# Kerodon

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

Proposition 3.2.6.2. Let $f: (X,x) \rightarrow (Y,y)$ be a morphism of pointed Kan complexes. The following conditions are equivalent:

$(1)$

The underlying morphism of simplicial sets $f: X \rightarrow Y$ is a homotopy equivalence (Definition 3.1.5.1): that is, there exists a morphism of simplicial sets $g: Y \rightarrow X$ such that $g \circ f$ and $f \circ g$ are homotopic to the identity maps $\operatorname{id}_{X}$ and $\operatorname{id}_{Y}$, respectively.

$(2)$

The map $f$ is a pointed homotopy equivalence: that is, there exists a morphism of pointed simplicial sets $g: (Y,y) \rightarrow (X,x)$ such that $g \circ f$ and $f \circ g$ are pointed homotopic to the identity maps $\operatorname{id}_{X}$ and $\operatorname{id}_{Y}$, respectively.

Proof. The implication $(2) \Rightarrow (1)$ is clear. For the converse, assume that $f$ is a homotopy equivalence; we will prove that $f$ induces an isomorphism in the pointed homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}_{\ast }$. Fix another pointed Kan complex $(Z,z)$, and consider the evaluation maps

$\operatorname{ev}_{x}: \operatorname{Fun}(X,Z) \rightarrow Z \quad \quad \operatorname{ev}_{y}: \operatorname{Fun}(Y,Z) \rightarrow Z.$

Since $Z$ is a Kan complex, both $\operatorname{ev}_{x}$ and $\operatorname{ev}_{y}$ are Kan fibrations (Corollary 3.1.3.3). Let $\operatorname{Fun}(X,Z)_{z} = \{ z\} \times _{Z} \operatorname{Fun}(X,Z)$ and $\operatorname{Fun}(Y,Z)_{z} = \{ z\} \times _{Z} \operatorname{Fun}(Y,Z)$ denote the fibers of $\operatorname{ev}_{x}$ and $\operatorname{ev}_{y}$ over the vertex $z$. We wish to show that precomposition with $f$ induces a bijection

$\theta : \pi _0 \operatorname{Fun}(Y,Z)_{z} = \operatorname{Hom}_{ \mathrm{h} \mathit{\operatorname{Kan}}_{\ast } }( (Y,y), (Z,z) ) \rightarrow \operatorname{Hom}_{ \mathrm{h} \mathit{\operatorname{Kan}}_{\ast } }( (X,x), (Z,z) ) = \pi _0 \operatorname{Fun}(X,Z)_{z} .$

Since $f$ is a homotopy equivalence, precomposition with $f$ induces a homotopy equivalence of Kan complexes $\operatorname{Fun}(Y,Z) \rightarrow \operatorname{Fun}(X,Z)$. It follows that the induced map of homotopy categories $\mathrm{h} \mathit{ \operatorname{Fun}(Y,Z) } \rightarrow \mathrm{h} \mathit{ \operatorname{Fun}(X,Z) }$ is an equivalence (Remark 3.1.5.4). In particular, the induced map $\pi _0( \operatorname{Fun}(Y,Z) ) \rightarrow \pi _0( \operatorname{Fun}(X,Z) )$ is bijective. We have a commutative diagram of pointed sets

$\xymatrix@R =50pt@C=50pt{ \pi _0( \operatorname{Fun}(Y,Z)_{z} ) \ar [r]^-{v} \ar [d]^-{ \theta } & \pi _0( \operatorname{Fun}(Y,Z) ) \ar [d] \ar [r] & \pi _0(Z) \ar@ {=}[d] \\ \pi _0( \operatorname{Fun}(X,Z)_{z} ) \ar [r]^-{u} & \pi _0( \operatorname{Fun}(X,Z) ) \ar [r] & \pi _0(Z) }$

where each row is exact (Corollary 3.2.5.3), so that the induced map $\operatorname{im}(v) \rightarrow \operatorname{im}(u)$ is a bijection. Using Variant 3.2.4.5, we see that the fundamental group $\pi _{1}(Z,z)$ acts on both $\pi _0( \operatorname{Fun}(Y,Z)_{z} )$ and $\pi _0( \operatorname{Fun}(X,Z)_{z} )$, and we can identify the images of $v$ and $u$ with the quotient sets $\pi _{1}(Z,z) \backslash \pi _0( \operatorname{Fun}(Y,Z)_{z} )$ and $\pi _1(Z,z) \backslash \pi _0( \operatorname{Fun}(X,Z)_ z)$, respectively (Corollary 3.2.5.5). Consequently, to show that $\theta$ is bijective, it will suffice to show that $\theta$ induces a bijection from each orbit of $\pi _{1}(Z,z)$ in $\pi _0( \operatorname{Fun}(Y,Z)_{z} )$ to the corresponding orbit in $\pi _0( \operatorname{Fun}(X,Z)_{z} )$. Equivalently, we must show that for every pointed map $g: (Y,y) \rightarrow (Z,z)$, the stabilizer of $[g] \in \pi _0( \operatorname{Fun}(Y,Z)_{z} )$ is equal to the stabilizer of $[g \circ f] \in \pi _0( \operatorname{Fun}(X,Z)_{z} )$. By virtue of Corollary 3.2.5.7, this is equivalent to the assertion that the maps

$\pi _{1}( \operatorname{Fun}(Y,Z), g) \rightarrow \pi _{1}(Z,z) \quad \quad \pi _{1}( \operatorname{Fun}(X,Z), g \circ f) \rightarrow \pi _{1}(Z,z)$

have the same image. This follows from the fact that $f$ induces an isomorphism of fundamental groups $\pi _{1}( \operatorname{Fun}(Y,Z), g) \rightarrow \pi _{1}( \operatorname{Fun}(X,Z), g \circ f)$ (because the functor of fundamental groupoids $\tau _{\leq 1}(\operatorname{Fun}(Y,Z)) \rightarrow \tau _{\leq 1}(\operatorname{Fun}(X,Z))$ is an equivalence, by virtue of Remark 3.1.5.4). $\square$