ScalingStacks

Proof. Since $\kappa$ is uncountable, ${\mathcal{F}}_{\kappa}{\mathcal{A}}$ has countable coproducts, and so the usual Eilenberg swindle argument implies that the identity map is null-homotopic on $K$-theory and so its $K$-theory vanishes. Specifically, the functor $F\colon{\mathcal{F}}_{\kappa}{\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{F}}_{\kappa}{\mathcal{A}}$ defined by $X\mapsto\coprod_{{\mathbb{N}}}X$ is exact. Moreover, there is a natural equivalence of exact functors $\id\coprod F\simeq F$ induced by the equivalence $X\coprod(\coprod_{{\mathbb{N}}}X)\simeq\coprod_{{\mathbb{N}}}X$. Applying $K$-theory, we can split off the $F$ component of the resulting equivalence of spectra and deduce that the identity of ${\mathcal{F}}_{\kappa}{\mathcal{A}}$ is null-homotopic. ∎