ScalingStacks

Proof. Let $f:A\to Y$ be a morphism from an affine derived scheme to the base, and $X_{A}\to A$ the base change of $X$. Since $Y$ has affine diagonal, $f$ is affine as is the base change morphism $f_{X}:X_{A}\to X$. In particular Proposition 3.21 (again in the basic case of an affine morphism) applies to $f_{X}$, so that the total space $X_{A}$ is perfect. ∎