### 3.2.6 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 3.2.6.1. 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.

Our first step will be to prove the easy direction of Theorem 3.2.6.1, 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 3.2.6.2. 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 3.1.6.1): 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 3.2.6.3. 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.

Definition 3.2.6.4. Let $X$ be a simplicial set. We will say that $X$ is *contractible* if the projection map $X \rightarrow \Delta ^{0}$ is a homotopy equivalence (Definition 3.1.6.1). We say that $X$ is *weakly contractible* if the projection map $X \rightarrow \Delta ^{0}$ is a weak homotopy equivalence (Definition 3.1.6.12).

Example 3.2.6.6. Let $\operatorname{\mathcal{C}}$ be a category. If $\operatorname{\mathcal{C}}$ has an initial object or a final object, then the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is contractible (this is a special case of Proposition 3.1.6.10).

Proposition 3.2.6.7. Let $X$ be a Kan complex. The following conditions are equivalent:

- $(1)$
The projection map $X \rightarrow \Delta ^{0}$ is a trivial Kan fibration.

- $(2)$
The Kan complex $X$ is contractible: that is, the projection map $X \rightarrow \Delta ^{0}$ is a homotopy equivalence.

- $(3)$
The Kan complex $X$ is nonempty. Moreover, for each vertex $x \in X$ and each $n \geq 0$, the set $\pi _{n}(X,x)$ has a single element.

- $(4)$
The Kan complex $X$ is connected. Moreover, there exists a vertex $x \in X$ such that the homotopy groups $\pi _{n}(X,x)$ are trivial for $n \geq 1$.

**Proof.**
The implication $(1) \Rightarrow (2)$ is a special case of Proposition 3.1.6.11, the implication $(2) \Rightarrow (3)$ follows from Corollary 3.2.6.3, and the implication $(3) \Rightarrow (4)$ is immediate. We will complete the proof by showing that $(4)$ implies $(1)$. Assume that $X$ is connected, and fix a vertex $x \in X$ for which the homotopy groups $\pi _{n}(X,x)$ vanish for $n \geq 1$. We first prove the following:

- $(\ast )$
Let $f: B \rightarrow X$ be a morphism of simplicial sets, let $A \subseteq B$ be a simplicial subset, and let $h: \Delta ^{1} \times A \rightarrow X$ be a homotopy from $f|_{A}$ to the constant map $A \rightarrow \{ x\} \subseteq X$. Then $h$ can be extended to a homotopy $\overline{h}: \Delta ^1 \times B \rightarrow X$ from $f$ to the constant map $B \rightarrow \{ x \} \subseteq X$.

To prove $(\ast )$, we may assume without loss of generality that $B = \Delta ^ n$ and $A = \operatorname{\partial \Delta }^ n$ for some $n \geq 0$. Since $X$ is a Kan complex, we can extend $h$ to a homotopy $\overline{h}': \Delta ^1 \times \Delta ^ n \rightarrow X$ from $f$ to some other map $f': \Delta ^{n} \rightarrow X$ for which $f'|_{ \operatorname{\partial \Delta }^{n} }$ is the constant map taking the value $x$ (Remark 3.1.5.3). Replacing $f$ by $f'$, we can reduce to the case where $f|_{ \operatorname{\partial \Delta }^{n} }$ and $h: \Delta ^1 \times \operatorname{\partial \Delta }^ n \rightarrow X$ are the constant maps taking the value $x$. In this case, the existence of the desired extension follows from the vanishing of the homotopy group $\pi _{n}(X,x)$ (or from the connectedness of $X$, in the special case $n=0$).

