Proposition 10.1.1.8. 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.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
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
\[ K \xrightarrow {\delta } K \times K \xrightarrow { \operatorname{id}_ K \times \delta _{J} } K \times K^{J}. \]
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. $\square$