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

a dotted arrow exists making the diagram commute.

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)