In the proof of Proposition 3.3.7.6, we made essential use of the fact that any Kan fibration of simplicial sets $f: X \rightarrow S$ is (fiberwise) homotopy equivalent to a pullback $S \times _{S'} X' \rightarrow S$, where $f': X' \rightarrow S'$ is a Kan fibration between Kan complexes. This can be achieved by taking $f'$ to be the Kan fibration $\operatorname{Ex}^{\infty }(f): \operatorname{Ex}^{\infty }(X) \rightarrow \operatorname{Ex}^{\infty }(S)$. Using a variant of this construction, one can obtain a more precise result.
Theorem 3.3.8.1. Let $f: X \rightarrow S$ be a Kan fibration between simplicial sets, and let $g: S \hookrightarrow S'$ be an anodyne map. Then there exists a pullback diagram of simplicial sets where $f'$ is a Kan fibration.
\[ \xymatrix@R =50pt@C=50pt{ X \ar [d]^{f} \ar [r] & X' \ar [d]^{f'} \\ S \ar [r]^-{g} & S', } \]
Remark 3.3.8.2. We refer the reader to [KLV] for a proof of Theorem 3.3.8.1 which is slightly different from the proof given below (it avoids the use of Kan's $\operatorname{Ex}^{\infty }$-functor by appealing instead to the theory of minimal Kan fibrations, which we will discuss in §
). See also [Shulman] and [Sattler].
Remark 3.3.8.3. If $f: X \rightarrow S$ is a Kan fibration of simplicial sets, then every vertex $s \in S$ determines a Kan complex $X_{s} = \{ s\} \times _{S} X$. One can think of the construction $s \mapsto X_{s}$ as supplying a map from $S$ to the “space” of all Kan complexes. Roughly speaking, one can think of Theorem 3.3.8.1 as asserting that this “space” itself behaves like a Kan complex. We will return to this idea in §5.6.
The proof of Theorem 3.3.8.1 is based on the following observation:
Lemma 3.3.8.4. Let $f: Y \rightarrow T$ be a Kan fibration of simplicial sets, and suppose we are given simplicial subsets $X \subseteq Y$ and $S \subseteq T$ satisfying the following conditions:
The morphism $f$ carries $X$ to $S$, and the restriction $f_0 = f|_{X}$ is a Kan fibration from $X$ to $S$.
For every vertex $s \in S$, the inclusion of fibers $X_{s} \hookrightarrow Y_{s}$ is a homotopy equivalence of Kan complexes.
Let $Y' \subseteq Y$ denote the simplicial subset spanned by those simplices $\sigma : \Delta ^{n} \rightarrow Y$ having the property that the restriction $\sigma |_{ S \times _{T} \Delta ^{n} }$ factors through $X$. Then the restriction $f|_{Y'}: Y' \rightarrow T$ is a Kan fibration.
Proof. Set $Y_{S} = S \times _{T} Y \subseteq Y$. It follows from assumption $(b)$ and Corollary 3.3.7.5 that the inclusion $X \hookrightarrow Y_{S}$ is a weak homotopy equivalence, and is therefore anodyne (Corollary 3.3.7.7). Since $f_0$ is a Kan fibration, the lifting problem
admits a solution: that is, there exists a retraction $r$ from $Y_{S}$ to the simplicial subset $X \subseteq Y_{S}$ which is compatible with projection to $S$. Since $f$ is a Kan fibration, Theorem 3.1.3.5 guarantees that the map
is a trivial Kan fibration. We can therefore choose a homotopy $H: \Delta ^{1} \times Y_{S} \rightarrow Y_{S}$ from $\operatorname{id}_{Y_ S} = H|_{ \{ 0\} \times Y_{S} }$ to $r = H|_{ \{ 1\} \times Y_{S} }$, such that $f \circ H$ is the constant homotopy from $f|_{Y_{S}}$ to itself.
Choose an anodyne map of simplicial sets $i: A \hookrightarrow B$. We wish to show that every lifting problem of the form
admits a solution. Since $f$ is a Kan fibration, we can choose a map $g': B \rightarrow Y$ satisfying $g'|_{A} = g_0$ and $f \circ g = \overline{g}$. Let $B_ S \subseteq B$ denote the simplicial subset given by the fiber product $S \times _{T} B$, and let $g_1: (A \cup B_ S) \rightarrow Y$ be the map of simplicial sets characterized by $g_{1}|_{A} = g_0$ and $g_{1} |_{B_ S} = r \circ g'|_{B_ S}$ (this map is well-defined, since $r \circ g'$ and $g_0$ agree on the intersection $A \cap B_{S}$). Note that $H$ induces a homotopy $h_0: \Delta ^{1} \times (A \cup B_{S} ) \rightarrow Y$ from $g'|_{ A \cup B_{S} }$ to $g_{1}$ (compatible with the projection to $S$). Since $f$ is a Kan fibration, we can lift $h_0$ to a homotopy $h: \Delta ^1 \times B \rightarrow Y$ from $g'$ to some map $g: B \rightarrow Y$, compatible with the projection to $S$ (Remark 3.1.5.3). It follows from the construction that $g$ takes values in the simplicial subset $Y' \subseteq Y$ and satisfies the requirements $g |_{A} = g_0$ and $f \circ g= \overline{g}$. $\square$
Proof of Theorem 3.3.8.1. Let $f: X \rightarrow S$ be a Kan fibration of simplicial sets. Let us abuse notation by identifying $X$ and $S$ with simplicial subsets of $Y = \operatorname{Ex}^{\infty }(X)$ and $T = \operatorname{Ex}^{\infty }(S)$, respectively (via the monomorphisms $\rho _{X}^{\infty }: X \hookrightarrow \operatorname{Ex}^{\infty }(X)$ and $\rho _{S}^{\infty }: S \hookrightarrow \operatorname{Ex}^{\infty }(S)$), and let $Y' \subseteq \operatorname{Ex}^{\infty }(X)$ be the simplicial subset defined in the statement of Lemma 3.3.8.4. Let $g: S \hookrightarrow S'$ be an anodyne morphism of simplicial sets. Since $\operatorname{Ex}^{\infty }(S)$ is a Kan complex (Proposition 3.3.6.9), the morphism $\rho ^{\infty }_{S}: S \rightarrow \operatorname{Ex}^{\infty }(S)$ extends to a map of simplicial sets $u: S' \rightarrow \operatorname{Ex}^{\infty }(S)$. Set $X' = S' \times _{ \operatorname{Ex}^{\infty }(S)} Y'$, so that we have a commutative diagram
where the right square and outer rectangle are pullback diagrams, so the left square is a pullback diagram as well. Since the projection map $Y' \rightarrow \operatorname{Ex}^{\infty }(S)$ is a Kan fibration (Lemma 3.3.8.4), it follows that $f'$ is also a Kan fibration. $\square$