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.


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