ScalingStacks

Proof. Let $\mathcal{M}\to\Delta^{1}$ be the Cartesian-coCartesian fibration associated with the adjunction $F\dashv G$. Since $\mathcal{C}$ and $\mathcal{D}$ are full subcategories of $\mathcal{M}$ we can think of the square $q$ as taking values in $\mathcal{M}$ and it does not change the space of lifts. Consider the diagram in $\mathcal{M}$ given by

where in the left square $q_{l}$ the horizontal arrows are coCartesian and the rest of the data is given by the lifting property of coCartesian edges. Since the inclusion of the spine $\Lambda_{1}^{2}\hookrightarrow\Delta^{2}$ is inner anodyne, so is $\Delta^{1}\times\Lambda_{1}^{2}\hookrightarrow\Delta^{1}\times\Delta^{2}$ (by T.2.3.2.4) and since $\mathcal{M}\to\Delta^{1}$ is an inner fibration, the diagram can be extended to $\Delta^{1}\times\Delta^{2}\to\mathcal{M}$ and we can denote the outer square by $r\colon\Delta^{1}\times\Delta^{1}\to\mathcal{M}$. We now claim that $q_{l}$ is a pushout square in $\mathcal{M}$. For every $Z\in\mathcal{M}$, consider the induced diagram

If $Z\in\mathcal{M}_{0}\simeq\mathcal{C}$, then the spaces on both left corners are empty and if $Z\in\mathcal{M}_{1}\simeq\mathcal{D}$, then both horizontal arrows are equivalences. Either way, this is a pullback square and hence $q_{l}$ is a pushout square. By 4.1.3 we get $L\left(q\right)\simeq L\left(r\right)$.

We can now factor the outer square $r\colon\Delta^{1}\times\Delta^{1}\to\mathcal{M}$ as

where the left square is $p$ and in the right square $q_{r}$ the horizontal arrows are Cartesian and the square is determined by the lifting property of Cartesian edges. Repeating the argument in the dual form we get that $q_{r}$ is a pullback square and using 4.1.3 again we get $L\left(p\right)\simeq L\left(r\right)$ and therefore $L\left(p\right)\simeq L\left(q\right)$. ∎