# Kerodon

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Proposition 3.2.8.1. Suppose we are given a commutative diagram of Kan complexes

$\xymatrix@R =50pt@C=50pt{ X \ar [r]^-{g} \ar [d]^{f} & X' \ar [d]^{f'} \\ S \ar [r]^-{h} & S', }$

where $f$ and $f'$ are Kan fibrations and $h$ is a homotopy equivalence. Then the following conditions are equivalent:

$(1)$

The morphism $g$ is a homotopy equivalence.

$(2)$

For each vertex $s \in S$ having image $s' = h(s)$ in $S'$, the map of fibers $g_{s}: X_{s} \rightarrow X'_{s'}$ is a homotopy equivalence.

Proof of Proposition 3.2.8.1. Assume first that $(1)$ is satisfied. Let $s$ be a vertex of $S$ having image $s' = h(s)$ in $S'$; we wish to show that the induced map $g_{s}: X_{s} \rightarrow X'_{s'}$ is a homotopy equivalence. By virtue of Remark 3.1.6.6, it will suffice to show that for every simplicial set $W$, the induced map $\operatorname{Fun}(W,X_{s} ) \rightarrow \operatorname{Fun}( W, X'_{h(s)} )$ is bijective on connected components. Replacing $X$ by $\operatorname{Fun}(W,X)$ (and making similar replacements for $X'$, $S$, and $S'$), we may reduce to the problem of showing that $g_{s}$ induces a bijection $\pi _0( X_ s) \rightarrow \pi _{0}( X'_{s'} )$. Let us regard $\pi _0(X_ s)$ and $\pi _0( X'_{s'} )$ as endowed with actions of the fundamental groups $\pi _{1}(S,s)$ and $\pi _{1}(S',s')$, respectively (Variant 3.2.4.5). Using our assumption that $g$ and $h$ are homotopy equivalences, we conclude that the induced maps

$\pi _0(X) \rightarrow \pi _0(X') \quad \quad \pi _0(S) \rightarrow \pi _0(S') \quad \quad \pi _{1}(S,s) \rightarrow \pi _{1}(S',s')$

are bijective. Applying Corollaries 3.2.5.3 and 3.2.5.5, we conclude that $g_{s}$ induces a bijection $\pi _{1}(S,s) \backslash \pi _0(X_ s) \rightarrow \pi _{1}(S',s') \backslash \pi _0( X'_{s'} )$. It will therefore suffice to show that, for every vertex $x \in X_{s}$, the stabilizer in $\pi _{1}(S,s)$ of the connected component $[x] \in \pi _0(X_ s)$ maps isomorphically to the stabilizer in $\pi _{1}(S',s')$ of the connected component $[g(x)] \in \pi _0(X'_{s'} )$. This follows from Corollary 3.2.5.7, since $g$ induces an isomorphism $\pi _{1}(X,x) \rightarrow \pi _{1}(X', g(x) )$.

We now show that $(2) \Rightarrow (1)$. Assume that, for each vertex $s \in S$ having image $s' = h(s)$ in $S'$, the induced map $g_{s}: X_{s} \rightarrow X'_{s'}$ is a homotopy equivalence. We wish to show that $g$ is a homotopy equivalence. We first show that the map $\pi _0(g): \pi _0(X) \rightarrow \pi _0(X')$ is bijective. Our assumption that $h$ is a homotopy equivalence guarantees that the map $\pi _0(h): \pi _0(S) \rightarrow \pi _0(S')$ is bijective. It will therefore suffice to show that, for each vertex $s \in S$ having image $s' = h(s)$, the induced map $\pi _0(X) \times _{ \pi _0(S)} \{ [s] \} \rightarrow \pi _0( X' ) \times _{ \pi _0(S') } \{ [s'] \}$ is bijective. Using Corollaries 3.2.5.3 and 3.2.5.5, we can identify this with the map of quotients $(\pi _{1}(S,s) \backslash \pi _0(X_ s)) \rightarrow ( \pi _1(S',s') \backslash \pi _0(X'_{s'} ) )$. The desired result now follows from the bijectivity of the map $\pi _0(g_ s): \pi _0( X_ s ) \rightarrow \pi _0( X'_{s'} )$ and of the group homomorphism $\pi _{1}(S,s) \rightarrow \pi _{1}(S',s')$.

To complete the proof that $g$ is a homotopy equivalence, it will suffice (by virtue of Theorem 3.2.7.1) to show that for every vertex $x \in X$ having image $x' = g(x)$ and every positive integer $n$, the group homomorphism $\pi _{n}(X,x) \rightarrow \pi _{n}(X',x')$ is an isomorphism. Setting $s = f(x)$ and $s' = f(x')$, we have a commutative diagram of exact sequences

$\xymatrix@R =50pt@C=30pt{ \pi _{n+1}(S,s) \ar [r] \ar [d]^{\sim } & \pi _{n}(X_ s, x) \ar [r] \ar [d]^{\sim } & \pi _{n}(X,x) \ar [r] \ar [d] & \pi _{n}(S,s) \ar [r] \ar [d]^{\sim } & \pi _{n-1}(X_ s,x) \ar [d]^{\sim } \\ \pi _{n+1}(S',s') \ar [r] & \pi _{n}(X'_{s'}, x') \ar [r] & \pi _{n}(X',x') \ar [r] & \pi _{n}(S',s') \ar [r] & \pi _{n-1}( X'_{s'}, x'). }$

Our assumptions that $g_{s}$ and $h$ are homotopy equivalences guarantee that the outer vertical maps are bijective, and elementary diagram chase shows that that the middle vertical map is an isomorphism. $\square$