# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

Proposition 3.3.9.6. Suppose we are given a commutative diagram of Kan complexes

$\xymatrix { \cdots \ar [r]^{ p_3} & X(3) \ar [d]^{ f_3} \ar [r]^{ p_2} & X(2) \ar [r]^{p_1} \ar [d]^{f_2} & X(1) \ar [r]^{ p_0} \ar [d]^{f_1} & X(0) \ar [d]^{f_0} \\ \cdots \ar [r]^{ q_3 } & Y(3) \ar [r]^{ q_2} & Y(2) \ar [r]^{q_1} & Y(1) \ar [r]^{ q_0 } & Y(0). }$

Assume that for each $n \geq 0$, the morphisms $p_ n$ and $q_ n$ are Kan fibrations, and the morphism $f_ n$ is a homotopy equivalence. Then the induced map $f: \varprojlim _{n \geq 0} X(n) \rightarrow \varprojlim _{n \geq 0} Y(n)$ is a homotopy equivalence of Kan complexes.

Proof. Set $X = \varprojlim _{n \geq 0} X(n)$ and $Y = \varprojlim _{n \geq 0} Y(n)$, so that $X$ and $Y$ are Kan complexes by virtue of Corollary 3.3.9.2. Let $Q(f) = X \times _{ \operatorname{Fun}( \{ 0\} , Y)} \operatorname{Fun}( \Delta ^1, Y)$ be as in Example 3.1.6.10, so that $f$ factors as a composition $X \xrightarrow {f'} Q(f) \xrightarrow {f''} Y$ where $f'$ is a homotopy equivalence and $f''$ is a Kan fibration. We will complete the proof by showing that $f''$ is a trivial Kan fibration. By virtue of Proposition 3.3.7.4, it will suffice to show that for each vertex $y \in Y$, the fiber $Q(f)_{y} = \{ y\} \times _{Y} Q(f)$ is contractible.

For each $n \geq 0$, define $Q(f_ n) = X(n) \times _{ \operatorname{Fun}( \{ 0\} , Y(n) )} \operatorname{Fun}(\Delta ^1, Y(n) )$ as in Example 3.1.6.10, so that $f_ n$ factors as a composition

$X(n) \xrightarrow {f'_{n} } Q(f_ n) \xrightarrow { f''_{n} } Y(n),$

where $f'_{n}$ is a homotopy equivalence and $f''_{n}$ is a Kan fibration. Since $f_ n$ is also a homotopy equivalence, the morphism $f''_{n}$ is a homotopy equivalence, and therefore a trivial Kan fibration (Proposition 3.3.7.4). Let $Q(f_ n)_{y} = \{ y\} \times _{ Y(n) } Q(f_ n)$ denote the fiber of $f''_{n}$ over the image of $y$, so that $Q(f_ n)_{y}$ is a contractible Kan complex. The simplicial set $Q(f)_{y}$ can then be realized as the inverse limit of the tower

$\cdots \xrightarrow {r_3} Q(f_3)_{y} \xrightarrow { r_2} Q(f_2)_{y} \xrightarrow {r_1} Q(f_1)_{y} \xrightarrow { r_0} Q(f_0)_{y}.$

By virtue of Corollary 3.3.9.5, it will suffice to show that each of the morphisms $r_ n: Q(f_{n+1})_{y} \rightarrow Q(f_{n})_{y}$ is a Kan fibration. To prove this, we observe that $r_{n}$ factors as a composition

\begin{eqnarray*} Q( f_{n+1} )_{y} & = & (X(n+1) \times \{ y\} ) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, Y(n+1)) } \operatorname{Fun}( \Delta ^1, Y(n+1) ) \\ & \xrightarrow {r'} & (X(n+1) \times \{ y\} ) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, Y(n)) } \operatorname{Fun}( \Delta ^1, Y(n) ) \\ & \xrightarrow {r''} & (X(n) \times \{ y\} ) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, Y(n)) } \operatorname{Fun}( \Delta ^1, Y(n) ) \\ & = & Q(f_ n)_{y}.\end{eqnarray*}

The morphism $r''$ is a pullback of $p_{n}$, and therefore a Kan fibration. Similarly, the morphism $r'$ is a pullback of the restriction map

$\operatorname{Fun}( \Delta ^1, Y(n+1) ) \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^1, Y(n+1) ) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, Y(n) ) } \operatorname{Fun}( \Delta ^1, Y(n) ),$

and is therefore a Kan fibration by virtue of Theorem 3.1.3.1. $\square$