## 14.30 Trivial Kan fibrations

Recall that for $n \geq 0$ the simplicial set $\Delta [n]$ is given by the rule $[k] \mapsto \mathop{\mathrm{Mor}}\nolimits _\Delta ([k], [n])$, see Example 14.11.2. Recall that $\Delta [n]$ has a unique nondegenerate $n$-simplex and all nondegenerate simplices are faces of this $n$-simplex. In fact, the nondegenerate simplices of $\Delta [n]$ correspond exactly to injective morphisms $[k] \to [n]$, which we may identify with subsets of $[n]$. Moreover, recall that $\mathop{\mathrm{Mor}}\nolimits (\Delta [n], X) = X_ n$ for any simplicial set $X$ (Lemma 14.11.3). We set

$\partial \Delta [n] = i_{(n - 1)!}\text{sk}_{n - 1}\Delta [n]$

and we call it the boundary of $\Delta [n]$. From Lemma 14.21.5 we see that $\partial \Delta [n] \subset \Delta [n]$ is the simplicial subset having the same nondegenerate simplices in degrees $\leq n - 1$ but not containing the nondegenerate $n$-simplex.

Definition 14.30.1. A map $X \to Y$ of simplicial sets is called a trivial Kan fibration if $X_0 \to Y_0$ is surjective and for all $n \geq 1$ and any commutative solid diagram

$\xymatrix{ \partial \Delta [n] \ar[r] \ar[d] & X \ar[d] \\ \Delta [n] \ar[r] \ar@{-->}[ru] & Y }$

a dotted arrow exists making the diagram commute.

A trivial Kan fibration satisfies a very general lifting property.

Lemma 14.30.2. Let $f : X \to Y$ be a trivial Kan fibration of simplicial sets. For any solid commutative diagram

$\xymatrix{ Z \ar[r]_ b \ar[d] & X \ar[d] \\ W \ar[r]^ a \ar@{-->}[ru] & Y }$

of simplicial sets with $Z \to W$ (termwise) injective a dotted arrow exists making the diagram commute.

Proof. Suppose that $Z \not= W$. Let $n$ be the smallest integer such that $Z_ n \not= W_ n$. Let $x \in W_ n$, $x \not\in Z_ n$. Denote $Z' \subset W$ the simplicial subset containing $Z$, $x$, and all degeneracies of $x$. Let $\varphi : \Delta [n] \to Z'$ be the morphism corresponding to $x$ (Lemma 14.11.3). Then $\varphi |_{\partial \Delta [n]}$ maps into $Z$ as all the nondegenerate simplices of $\partial \Delta [n]$ end up in $Z$. By assumption we can extend $b \circ \varphi |_{\partial \Delta [n]}$ to $\beta : \Delta [n] \to X$. By Lemma 14.21.7 the simplicial set $Z'$ is the pushout of $\Delta [n]$ and $Z$ along $\partial \Delta [n]$. Hence $b$ and $\beta$ define a morphism $b' : Z' \to X$. In other words, we have extended the morphism $b$ to a bigger simplicial subset of $Z$.

The proof is finished by an application of Zorn's lemma (omitted). $\square$

Lemma 14.30.3. Let $f : X \to Y$ be a trivial Kan fibration of simplicial sets. Let $Y' \to Y$ be a morphism of simplicial sets. Then $X \times _ Y Y' \to Y'$ is a trivial Kan fibration.

Proof. This follows immediately from the functorial properties of the fibre product (Lemma 14.7.2) and the definitions. $\square$

Lemma 14.30.4. The composition of two trivial Kan fibrations is a trivial Kan fibration.

Proof. Omitted. $\square$

Lemma 14.30.5. Let $\ldots \to U^2 \to U^1 \to U^0$ be a sequence of trivial Kan fibrations. Let $U = \mathop{\mathrm{lim}}\nolimits U^ t$ defined by taking $U_ n = \mathop{\mathrm{lim}}\nolimits U_ n^ t$. Then $U \to U^0$ is a trivial Kan fibration.

Proof. Omitted. Hint: use that for a countable sequence of surjections of sets the inverse limit is nonempty. $\square$

Lemma 14.30.6. Let $X_ i \to Y_ i$ be a set of trivial Kan fibrations. Then $\prod X_ i \to \prod Y_ i$ is a trivial Kan fibration.

Proof. Omitted. $\square$

Lemma 14.30.7. A filtered colimit of trivial Kan fibrations is a trivial Kan fibration.

Proof. Omitted. Hint: See description of filtered colimits of sets in Categories, Section 4.19. $\square$

Lemma 14.30.8. Let $f : X \to Y$ be a trivial Kan fibration of simplicial sets. Then $f$ is a homotopy equivalence.

Proof. By Lemma 14.30.2 we can choose an right inverse $g : Y \to X$ to $f$. Consider the diagram

$\xymatrix{ \partial \Delta [1] \times X \ar[d] \ar[r] & X \ar[d] \\ \Delta [1] \times X \ar[r] \ar@{-->}[ru] & Y }$

Here the top horizontal arrow is given by $\text{id}_ X$ and $g \circ f$ where we use that $(\partial \Delta [1] \times X)_ n = X_ n \amalg X_ n$ for all $n \geq 0$. The bottom horizontal arrow is given by the map $\Delta [1] \to \Delta [0]$ and $f : X \to Y$. The diagram commutes as $f \circ g \circ f = f$. By Lemma 14.30.2 we can fill in the dotted arrow and we win. $\square$

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