Homotopy Theory on Simplicial Sets

Let f,g:X\to Y be simplicial maps. We say that f is homotopic to g (denoted by f\simeq g) if there exists a simplicial map F:X\times I\to Y such that F(x,0)=f(x) and F(x,1)=g(x) for all x\in X. If A is a simplicial subset of X and f,g:X\to Y are simplicial maps such that f|_A=g|_A, we say that f\simeq g\ \text{rel}\ A if there is a homotopy F:X\times I\to Y such that F(x,0)=f(x), F(x,1)=g(x) and F(a,t)=f(a) for all x\in X, a\in A, t\in I.

Let X be a simplicial set. The elements x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1} are said to be matching faces with respect to i if d_jx_k=d_kx_{j+1} for j\geq k and k,j+1\neq i.

A simplicial set X is said to be fibrant (or Kan complex) if it satisfies the following homotopy extension condition for each i:

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 there exists an element w\in X_n such that d_jw=x_j for j\neq i.

We define \Lambda^i[n] as the simplicial subset of \Delta[n] generated by all d_j\sigma_n for j\neq i, where \sigma_n=(0,1,\dots\,n)\in\Delta[n]_n is the nondegenerate element.

Proposition: 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.

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