Definition 3.6.3.1. Let $X$ and $Y$ be topological spaces. We say that a continuous function $f: X \rightarrow Y$ is a weak homotopy equivalence if the induced map of singular simplicial sets $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(Y)$ is a homotopy equivalence (Definition 3.1.6.1).
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3.6.3 Weak Homotopy Equivalences in Topology
Let $X$ and $Y$ be topological spaces, and let $f: X \rightarrow Y$ be a continuous function. Recall that $f$ is a homotopy equivalence if there exists a continuous function $g: Y \rightarrow X$ such that $g \circ f$ and $f \circ g$ are homotopic to the identity maps $\operatorname{id}_{X}$ and $\operatorname{id}_{Y}$, respectively. In other words, $f$ is a homotopy equivalence if its homotopy class $[f]$ is invertible when regarded as a morphism in the homotopy category of topological spaces $\mathrm{h} \mathit{\operatorname{Top}}$ (see Example 2.4.6.6). For some purposes, it is convenient to consider a somewhat weaker condition.
Remark 3.6.3.2. Let $f: X \rightarrow Y$ be a continuous function between topological spaces. Then $f$ is a weak homotopy equivalence of topological spaces if and only if $\operatorname{Sing}_{\bullet }(f)$ is a weak homotopy equivalence of simplicial sets. This is a special case of Proposition 3.1.6.13, since the simplicial sets $\operatorname{Sing}_{\bullet }(X)$ and $\operatorname{Sing}_{\bullet }(Y)$ are Kan complexes (Proposition 1.2.5.8).
Example 3.6.3.3. Let $X$ and $Y$ be topological spaces, and let $f: X \rightarrow Y$ be a homotopy equivalence. Then $f$ is a weak homotopy equivalence. This is a reformulation of Example 3.1.6.3.
Remark 3.6.3.4. Let $f: X \rightarrow Y$ be a continuous function between topological spaces. Then $f$ is a weak homotopy equivalence if and only if it satisfies the following pair of conditions:
The induced map of path components $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection.
For every point $x \in X$ and every $n \geq 1$, the map of homotopy groups $\pi _{n}(f): \pi _{n}( X, x) \rightarrow \pi _{n}(Y, f(x) )$ is an isomorphism.
This follows by applying Theorem 3.2.7.1 to the map of Kan complexes $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(Y)$ (see Example 3.2.2.7).
Example 3.6.3.5. We say that a topological space $X$ is weakly contractible if the projection map $f: X \rightarrow \ast $ is a weak homotopy equivalence (in other words, $X$ is weakly contractible if the singular simplicial set $\operatorname{Sing}_{\bullet }(X)$ is a contractible Kan complex). Using Remark 3.6.3.4, we see that $X$ is weakly contractible if and only if it is path connected (that is, the set $\pi _0(X)$ is a singleton) and the homotopy groups $\pi _{n}(X,x)$ are trivial for $n > 0$ and any choice of base point $x \in X$ (assuming that $X$ is path connected, this condition is independent of the choice of base point).
Remark 3.6.3.6. Recall that a topological space $X$ is contractible if the projection map $X \rightarrow \ast $ is a homotopy equivalence. Equivalently, $X$ is contractible if the identity map $\operatorname{id}_{X}: X \rightarrow X$ is homotopic to the constant function $X \rightarrow \{ x\} \hookrightarrow X$, for some base point $x \in X$. It follows from Example 3.6.3.3 that every contractible topological space is weakly contractible. In particular, for each $n \geq 0$, the standard simplex $| \Delta ^ n |$ is weakly contractible.
Example 3.6.3.7. Let $X$ be a topological space with the property that every continuous path $p: [0,1] \rightarrow X$ is constant (this condition is satisfied, for example, if $X$ is totally disconnected). Let $X'$ denote the topological space whose underlying set coincides with $X$, but endowed with the discrete topology. Then the identity map $f: X' \rightarrow X$ induces an isomorphism of singular simplicial sets $\operatorname{Sing}_{\bullet }(X') \rightarrow \operatorname{Sing}_{\bullet }(X)$, and is therefore a weak homotopy equivalence of topological spaces. However, $f$ is a homotopy equivalence if and only if the topology on $X$ is discrete (since any homotopy inverse of $f$ must coincide with the identity map $f^{-1}: X \rightarrow X'$).
Example 3.6.3.7 illustrates that the notions of homotopy equivalence and weak homotopy equivalence are not the same in general. However, they agree for sufficiently nice topological spaces.
Proposition 3.6.3.8. Let $f: X \rightarrow Y$ be a weak homotopy equivalence of topological spaces. Assume that both $X$ and $Y$ have the homotopy type of a CW complex (that is, there exist homotopy equivalences $X' \rightarrow X$ and $Y' \rightarrow Y$, where $X'$ and $Y'$ are CW complexes). Then $f$ is a homotopy equivalence.
Warning 3.6.3.9. In the formulation of Proposition 3.6.3.8, the hypothesis that $X$ and $Y$ have the homotopy type of a CW complex cannot be omitted. For any topological space $Y$, the counit map $v: | \operatorname{Sing}_{\bullet }(Y) | \rightarrow Y$ is a weak homotopy equivalence (Corollary 3.6.4.2), whose domain is a CW complex (Remark 1.2.3.12). If $Y$ satisfies the conclusion of Proposition 3.6.3.8, then $v$ is a homotopy equivalence, so $Y$ has the homotopy type of a CW complex.
