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 3.6.3.10). 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:
The map of sets $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection.
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 an isomorphism.
We begin by proving Theorem 3.2.7.1 in the case of a Kan fibration.
Proposition 3.2.7.2. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes. The following conditions are equivalent:
The morphism $f$ is a trivial Kan fibration.
The morphism $f$ is a homotopy equivalence.
The map of sets $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection and, for every vertex $x \in X$ and every integer $n > 0$, the map of homotopy groups $\pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is an isomorphism.
For each vertex $y \in Y$, the fiber $X_{y} = \{ y\} \times _{Y} X$ is a contractible Kan complex.
Proof. The implication $(1) \Rightarrow (2)$ follows from Proposition 3.1.6.10 and the implication $(2) \Rightarrow (3)$ from Remark 3.2.2.17. Using Corollary 3.2.6.8 and Variant 3.2.6.9, we can reformulate $(3)$ as follows:
The map $\pi _0(f)$ is surjective and, for every vertex $x \in X$ having image $y = f(x)$, the homotopy groups $\pi _{n}(X_{y}, x)$ vanish for $n > 0$.
The equivalence of $(3') \Leftrightarrow (4)$ follows from Theorem 3.2.4.3. We will complete the proof by showing that $(4)$ implies $(1)$. Assume that condition $(4)$ is satisfied; we wish to show that every lifting problem
admits a solution. Since $\sigma $ is nullhomotopic (Exercise 3.2.4.8), we can use the homotopy extension lifting property to to reduce to the special case where $\sigma $ is the constant map $\Delta ^{m} \rightarrow \{ y\} \hookrightarrow Y$ for some vertex $y \in Y$. In this case, the desired result follows from the contractibility of the fiber $X_{y}$ (Theorem 3.2.4.3). $\square$
Remark 3.2.7.3. For the equivalences $(1) \Leftrightarrow (2) \Leftrightarrow (4)$ of Proposition 3.2.7.2, it is not necessary to assume that $X$ and $Y$ are Kan complexes. See Proposition 3.3.7.6.
Proof of Theorem 3.2.7.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Suppose that $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection and that the induced map $\pi _{n}(X,x) \rightarrow \pi _{n}(Y, f(x) )$ is an isomorphism for every base point $x \in X$ and every positive integer $n$; we wish to show that $f$ is a homotopy equivalence (the converse follows from Remark 3.2.2.17). Using Proposition 3.1.7.1 (or Example 3.1.7.10), we can factor $f$ as a composition $X \xrightarrow {f'} X' \xrightarrow {f''} Y$, where $f'$ is anodyne and $f''$ is a Kan fibration. Then $X'$ is Kan complex (Remark 3.1.1.11), so that $f'$ is a homotopy equivalence (Proposition 3.1.6.13). It will therefore suffice to show that the Kan fibration $f''$ is a homotopy equivalence, which follows from Proposition 3.2.7.2. $\square$
Corollary 3.2.7.4. Let $C_{\ast }$ and $D_{\ast }$ be chain complexes of abelian groups and let $f: C_{\ast } \rightarrow D_{\ast }$ be a morphism of chain complexes. The following conditions are equivalent:
The induced map of generalized Eilenberg-MacLane spaces $\mathrm{K}( C_{\ast } ) \rightarrow \mathrm{K}( D_{\ast } )$ is a homotopy equivalence (see Construction 2.5.6.3).
For every integer $n \geq 0$, the induced map of homology groups $\mathrm{H}_{n}( C ) \rightarrow \mathrm{H}_{n}(D)$ is an isomorphism.
Proof. Remark 2.5.6.4 guarantees that the simplicial sets $\mathrm{K}( C_{\ast } )$ and $\mathrm{K}(D_{\ast })$ are Kan complexes. By virtue of Theorem 3.2.7.1, $(1)$ is equivalent to the following pair of assertions:
The chain map $f$ induces a bijection $\pi _0( \mathrm{K}( C_{\ast } ) ) \rightarrow \pi _0( \mathrm{K}( D_{\ast } ) )$.
For every vertex $x$ of $\mathrm{K}( C_{\ast } )$ having image $y \in \mathrm{K}( D_{\ast } )$ and every integer $n > 0$, the induced of homotopy groups
is an isomorphism.
Note that we have a commutative diagram of pointed Kan complexes
where the vertical isomorphisms are given by translation by $x$ and $y$, respectively (using the group structure on the Kan complexes $\mathrm{K}( C_{\ast } )$ and $\mathrm{K}( D_{\ast } )$. Consequently, to verify $(1'')$, we may assume without loss of generality that $x = 0$. Applying Exercise 3.2.2.22, we see that $(1')$ and $(1'')$ can be reformulated as follows:
The chain map $f$ induces an isomorphism $\mathrm{H}_{0}( C ) \rightarrow \mathrm{H}_0(D)$.
For every integer $n > 0$, the chain map $f$ induces an isomorphism $\mathrm{H}_{n}(C) \rightarrow \mathrm{H}_ n(D)$.