ScalingStacks

Proof. For any infinite cardinal $\kappa>\omega$, there is a natural inclusion map

$$\Psi_{\perf}(F_{R}^{\infty})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}(\Ind(\Psi_{\perf}(R)))^{\kappa}$$

induced by the fact that any countable wedge of copies of $R$ is in $(\Ind(\Psi(R)_{\perf}))^{\kappa}$, and the latter is closed under retracts and stable under finite colimits. Since the inclusion $\Psi_{\perf}(F_{R})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Psi_{\perf}(F_{R}^{\infty})$ is compatible with the (Yoneda) inclusion $\Psi_{\perf}(R)\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}(\Ind(\Psi_{\perf}(R)))^{\kappa}$, we have a commutative diagram

Combining this with $\Psi_{\perf}(\tilde{F}_{R})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Psi_{\perf}(\tilde{F}^{\infty}_{R})$ we obtain the commutative diagram

(9.52)

and hence an induced composite map of quotients

	$\displaystyle\alpha\colon\Psi_{\perf}(\tilde{F}_{R}^{\infty})/\Psi_{\perf}(F_{R})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Psi_{\perf}(F_{R}^{\infty})/\Psi_{\perf}(F_{R})$
	$\displaystyle\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}(\Ind(\Psi_{\perf}(R)))^{\kappa}/\Psi_{\perf}(R).$

By the work above, $\alpha$ can be described as a map

$$\Psi_{\tri}(\hat{\mu}R)\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}(\Ind(\Psi_{\perf}(R)))^{\kappa}/\Psi_{\perf}(R).$$

Finally, since $F_{R}^{\infty}$ has countable coproducts, the usual Eilenberg swindle implies that $K(F_{R}^{\infty})$ is contractible. We also know that $K(\Psi_{\perf}(\tilde{F}_{R}^{\infty})$ is contractible [33, 6.1,6.3]. Therefore, applying $\mathrm{Map}(\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}({\mathcal{S}}_{\infty}^{\omega}),\,\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(-))$ to the commutative diagram, the fact that all of the horizontal sequences are strict-exact allows us to apply theorem 9.10 to conclude that $\alpha$ induces an equivalence on $K$-theory spectra. ∎