Definition 7.2.8.1. Let $K$ be a simplicial set. We say that $K$ is *sifted* if, for every finite set $I$, the diagonal map $K \rightarrow K^{I}$ is right cofinal. If $\operatorname{\mathcal{C}}$ is an $\infty $-category, we say that a diagram $K \rightarrow \operatorname{\mathcal{C}}$ is *sifted* if the simplicial set $K$ is sifted.

### 7.2.8 Sifted Simplicial Sets

We now introduce a useful enlargement of the class of filtered $\infty $-categories.

Warning 7.2.8.2. Definition 7.2.8.1 has a counterpart in classical category theory. In [MR1815045], Adámek and Rosický define a *sifted category* to be a nonempty category $\operatorname{\mathcal{C}}$ which satisfies the following condition:

- $(\ast )$
For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the nerve of the category $\operatorname{\mathcal{C}}_{X/} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_{Y/}$ is connected.

It follows from Corollary 7.2.8.9 below that if the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is sifted (in the sense of Definition 7.2.8.1), then the category $\operatorname{\mathcal{C}}$ satisfies condition $(\ast )$. Beware that the converse is false (see Exercise 7.2.8.11). In other words, Definition 7.2.8.1 is not a generalization of the classical notion of a sifted category (instead, it generalizes the notion of a *homotopy sifted* category, introduced by Rosický in [MR2349711]).

Variant 7.2.8.3. Let $K$ be a simplicial set. We say that $K$ is *cosifted* if, for every finite set $I$, the diagonal map $K \rightarrow K^{I}$ is left cofinal. Equivalently, $K$ is cosifted if and only if the opposite simplicial set $K^{\operatorname{op}}$ is sifted.

Example 7.2.8.4. Every filtered $\infty $-category $\operatorname{\mathcal{C}}$ is sifted (see Proposition 7.2.4.10). In particular, if $\operatorname{\mathcal{C}}$ is an $\infty $-category which contains a final object, then $\operatorname{\mathcal{C}}$ is sifted (see Example 7.2.4.5).

Proposition 7.2.8.5. Let $f: K \rightarrow K'$ be a right cofinal morphism of simplicial sets. If $K$ is sifted, then $K'$ is also sifted.

**Proof.**
Fix a finite set $I$. We have a commutative diagram of simplicial sets

where the vertical maps are right cofinal (Corollary 7.2.1.20). Our assumption that $K$ is sifted guarantees that $\delta _{K}$ is right cofinal, so that $\delta _{K'}$ is also right cofinal (Proposition 7.2.1.6). $\square$

Proposition 7.2.8.6. Let $f: K \rightarrow K'$ be a categorical equivalence of simplicial sets. Then $K$ is sifted if and only if $K'$ is sifted.

**Proof.**
It will suffice to show that, for every finite set $I$, the diagonal map $\delta _{K}: K \rightarrow K^{I}$ is right cofinal if and only if the diagonal map $\delta _{K'}: K \rightarrow K'^{I}$ is right cofinal. This follows by applying Corollary 7.2.1.22 to the commutative diagram

Proposition 7.2.8.7. Every sifted simplicial set is weakly contractible.

**Proof.**
Let $K$ be a sifted simplicial set. Taking $I = \emptyset $ in Definition 7.2.8.1, we conclude that the projection map $K \rightarrow \Delta ^0$ is right cofinal, so that $K$ is weakly contractible by virtue of Proposition 7.2.1.5.
$\square$

Proposition 7.2.8.8. Let $K$ be a simplicial set. Then $K$ is sifted if and only if it is nonempty and the diagonal map $\delta : K \hookrightarrow K \times K$ is right cofinal.

**Proof.**
It follows immediately from the definition that if $K$ is sifted, then the diagonal map $\delta : K \hookrightarrow K \times K$ is right cofinal. Moreover, Proposition 7.2.1.5 guarantees that $K$ is weakly contractible, and therefore nonempty.

For the converse, assume that $K$ is nonempty and that $\delta $ is right cofinal. We wish to prove that, for every finite set $I$, the map $\delta _{I}: K \rightarrow K^{I}$ is right cofinal. The proof proceeds by induction on the cardinality of $I$. We first treat the case where $I = \emptyset $. Note that our assumption that $\delta $ is right cofinal guarantees in particular that it is a weak homotopy equivalence (Proposition 7.2.1.5). Since $K$ is nonempty, it follows that $K$ is weakly contractible (Corollary 3.5.1.33). Applying Proposition 7.2.1.5 again, we deduce that the projection map $K \rightarrow \Delta ^0$ is right cofinal, as desired.

