Corollary 6.124.
Let \(T\) be any topos. Then there exists a surjection \(\Shv(\Lambda) \twoheadrightarrow T\), where \(\Lambda\) is a complete Boolean algebra.
Proof
We may write \(T^{\hyp} \simeq \Shv_{\tau}(C)^{\hyp}\) for some \(C\). In light of the surjection \(T^{\hyp} \twoheadrightarrow T\), we may assume \(T = \Shv_{\tau}(C)^{\hyp}\). By Proposition 6.123, there is a surjection \(\Shv_{\rho}(D) \twoheadrightarrow \Shv_{\tau}(C)\), which induces a surjection \(\Shv_{\rho}(D)^{\hyp} \twoheadrightarrow \Shv_{\tau}(C)^{\hyp}\). Consider now the 0-localic reflection \(L_0\Shv_{\rho}(D)\) of this sheaf category. By Proposition 6.122, this receives a surjection \(\Shv(\Lambda) \twoheadrightarrow L_0\Shv_{\rho}(D)\). Since \(\Shv(\Lambda)\) is hypercomplete, this functor factors through the hypercompletion \((L_0 \Shv_{\rho}(D))^{\hyp}\), which is equivalent to \(\Shv_{\rho}(D)^{\hyp}\) by Lemma 6.58. It follows that the resulting functor \(\Shv(\Lambda) \twoheadrightarrow \Shv_{\rho}(D)^{\hyp}\) is surjective. All in all, we obtain surjections
\[\Shv(\Lambda) \twoheadrightarrow (\Shv_{\rho}(D))^{\hyp} \twoheadrightarrow \Shv_{\tau}(C)^{\hyp} \simeq T^{\hyp} \twoheadrightarrow T\]
as desired.