Theorem 5.92.
Let \(K\) be the kernel of the colimit functor \(\colim \colon \Fun(C,T) \to T\). Then there is an equivalence
\[\Fun(C,T)/K^{n+1} \simeq \Exc^n(C,T).\]
In particular, if \(T\) is hypercomplete, we have an equivalence \(\Shv(C\catop; T)^{(n)} \simeq \Exc^n(C,T)\).
Proof
We use the set \(\Sigma_T\) from Construction 5.89. Since \(K=\Sigma_T^s\) is the kernel of a morphism of logoi, it is a congruence. It follows from the two universal properties that
\[K=\Sigma_T^c:\]
the inclusion \(\Sigma_T^c\subseteq K\) holds because \(K\) is a congruence containing \(\Sigma_T\), while \(K\subseteq\Sigma_T^c\) holds because \(\Sigma_T^c\) is strongly saturated. Moreover, \(\Sigma_T\) is closed under diagonals up to isomorphisms. Indeed, the diagonal of \(y_G(f)\) is obtained by applying \(y_G\) to the codiagonal of \(f\), since \(C\) has pushouts; the same description applies to every iterated diagonal. The ABFJ formula now gives \[K=\Sigma_T^c=\Sigma_T^m.\]
By Lemma 5.74, the power \(K^{n+1}\) is generated as an acyclic class by the \((n+1)\)-fold pushout products of maps in \(\Sigma_T\).For maps \(f_i\colon A_i\to B_i\) in \(C\), let \[f_0\boxplus\dots\boxplus f_n\colon[1]^{n+1}\longrightarrow C\]
be their external coproduct cube. It is strongly cocartesian. The iterated pushout product of the corresponding opposite Yoneda maps is the cocartesian gap map of the Yoneda image of this cube. Every strongly cocartesian cube in \(C\) is a cobase change of such a free cocartesian cube, and base changes of its Yoneda gap map correspond to such cobase changes. Consequently, locality with respect to the \((n+1)\)-fold pushout products is equivalent to carrying every strongly cocartesian \((n+1)\)-cube to a cartesian cube. This is the content of [Anel et al. 2025, Lemmas 4.4.2--4.4.4]; tensoring with the generators \(G\in\mathcal G\) makes the same argument detect cartesian cubes in \(T\). Thus the \(K^{n+1}\)-local objects are precisely the \(n\)-excisive functors, proving the first equivalence.Now suppose that \(T\) is hypercomplete, and put \(E:=\Shv(C\catop;T)=\Fun(C,T)/K^{\mono}\). Since \(\Fun(C,T)/K\iso T\) is hypercomplete, Lemma 5.67 gives \[K=(K^{\mono})^{\hyp}.\]
It follows that the image of \(K\) in \(E\) is precisely the congruence \(\Conn_\infty(E)\). Quotient functors preserve acyclic products, see [Anel et al. 2025, Proposition 3.6.1], so the image of \(K^{n+1}\) is \(\Conn_\infty(E)^{n+1}\). The first part now identifies \[E^{(n)}=E/\Conn_\infty(E)^{n+1}\iso\Exc^n(C,T).\]
References
- Mathieu Anel, Georg Biedermann, Eric Finster, André Joyal. Left-exact localizations of $\infty$-topoi III: The acyclic product. 2025.