ScalingStacks

Proof. We denote by $G\colon\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)\to\underline{\operatorname{Alg}}_{\mathcal{P}}\left(\mathcal{C}\right)$ the forgetful functor induced by the map $g$. Let

$$f\colon X\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)\simeq U_{\mathcal{P}}GF_{\mathcal{Q}}\left(X\right)$$

be the unit map. It induces a cone diagram

$$\mathcal{P}_{\mathbf{SSeq}}\left(f\right)\colon\mathcal{P}_{\mathbf{SSeq}}\to\mathcal{\mathcal{C}}_{/U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)}$$

and, by 2.4.5, the associated map $\tilde{f}\colon U_{\mathcal{P}}F_{\mathcal{P}}\left(X\right)\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ is equivalent to the map $\operatorname*{colim}\mathcal{P}_{\mathbf{SSeq}}\left(X\right)\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ specified by the cone diagram $\mathcal{P}_{\mathbf{SSeq}}\left(f\right)$. On the other hand, inspecting Construction A.3.1.3.1, it can be seen that the diagram $\mathcal{P}_{\mathbf{SSeq}}\left(f\right)$ is obtained from the diagram

$$\mathcal{Q}_{\mathbf{SSeq}}\left(f\right)\colon\mathcal{Q}_{\mathbf{SSeq}}\to\mathcal{\mathcal{C}}_{/U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)}$$

by pre-composition with $g_{\mathbf{SSeq}}\colon\mathcal{P}_{\mathbf{SSeq}}\to\mathcal{Q}_{\mathbf{SSeq}}$ and that $\mathcal{Q}_{\mathbf{SSeq}}\left(f\right)$ exhibits $U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ as the colimit of $\mathcal{Q}_{\mathbf{SSeq}}\left(X\right)$. Thus, we get the desired equivalence. ∎