ScalingStacks

For any algebra homomorphism \(A\rightarrow B\), it follows from (2) that we can view \(B\) as an object in \(\mathrm{CAlg}(\mathrm{Mod}_A(\mathcal C))\). Forgetting the \(A\)-action induces a symmetric monoidal equivalence: \[\mathrm{Mod}_B(\mathrm{Mod}_A(\mathcal C)) \xrightarrow{\simeq} \mathrm{Mod}_B(\mathcal C)\]