ScalingStacks

Proof. By symmetry, it is enough to prove (1). Observe that the prism $\Delta^{1}\times\Delta^{2}$ is a left cone on the simplicial set obtained by removing the initial vertex. Formally,

$$\Delta^{1}\times\Delta^{2}\simeq\left(\Delta^{2}\times\Delta^{\left\{1\right\}}\sqcup_{\Delta^{\left\{1,2\right\}}\times\Delta^{\left\{1\right\}}}\Delta^{\left\{1,2\right\}}\times\Delta^{1}\right)^{\triangleleft}.$$

We can therefore interpret the rectangle as a diagram in $\mathcal{C}_{A/}$ (and hence ignore $A$). Since the projection $\mathcal{C}_{A/}\to\mathcal{C}$ preserves and reflects limits (dual of T.1.2.13.8), the square $q_{r}$ is a pullback square in $\mathcal{C}_{A/}$. The universal property of the pullback implies that we have a homotopy Cartesian square

which in turn induces a homotopy equivalence of homotopy fibers of the vertical maps. Considering the given map $B\to Y$ as a point in $\operatorname{Map}_{\mathcal{C}_{A/}}\left(B,Y\right)$ and considering the induced equivalence on the homotopy fibers of the vertical maps, we obtain by T.5.5.5.12 an equivalence

$$\operatorname{Map}_{\mathcal{C}_{A//Y}}\left(\overline{B},\overline{X}\right)\overset{\sim}{\longrightarrow}\operatorname{Map}_{\mathcal{C}_{A//W}}\left(\overline{\overline{B}},\overline{\overline{Z}}\right),$$

where $\overline{B}$ and $\overline{X}$ are $A\to B\to Y$ and $A\to X\to Y$ viewed as objects of $\mathcal{C}_{A//Y}$ and $\overline{\overline{B}}$ and $\overline{\overline{Z}}$ are $A\to B\to W$ and $A\to Z\to W$ viewed as objects of $\mathcal{C}_{A//W}$. By the definition of the space of lifts, this is precisely the equivalence $L\left(q_{l}\right)\simeq L\left(q\right)$. ∎