We now carry out the inductive step. Assume that the set $I$ is nonempty. Choose an element $i \in I$, and set $J = I \setminus \{ i\} $. Unwinding the definitions, we see that $\delta _{I}$ can be identified with the composition

Our inductive hypothesis guarantees that $\delta _{J}$ is right cofinal, so that the product map $\operatorname{id}_{K} \times \delta _{J}$ is also right cofinal (Corollary 7.2.1.19). Since the collection of right cofinal morphisms is closed under composition (Proposition 7.2.1.6), it follows that $\delta _{I}$ is also right cofinal. $\square$

Corollary 7.2.8.9. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category. Then $\operatorname{\mathcal{C}}$ is sifted if and only if it is nonempty and, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the $\infty $-category $\operatorname{\mathcal{C}}_{X/} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_{Y/}$ is weakly contractible.

**Proof.**
Combine Proposition 7.2.8.8 with Theorem 7.2.3.1.
$\square$

We now consider an important example.

Proposition 7.2.8.10. Let $\operatorname{{\bf \Delta }}$ be the simplex category (Definition 1.1.0.2). Then the $\infty $-category $\operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }})$ is cosifted.

**Proof.**
We use the criterion of Corollary 7.2.8.9. Since the category $\operatorname{{\bf \Delta }}$ is nonempty, it will suffice to show that for every pair of nonnegative integers $m,n \geq 0$, the simplicial set

is weakly contractible. Unwinding the definitions, we can identify $\operatorname{{\bf \Delta }}_{/ [m]} \times _{\operatorname{{\bf \Delta }}} \operatorname{{\bf \Delta }}_{ / [n] }$ with the category of simplices $\operatorname{{\bf \Delta }}_{S}$ of Construction 1.1.3.9, where $S$ is the product $\Delta ^{m} \times \Delta ^ n$. Note that $S$ can be identified with the nerve of a partially ordered set, and is therefore a braced simplicial set (Exercise 3.3.1.2). Let $\operatorname{{\bf \Delta }}_{S}^{\mathrm{nd}}$ denote the full subcategory of $\operatorname{{\bf \Delta }}_{S}$ spanned by the nondegenerate simplices of $S$ (Notation 3.3.3.11), so that the inclusion $\operatorname{{\bf \Delta }}_{S}^{\mathrm{nd}} \hookrightarrow \operatorname{{\bf \Delta }}_{S}$ admits a left adjoint (Exercise 3.3.3.15). It follows that the inclusion map $\operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{S}^{\mathrm{nd}} ) \hookrightarrow \operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{S} )$ is a homotopy equivalence of simplicial sets (Proposition 3.1.6.9). It will therefore suffice to show that the nerve $\operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{S}^{\mathrm{nd} })$ is weakly contractible. Using Proposition 3.3.3.16, we can identify $\operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{S}^{\mathrm{nd}} )$ with the subdivision $\operatorname{Sd}(S)$, so that Construction 3.3.4.3 supplies a weak homotopy equivalence $\lambda _{S}: \operatorname{N}_{\bullet }(\operatorname{{\bf \Delta }}^{\mathrm{nd}}_{S} ) \rightarrow S$. We conclude by observing that the simplicial set $S = \Delta ^ m \times \Delta ^ n$ is weakly contractible (in fact, it is contractible, since it is the nerve of a partially ordered set having a smallest element). $\square$

Exercise 7.2.8.11. Let $\operatorname{{\bf \Delta }}_{\leq 1}$ denote the full subcategory of $\operatorname{{\bf \Delta }}$ spanned by the objects $[0]$ and $[1]$, which we depict informally as a diagram

Show that:

The opposite category $\operatorname{{\bf \Delta }}_{\leq 1}^{\operatorname{op}}$ satisfies condition $(\ast )$ of Warning 7.2.8.2 (that is, it is a sifted category in the sense of [MR1815045]).

The simplicial set $\operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{\leq 1}^{\operatorname{op}})$ is not sifted.