Let $A$ be an $\mathbb{E}_\infty$-ring spectrum, and let $R_1$, $R_2$ and $R_3$ be $\mathbb{E}_\infty$-$A$-algebras. We assume there is a homotopy fibre sequence $$ R_1\to R_2 \to R_3 $$ in the stable infinity category $\text{Mod}_{A}(\text{Sp})$. Then, is there a homotopy fibre sequence $$ K(R_1) \to K(R_2) \to K(R_3) $$ in $\text{Mod}_{K(A)}(\text{Sp})$?
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5$\begingroup$ The problem is that the composite $R_1\to R_3$ must then be zero and, assuming you mean unital $E_\infty$-algebras, that implies that $R_3=0$ so $R_1\to R_2$ is an equivalence and you do get a homotopy fibre sequence in $K$-theory, but a trivial one. $\endgroup$– Fernando MuroCommented May 31, 2023 at 6:30
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No, this is not true, and nothing like this is expected.
The exact sequence $\Bbb Z/2 \to \Bbb Z/4 \to \Bbb Z/2$ gives rise to a fiber sequence $H\Bbb Z/2 \to H\Bbb Z/4 \to H\Bbb Z/2$ of Eilenberg--Mac Lane spectra, and so your request is for the existence of a fiber sequence $K\Bbb Z/2 \to K\Bbb Z/4 \to K\Bbb Z/2$. However, by Quillen's work all the K-groups of $\Bbb Z/2$ are odd in positive degrees, while the K-groups of $\Bbb Z/4$ have even torsion (eg $K_1$ contains a $\Bbb Z/2 \cong \Bbb Z/4^\times$).
The closest thing to what you're asking is probably work of Land--Tamme, which systematically studies the extent to which K-theory fails to preserve pullbacks.
answered May 31, 2023 at 13:16