_n)=g(d_j\sigma_n)=f(d_j\sigma_n)=x_j$ for $j\neq i$ . Thus $X$ is fibrant.