### 3.2.7 Whitehead's Theorem for Kan Complexes

Let $f: X \rightarrow Y$ be a continuous function between nonempty topological spaces. If $X$ and $Y$ are CW complexes, then a classical theorem of Whitehead (see [MR30759]) asserts that $f$ is a homotopy equivalence if and only if it induces a bijection $\pi _0(X) \simeq \pi _0(Y)$ and, for every base point $x \in X$, the induced map of homotopy groups $\pi _{n}(X,x) \rightarrow \pi _{n}(Y, f(x))$ is an isomorphism for $n > 0$ (Corollary 3.5.3.10). Our goal in this section is to prove an analogous statement in the setting of Kan complexes.

Theorem 3.2.7.1. 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.

Our first step will be to prove the easy direction of Theorem 3.2.7.1, which asserts that every homotopy equivalence $f: X \rightarrow Y$ induces an isomorphism on homotopy groups. Here we encounter a slight annoyance: the hypothesis that $f$ is a homotopy equivalence guarantees that it induces an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$ of Kan complexes, but the homotopy groups of $X$ and $Y$ (with respect to base points $x \in X$ and $y \in Y$) are computed by viewing $(X,x)$ and $(Y,y)$ as objects of the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}_{\ast }$ of *pointed* Kan complexes. To address this point, we prove the following:

Proposition 3.2.7.2. 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 3.1.6.1): 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 3.1.6.5). 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 3.1.6.5).
$\square$

Corollary 3.2.7.3. Let $f: (X,x) \rightarrow (Y,y)$ be a morphism of pointed Kan complexes, and suppose that the underlying morphism of simplicial sets $X \rightarrow Y$ is a homotopy equivalence. Then, for every nonnegative integer $n \geq 0$, the induced map $\pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is a bijection.

Corollary 3.2.7.4. Let $f: X \rightarrow S$ be a Kan fibration between Kan complexes. The following conditions are equivalent:

- $(1)$
The morphism $f$ is a trivial Kan fibration.

- $(2)$
The morphism $f$ is a homotopy equivalence.

- $(3)$
The map $f$ induces a bijection $\pi _0(f): \pi _0(X) \rightarrow \pi _0(S)$. Moreover, for each vertex $x \in X$ having image $s = f(x)$ in $S$, the induced map $\pi _{n}(f): \pi _{n}(X,x) \rightarrow \pi _{n}(S,s)$ is an isomorphism for $n > 0$.

- $(4)$
For each vertex $s \in S$, the fiber $X_{s}$ is connected. Moreover, the homotopy groups $\pi _{n}(X_ s, x)$ vanish for each vertex $x \in X_{s}$ and each $n > 0$.

**Proof.**
The implication $(1) \Rightarrow (2)$ follows from Proposition 3.1.6.10, the implication $(2) \Rightarrow (3)$ from Corollary 3.2.7.3, and the implication $(4) \Rightarrow (1)$ from Proposition 3.2.6.15. The implication $(3) \Rightarrow (4)$ follows from the long exact sequence of Theorem 3.2.5.1.
$\square$

Corollary 3.2.7.6. A simplicial set $X$ is weakly contractible if and only if it is nonempty and the diagonal map $\delta _{X}: X \hookrightarrow X \times X$ is a weak homotopy equivalence.

**Proof.**
By virtue of Proposition 3.1.7.1, there exists an anodyne morphism $X \hookrightarrow Y$, where $Y$ is a Kan complex. We have a commutative diagram

\[ \xymatrix { X \ar [r]^{\delta _{X} } \ar [d] & X \times X \ar [d] \\ Y \ar [r]^{\delta _{Y}} & Y \times Y. } \]

where the vertical maps are weak homotopy equivalences. It follows that $\delta _{X}$ is a weak homotopy equivalence if and only if $\delta _{Y}$ is a weak homotopy equivalence. We may therefore replace $X$ by $Y$, and thereby reduce to the case where $X$ is a Kan complex.

Let $q: X \times X \rightarrow X$ be the map given by projection onto the first factor. Since $q \circ \delta _{X}$ is the identity morphism $\operatorname{id}_{X}$, it follows from Remark 3.1.6.15 that $\delta _{X}$ is weak homotopy equivalence if and only if $q$ is a weak homotopy equivalence. Since $q$ is a Kan fibration between Kan complexes, this is equivalent to the requirement that for each vertex $x \in X$, the fiber $q^{-1} \{ x\} \simeq \{ x\} \times X$ is a contractible Kan complex (Corollary 3.2.7.4). Provided that $X$ is nonempty, this is equivalent to the contractibility of $X$.
$\square$

**Proof of Theorem 3.2.7.1.**
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 3.2.7.3). By virtue of Proposition 3.1.7.1, 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 3.1.6.13) between Kan complexes, and is therefore a homotopy equivalence (Proposition 3.1.6.12). Consequently, to show that $f$ is a homotopy equivalence, it will suffice to show that $f''$ is a homotopy equivalence (Remark 3.1.6.7). In fact, we claim that $f''$ is a trivial Kan fibration. By virtue of Proposition 3.2.6.15, 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 (see Theorem 5.2.2.15). We may therefore assume without loss of generality that $y = f(x)$. Set $q = f'(x) \in Q$. Using the criterion of Proposition 3.2.6.14, 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 3.2.7.3.
$\square$