## 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

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.