Lemma 6.58.
Let \(n \geq -1\), and let \(T\) be a topos such that every object is covered by \((n-1)\)-truncated objects. Then the map of topoi \(T \to L_n(T)\) induces an equivalence on hypercompletions:
\[T^{\hyp} \; \iso \; (L_nT)^{\hyp}.\]
Proof
By Proposition 6.54, there is an \((n,1)\)-category \(C\) with finite limits and a Grothendieck topology \(\tau\) such that \(L_nT \simeq \Shv_{\tau}(C)\). Using the hypothesis on \(T\), we may choose \(C\subseteq T_{\leq n-1}\) so that every object of \(T\) is covered by objects of \(C\). Let \(u\colon C\hookrightarrow T\) denote the inclusion. Since \(u\) preserves finite limits, its colimit extension \(u_!\colon\PSh(C)\to T\) is left exact by Proposition 2.43. By Lemma 6.57, the counit of \(u_!\dashv u^*\) is \(\infty\)-connected. It follows that the composite
\[\PSh(C) \xrightarrow{u_!} T \twoheadrightarrow T^{\hyp}\]
is a left exact localization.Let \(K\) be its kernel. We claim that the monogenic part of \(K\) is \(\tau^c\). By Proposition 6.9, it suffices to test monomorphisms \(R\hookrightarrow y(X)\) which are sieves on representables. Such a sieve is inverted by the displayed localization if and only if \(u_!(R)\to u(X)\) becomes an isomorphism after hypercompletion. Since this map is a monomorphism, this happens if and only if it is already an isomorphism in \(T\), or equivalently if \(u_!(R)\to u(X)\) is an effective epimorphism. By the definition of the effective epimorphism topology \(\tau\), this is precisely the condition that \(R\) be a covering sieve. Hence \(K^{\mono}=\tau^c\).The congruence \(K\) is hypercomplete because its quotient is \(T^{\hyp}\). By Lemma 5.67, a hypercomplete congruence is determined by its monogenic part, so \[K=(K^{\mono})^{\hyp}=(\tau^c)^{\hyp}.\]
But \((\tau^c)^{\hyp}\) is also the kernel of the localization \(\PSh(C)\to\Shv_{\tau}(C)^{\hyp}=(L_nT)^{\hyp}\). The two hypercomplete localizations therefore have the same kernel, which proves the asserted equivalence.