Proposition 3.3.6.4. The functor $X \mapsto \operatorname{Ex}^{\infty }(X)$ preserves filtered colimits and finite limits.

$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

**Proof.**
It will suffice to show that, for every nonnegative integer $n$, the functor $X \mapsto \operatorname{Ex}^{n}(X)$ preserves filtered colimits and finite limits. Proceeding by induction on $n$, we can reduce to the case $n = 1$. We now observe that the functor $\operatorname{Ex}$ preserves *all* limits of simplicial sets (either by construction, or because it admits a left adjoint $X \mapsto \operatorname{Sd}(X)$), and preserves filtered colimits by virtue of Remark 3.3.2.7.
$\square$