Corollary 15.2.18. The following conditions are equivalent for an \(\infty \)-operad \(\Oo \):
- (1)
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The \(\infty \)-operad \(\Oo \) is cocartesian;
- (2)
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The \(\infty \)-category \(\Oo ^{\otimes }\) is semiadditive;
- (3)
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The following two conditions are satisfied:
- (a)
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For every color \(x \in \Oo ^{\simeq }\), the anima \(\Oo (\emptyset ; x)\) of nullary operations is contractible;
- (b)
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For objects \(\{x_i\}_{i \in I}, \{y_j\}_{j \in J}\) of \(\Oo ^{\otimes }\) and a color \(z \in \Oo ^{\simeq }\), composition with the morphisms \(\{x_i\}_{i \in I} \to \{x_i\}_{i \in I} \sqcup \{y_j\}_{j \in J}\) and \(\{y_j\}_{j \in J} \to \{x_i\}_{i \in I} \sqcup \{y_j\}_{j \in J}\) obtained from condition (a) provides an equivalence \[ \Oo (\{x_i\}_{i \in I} \sqcup \{y_j\}_{j \in J}; z) \simeq \Oo (\{x_i\}_{i \in I};z) \times \Oo (\{y_j\}_{j \in J};z). \]
Proof. The equivalence between (2) and (3) is a simple unwinding of definitions: condition (a) says that the terminal object of \(\Oo ^{\otimes }\) is also initial, and condition (b) says that the maps \(X \simeq X \times * \to X \times Y\) and \(Y \simeq * \times Y \to X \times Y\) in \(\Oo ^{\otimes }\) exhibit the product also as a coproduct.
Since \(C^{\amalg } = \Span _{\ct ,\all }(\Fin (C))\) is semiadditive, it is clear that (1) implies (2). To see that (2) implies (1), let \(\Oo \) be such an \(\infty \)-operad, and write \(C := \Oo ^{\otimes }_{\lra {1}}\) for its underlying \(\infty \)-category. By Proposition 15.2.16, the identity \(\id _C\colon C \to \Oo _{\lra {1}}\) extends to a morphism of \(\infty \)-operads \(\OpCocart _C \to \Oo \). It is clear that it induces an equivalence on colors, so it remains to show that it induces equivalences on all multimorphism animae: \[ \OpCocart _C(\{x_i\}_{i \in I}; y) \simeq \Oo (\{x_i\}_{i \in I}; y). \] But using condition (3), we may inductively reduce this to the case of the one-point set \(I = \lra {1}\), where it is simply the equality \(C = \Oo ^{\otimes }_{\lra {1}}\). The resulting fully faithful operad map is an equivalence by Lemma 14.1.8. □
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