Let \(k \geq 0\) and \(n\geq -2\), and consider composable functors \(\mathcal A\xrightarrow{F} \mathcal B\xrightarrow{G} \mathcal C\) of \((\infty,k)\)-categories. Then, if \(G\) is \((n+1)\)-faithful and \(GF\) is \(n\)-faithful, then \(F\) is \(n\)-faithful.
Proof.
The case \(n=-2\) is the straight-forward statement that a section of a fully faithful functor is an equivalence. For \(n\geq -1\), we induct on \(k \geq 0\). The base case \(k=0\) is Lemma 5.2.4. For \(k \geq 1\), \(F\) being \(n\)-faithful is equivalent to proving that for \(a, a' \in \mathcal A\) the induced functor of \((\infty,k-1)\)-categories \(\underline{\mathrm{Hom}}_{\mathcal A}(a,a') \rightarrow\underline{\mathrm{Hom}}_{\mathcal B}(Fa, Fa')\) is \((n-1)\)-faithful. Since the composable sequence of functors of \((\infty,k-1)\)-categories \(\underline{\mathrm{Hom}}_{\mathcal A}(a,a') \rightarrow\underline{\mathrm{Hom}}_{\mathcal B}(Fa, Fa') \rightarrow\underline{\mathrm{Hom}}_{\mathcal C}(GFa, GFa')\) the last functor is \(n\)-faithful, and the composite is \((n-1)\)-faithful by assumption, the first functor is \((n-1)\)-faithful by induction. ◻
Original source: arXiv:2401.02956v2
Original source · 2401.02956v2