Remark 5.91.
For \(n = 0\), we see that a \(0\)-excisive functor \(F\) is one that sends any map \(X \to Y\) in \(C\) to a diagram \(F(X) \to F(Y)\) exhibiting \(F(X)\) as the limit of \(\{F(Y)\}\). This is precisely saying that \(F\) is a constant functor.