We now prove that the map $X \rightarrow \Delta ^{0}$ is a trivial Kan fibration. Fix an integer $n \geq 0$ and a morphism of simplicial sets $f: \operatorname{\partial \Delta }^{n} \rightarrow X$; we wish to show that $f$ can be extended to an $n$-simplex of $X$. Applying $(\ast )$ in the case $B = \operatorname{\partial \Delta }^{n}$ and $A = \emptyset $, we conclude that there exists a homotopy $h: \Delta ^1 \times \operatorname{\partial \Delta }^ n \rightarrow X$ from $f$ to the constant map $\operatorname{\partial \Delta }^{n} \rightarrow \{ x\} \subseteq X$. Because $X$ is a Kan complex, we can extend $h$ to a map $\overline{h}: \Delta ^1 \times \Delta ^ n \rightarrow X$ for which $\overline{h}|_{\{ 1\} \times \Delta ^ n}$ is the constant map taking the value $x$ (Remark 3.1.5.3). The restriction $\overline{h}|_{ \{ 0\} \times \Delta ^ n }$ then provides the desired extension of $f$.
$\square$

We establish a relative version of Proposition 3.2.6.7.

Proposition 3.2.6.8. Let $f: X \rightarrow S$ be a Kan fibration of simplicial sets. The following conditions are equivalent:

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

- $(2)$
For each vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a contractible Kan complex.

- $(3)$
For each vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is connected. Moreover, for each vertex $x \in X_{s}$, the homotopy groups $\pi _{n}(X_ s, x)$ vanish for $n > 0$.

**Proof.**
The implication $(1) \Rightarrow (2)$ and the equivalence $(2) \Leftrightarrow (3)$ follow by applying Proposition 3.2.6.7 to the fibers of $f$. We will complete the proof by showing that $(2)$ implies $(1)$. Assume that $(2)$ is satisfied; we wish to show that every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^ n \ar [r]^-{\sigma _0} \ar [d] & X \ar [d]^{f} \\ \Delta ^ n \ar@ {-->}[ur]^{\sigma } \ar [r]^-{ \overline{\sigma } } & S } \]

admits a solution. Let $q: \Delta ^1 \times \Delta ^ n \rightarrow \Delta ^ n$ be the map given on vertices by the formula $q(i,j) = \begin{cases} j & \text{ if } i = 0 \\ n & \text{ if } i = 1. \end{cases}$ Then we can regard $\overline{\sigma } \circ q$ as a homotopy from $\overline{\sigma }: \Delta ^ n \rightarrow S$ to the constant map $\Delta ^ n \rightarrow \{ s\} \subseteq S$, where $s$ denotes the vertex $\overline{\sigma }(n) \in S$. Since $f$ is a Kan fibration, the restriction $(\overline{\sigma } \circ q)|_{ \Delta ^1 \times \operatorname{\partial \Delta }^ n}$ can be lifted to a homotopy $h: \Delta ^1 \times \operatorname{\partial \Delta }^ n \rightarrow X$ from $\sigma _0$ to some map $\sigma '_0: \operatorname{\partial \Delta }^ n \rightarrow X_{s}$ (Remark 3.1.5.3). It follows from assumption $(2)$ that we can extend $\sigma '_0$ to an $n$-simplex $\sigma ': \Delta ^{n} \rightarrow X_{s}$. Invoking our assumption that $f$ is a Kan fibration again, we see that $h$ can be extended to a homotopy $\widetilde{h}: \Delta ^1 \times \Delta ^ n \rightarrow X$ from $\sigma $ to $\sigma '$, where $\sigma : \Delta ^ n \rightarrow X$ is an extension of $\sigma _0$ satisfying $f( \sigma ) = \overline{\sigma }$.
$\square$

Corollary 3.2.6.9. 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 3.1.6.11, the implication $(2) \Rightarrow (3)$ from Corollary 3.2.6.3, and the implication $(4) \Rightarrow (1)$ from Proposition 3.2.6.8. The implication $(3) \Rightarrow (4)$ follows from the long exact sequence of Theorem 3.2.5.1.
$\square$

**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 3.1.7.1, 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 3.1.6.14) between Kan complexes, and is therefore a homotopy equivalence (Proposition 3.1.6.13). Consequently, to show that $f$ is a homotopy equivalence, it will suffice to show that $f''$ is a homotopy equivalence (Remark 3.1.6.8). 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 (see Theorem 5.2.2.13). 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.
$\square$