Equivalent characterization operads

Proof. Throughout the proof we write \(p = p_{\Oo }\) for simplicity. First assume that \(\Oo ^{\otimes }\) is an \(\infty \)-operad. We show that (i)–(iii) are satisfied:

(i)

Consider a morphism \(f\colon J \to I\) in \(\Fin \) and consider \(X \in \Oo ^{\otimes }_I\). By condition (2), we may uniquely write \(X\) as a product \(\prod _{i \in I} X_i\) of objects \(X_i \in \Oo ^{\otimes }_{\{i\}}\). We then define \(Y := \prod _{j \in J} X_{f(j)}\) and define \(\widetilde {f}\colon X \to Y\) to be the map whose \(j\)-th component \(X = \prod _{i \in I} X_i \to X_{f(j)}\) is the projection map onto \(X_{f(j)}\). Then assumption (3) tells us that \(\widetilde {f}\) is a \(p\)-cocartesian lift of the span \(I \xleftarrow {f} J \xrightarrow {=} J\).

(iii)

Using (2), we may write the object \(Y \in \Oo ^{\otimes }_J\) uniquely as a product \(\prod _{j \in J} Y_j\). By assumption (3), the projection maps \(Y = \prod _{j \in J} Y_j \to Y_j\) are cocartesian lifts of the maps \(\rho _j\colon J \hookleftarrow \{j\} \xrightarrow {=} \{j\}\) in \(\Span (\Fin )\), verifying (iii).

(ii)

To show that the map \(\Oo ^{\otimes }_I \to \prod _{i \in I} \Oo ^{\otimes }_{\{i\}}\) is an equivalence, it suffices to show it is a left-inverse to the product functor \(\prod _{i \in I} \Oo ^{\otimes }_{\{i\}} \iso \Oo ^{\otimes }_I\), as this was assumed to be an equivalence. This is again a consequence of the fact that the projection maps \(\prod _{i \in I} X_i \to X_i\) are cocartesian lifts of \(\rho _i\).

We now show that conditions (i)-(iii) imply that \(\Oo ^{\otimes }\) is an \(\infty \)-operad. We check conditions (1)-(3):

(1)

For finite products in \(\Oo ^{\otimes }\), consider objects \(Y^1, \dots , Y^n\) in \(\Oo ^{\otimes }\) and set \(T_i := p(Y^i)\) and \(T := \bigsqcup _i T_i\). Picking cocartesian lifts \(Y^i \to Y^i_t\) over each map \(\rho _t\), condition (iii) tells us that these maps exhibit each \(Y^i\) as a product \(\prod _{t \in T_i} Y^i_t\). Using (ii), we may pick some object \(Y \in \Oo ^{\otimes }_T\) whose cocartesian transport along each \(\rho _t\) is \(Y^i_t\), and we similarly see that the maps \(Y \to Y^i_t\) exhibit \(Y\) as a product \(\prod _{i=1}^n \prod _{t \in T_i} Y^i_t\). Consequently \[ \Hom _{\Oo ^{\otimes }}(Z,Y) \simeq \prod _{i=1}^n\prod _{t\in T_i}\Hom _{\Oo ^{\otimes }}(Z,Y^i_t) \simeq \prod _{i=1}^n\Hom _{\Oo ^{\otimes }}(Z,Y^i) \] for every \(Z\in \Oo ^{\otimes }\), so \(Y \simeq \prod _{i=1}^n Y^i\). Thus \(\Oo ^{\otimes }\) admits finite products. It is also clear from this calculation that \(p \colon \Oo ^{\otimes } \to \Span (\Fin )\) preserves products.

(2)

It will suffice to argue that the product functor \(\prod _{s \in S} \Oo ^{\otimes }_s \to \Oo ^{\otimes }_S\) is a right-inverse to the assumed equivalence \(\Oo ^{\otimes }_S \iso \prod _{s \in S} \Oo ^{\otimes }_{\{s\}}\) from (ii). But (iii) implies that for \(X_s \in \Oo ^{\otimes }_{\{s\}}\) the projection maps \(\prod _{s \in S} X_s \to X_s\) are the cocartesian lifts of the \(\rho _s\), which gives the claim.

(3)

Consider a morphism \(f\colon J \to I\) and let \(X_i \in \Oo ^{\otimes }\). We have to show that the map \(\widetilde {f}\colon \prod _{i \in I} X_i \to \prod _{j \in J} X_{f(j)}\) is \(p\)-cocartesian.

Let us first prove the special case where \(p(X_i) = *\) for all \(i \in I\). Let \(\overline {f}\colon \prod _{i \in I} X_i \to Y\) be a cocartesian lift of \(f\). By the universal property of cocartesian lifts, there is a morphism \(Y \to \prod _{j \in J} X_{f(j)}\) which we wish to show is an equivalence. Note that this map lives in the fiber over \(J\), so by condition (2) it suffices to check this after applying cocartesian transport along each of the maps \(\rho _j\). Note that we have \((\rho _j)_!Y \simeq X_{f(j)}\) since the composites of the spans \[ I \xleftarrow {f} J \xrightarrow {=} J \qquadtext { and } \rho _j\colon J \hookleftarrow \{j\} \to * \] is the span \(\rho _{f(j)}\colon I \hookleftarrow \{f(j)\} \xrightarrow {=} \{f(j)\}\). The projections from \(\prod _{j \in J}X_{f(j)}\) are cocartesian lifts of the maps \(\rho _j\) by the construction of products in part (1), so also \((\rho _j)_!(\prod _{j \in J} X_{f(j)}) \simeq X_{f(j)}\). This proves the claim in the special case.

The general case is an immediate consequence by writing each \(X_i\) as a product \(\prod _{t \in T_i} X^t_i\) and applying the previous argument to \(\prod _{i \in I} \prod _{t \in T_i} X^t_i\). □