# Kerodon

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

Theorem 3.2.6.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then $f$ is a homotopy equivalence if and only if it satisfies the following pair of conditions:

$(a)$

The map of sets $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection.

$(b)$

For every vertex $x \in X$ having image $y = f(x)$ in $Y$ and every positive integer $n$, the map of homotopy groups $\pi _{n}(f): \pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is bijective.

Proof of Theorem 3.2.6.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Assume that the induced map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection and that, for every vertex $x \in X$ having image $y = f(x)$ in $Y$, the induced map $\pi _{n}(f): \pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is an isomorphism. We wish to prove that $f$ is a homotopy equivalence (the converse implication follows from Corollary 3.2.6.3). By virtue of Proposition 3.1.6.1, we can assume that $f$ factors as a composition $X \xrightarrow {f'} Q \xrightarrow {f''} Y$, where $f'$ is anodyne and $f''$ is a Kan fibration. Note that $Q$ is automatically a Kan complex (since $Y$ is a Kan complex and $f''$ is a Kan fibration). Moreover, the anodyne morphism $f'$ is a weak homotopy equivalence (Proposition 3.1.5.13) between Kan complexes, and is therefore a homotopy equivalence (Proposition 3.1.5.12). Consequently, to show that $f$ is a homotopy equivalence, it will suffice to show that $f''$ is a homotopy equivalence (Remark 3.1.5.7). In fact, we claim that $f''$ is a trivial Kan fibration. By virtue of Proposition 3.2.6.8, it will suffice to show that for every vertex $y \in Y$, the Kan complex $Q_{y} = \{ y\} \times _{Y} Q$ is contractible. Since $\pi _0(f)$ is a bijection, there exists a vertex $x \in X$ such that $f(x)$ and $y$ belong to the same connected component of $Y$. Since $f''$ is a Kan fibration, the Kan complexes $Q_{y}$ and $Q_{f(x)}$ are homotopy equivalent (Example ). We may therefore assume without loss of generality that $y = f(x)$. Set $q = f'(x) \in Q$. Using the criterion of Proposition 3.2.6.7, we are reduced to proving that the set $\pi _{n}( Q_{f(x)}, q)$ is a singleton for each $n \geq 0$. Using the exact sequence

$\cdots \rightarrow \pi _{2}(Y,y) \xrightarrow {\partial } \pi _{1}(Q_ y, q) \rightarrow \pi _{1}( Q, q) \rightarrow \pi _1(Y,y) \xrightarrow {\partial } \pi _{0}(Q_ y, q) \rightarrow \pi _0( Q,q) \rightarrow \pi _0(Y,y)$

of Theorem 3.2.5.1, we are reduced to proving that each of the maps $\pi _{n}(f''): \pi _{n}(Q,q) \rightarrow \pi _{n}(Y,y)$ is bijective. This follows from the commutativity of the diagram

$\xymatrix@R =50pt@C=50pt{ \pi _{n}(X,x) \ar [rr] \ar [dr] & & \pi _{n}(Q,q) \ar [dl] \\ & \pi _{n}(Y,y), & }$

since the left vertical map is bijective by assumption and the upper horizontal map is bijective by virtue of Corollary 3.2.6.3. $\square$