Corollary 3.6.3.10 (Whitehead's Theorem for Topological Spaces). Let $X$ and $Y$ be topological spaces having the homotopy type of CW complexes, and let $f: X \rightarrow Y$ be a continuous function. Then $f$ is a homotopy equivalence if and only if it satisfies the following pair of conditions:
The induced map of path components $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection.
For every point $x \in X$ and every $n \geq 1$, the map of homotopy groups $\pi _{n}(f): \pi _{n}( X, x) \rightarrow \pi _{n}(Y, f(x) )$ is an isomorphism.
Proof. Combine Remark 3.6.3.4 with Proposition 3.6.3.8 (and Example 3.6.3.3). $\square$
We will deduce Proposition 3.6.3.8 from the following:
Lemma 3.6.3.11. Let $f: X \rightarrow Y$ be a weak homotopy equivalence of topological spaces, let $K$ be a CW complex, and let $g: K \rightarrow Y$ be a continuous function. Then there exists a continuous function $\overline{g}: K \rightarrow X$ such that $g$ is homotopic to $f \circ \overline{g}$.
Proof. For each $n \geq -1$, let $\operatorname{sk}_{n}(K)$ denote the $n$-skeleton of $K$ (with respect to some fixed cell decomposition), so that $\operatorname{sk}_{-1}(K) = \emptyset $. To prove Lemma 3.6.3.11, it will suffice to construct a compatible sequence of continuous functions $\overline{g}_ n: \operatorname{sk}_{n}(K) \rightarrow X$ and homotopies $h_{n}: [0,1] \times \operatorname{sk}_{n}(K) \rightarrow Y$ from $\overline{g}_ n$ to $g|_{\operatorname{sk}_ n(K)}$. We proceed by recursion. Assume that $n \geq 0$ and that the pair $(\overline{g}_{n-1}, h_{n-1})$ has already been constructed. Let $S$ denote the collection of $n$-cells of $K$. For each $s \in S$, let $b_{s}: | \operatorname{\partial \Delta }^ n | \rightarrow \operatorname{sk}_{n-1}(K)$ denote the corresponding attaching map. To construct the pair $( \overline{g}_{n}, h_{n} )$, it will suffice to show that each composition $\overline{g}_{n-1} \circ b_{s}$ can be extended to a continuous map $u_ s: | \Delta ^ n | \rightarrow X$ and that each composition $h_{n-1} \circ ( b_ s \times \operatorname{id}_{[0,1]} )$ can be extended to a homotopy from $u_ s$ to $g|_{ | \Delta ^ n |}$. Unwinding the definitions, we can rephrase this as a lifting problem
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^ n \ar [r] \ar [d] & \operatorname{Sing}_{\bullet }( X) \times _{ \operatorname{Fun}( \{ 0\} , \operatorname{Sing}_{\bullet }(Y) ) } \operatorname{Fun}( \Delta ^1, \operatorname{Sing}_{\bullet }(Y) ) \ar [d]^{\theta } \\ \Delta ^ n \ar [r] \ar@ {-->}[ur] & \operatorname{Fun}( \{ 1\} , \operatorname{Sing}_{\bullet }(Y) ) } \]
in the category of simplicial sets. Here the morphism $\theta $ is the path fibration of Example 3.1.7.10 (associated to the map of Kan complexes $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(Y)$). Our assumption that $f$ is a weak homotopy equivalence guarantees that $\operatorname{Sing}_{\bullet }(f)$ is a homotopy equivalence of Kan complexes, so that $\theta $ is also a homotopy equivalence. Applying Proposition 3.2.7.2, we deduce that $\theta $ is a trivial Kan fibration, so that the lifting problem admits a solution as desired. $\square$
Proof of Proposition 3.6.3.8. In what follows, we denote the homotopy class of a continuous function $f: X \rightarrow Y$ by $[f]$. Let $f: X \rightarrow Y$ be a weak homotopy equivalence of topological spaces, and suppose that there exists a homotopy equivalence $u: Y' \rightarrow Y$, where $Y'$ is a CW complex. Using Lemma 3.6.3.11, we deduce that $[u] = [f] \circ [ \overline{u} ]$ for some continuous function $\overline{u}: Y' \rightarrow X$. Let $v: Y \rightarrow Y'$ be a homotopy inverse to $u$ and set $g = \overline{u} \circ v$. Then
\[ [f] \circ [g] = [f \circ \overline{u}] \circ [v] = [u] \circ [v] = [ \operatorname{id}_{Y} ], \]
so $g$ is a right homotopy inverse to $f$. Since $f$ is a weak homotopy equivalence, it follows that $g$ is also a weak homotopy equivalence. If $X$ also has the homotopy type of a CW complex, then we can apply the same reasoning to deduce that $g$ admits a right homotopy inverse $f': X \rightarrow Y$. Then
\[ [g] \circ [f] = [g] \circ [f] \circ [\operatorname{id}_{X}] = [g] \circ [f] \circ [g] \circ [f'] = [g] \circ [ \operatorname{id}_ Y ] \circ [f'] = [g] \circ [f'] = [ \operatorname{id}_{X} ]. \]
It follows that $g$ is also a left homotopy inverse to $f$, so that $f$ is a homotopy equivalence (with homotopy inverse $g$). $\square$