Kerodon

Proof. We first treat the case $n = 1$. Fix a pair of objects $X,Y \in \operatorname{\mathcal{C}}$ having images $\overline{X}$ and $\overline{Y}$ in $\operatorname{\mathcal{D}}$. We wish to show that every morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ is homotopic to $F(v)$, for some morphism $v: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. Our assumption that $(G \circ F)$ is $1$-full guarantees that we can choose $v$ such that $(G \circ F)(v)$ is homotopic to $G( \overline{u} )$. Since $G$ is $2$-full, the map $\pi _0( \operatorname{Hom}_{\operatorname{\mathcal{D}}}( \overline{X}, \overline{Y} ) ) \rightarrow \pi _0( \operatorname{Hom}_{\operatorname{\mathcal{E}}}( G( \overline{X}), G( \overline{Y} ) )$ is injective, so that $F(v)$ is homotopic to $\overline{u}$ as desired.

We now treat the case $n \geq 2$. Without loss of generality, we may assume that $F$ and $G$ are inner fibrations. Using Proposition 4.8.5.27, we can further reduce to the case where $\operatorname{\mathcal{E}}$ is a standard simplex. Fix a morphism $u: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ having image $\overline{u}: \overline{X} \rightarrow \overline{Y}$ in $\operatorname{\mathcal{D}}$. We wish to show that the map

is injective for $m = n-2$ and surjective for $m = n-1$. This is clear: our assumption that $G \circ F$ is $n$-full guarantees that the set $\pi _{n-2}( \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y), u )$ consists of a single element (so $\theta _{n-2}$ is automatically injective), and our assumption that $G$ is $(n+1)$-full guarantees that the set $\pi _{n-1}( \operatorname{Hom}_{\operatorname{\mathcal{D}}}(\overline{X}, \overline{Y}), \overline{u} )$ consists of a single element (so that $\theta _{n-1}$ is automatically surjective). $\square$