# Kerodon

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

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. $\square$