Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 4.8.7.12 (Transitivity). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ and $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be functors of $\infty $-categories and let $n$ be an integer. Then:

$(1)$

If $F$ and $G$ are categorically $n$-connective, then the composite functor $G \circ F$ is categorically $n$-connective.

$(2)$

If $G \circ F$ is categorically $n$-connective, $G$ is categorically $(n+1)$-connective, and $n \geq 1$, then $F$ is categorically $n$-connective.

$(3)$

If $G \circ F$ is categorically $n$-connective and $F$ is categorically $(n-1)$-connective, then $G$ is categorically $n$-connective.

Proof. Assertions $(1)$ and $(3)$ follow by combining Remark 4.8.5.15 with Proposition 4.8.5.32 (supplemented by Remark 4.8.5.33), respectively. It will therefore suffice to prove $(2)$. Assume that $n \geq 1$, that $G \circ F$ is categorically $n$-connective, and that $G$ is categorically $(n+1)$-connective; we wish to prove that $F$ is categorically $n$-connective: that is, that $F$ is $m$-full for $m \leq n$. If $m > 0$, this follows from Proposition 4.8.5.30. It will therefore suffice to treat the case $m = 0$: that is, to show that $F$ is essentially surjective. This follows from the essential surjectivity of $G \circ F$, since $G$ induces an equivalence of homotopy categories (Remark 4.8.7.9). $\square$