Definitions and examples

Proof. Set \(X:=\{x_i\}_{i\in I}\), \(X_K:=\{x_{u(k)}\}_{k\in K}\) and \(Y:=\{y_j\}_{j\in J}\), and factor \(\alpha \) as the backwards span \(I\xleftarrow {u}K\xrightarrow {=}K\) followed by the active span \(\gamma \colon K\xleftarrow {=}K\xrightarrow {v}J\). By condition (3) of Definition 14.1.1, the projection-induced map \(X\to X_K\) is \(p_{\Oo }\)-cocartesian. Its universal property gives an equivalence \[ \Hom _{\Oo ^{\otimes }}^{\alpha }(X,Y) \simeq \Hom _{\Oo ^{\otimes }}^{\gamma }(X_K,Y). \] Since \(Y=\prod _{j\in J}y_j\) and \(p_{\Oo }\) preserves products, the right-hand side is equivalent to \[ \prod _{j\in J}\Hom _{\Oo ^{\otimes }}^{\rho _j\circ \gamma }(X_K,y_j), \] where \(\rho _j\colon J\hookleftarrow \{j\}\xrightarrow {=}\{j\}\) is the corresponding projection. The composite \(\rho _j\circ \gamma \) is represented by the span \(K\hookleftarrow v^{-1}(j)\to \{j\}\). A second application of condition (3), now to \(v^{-1}(j)\hookrightarrow K\), therefore identifies its \(j\)-th factor with \[ \Oo (\{x_{u(k)}\}_{k\in v^{-1}(j)};y_j). \] The resulting equivalence is induced by the top map in the displayed square, so that square is a pullback. Every map used in its construction is induced by products and their projections, and is therefore preserved by morphisms of \(\infty \)-operads. This proves naturality. □

Proof. For every finite set \(I\), the fiber \(\Oo ^{\otimes }_I\) is equivalent to \(\Oo _{\lra {1}}^I\), and similarly for \(\Pp \). Under these equivalences, the functor induced by \(f^{\otimes }\) is the product of \(I\) copies of the underlying functor of \(f\), so it is essentially surjective. Thus \(f^{\otimes }\) is essentially surjective on objects.

To prove full faithfulness, let \(X=\{x_i\}_{i\in I}\) and \(Y=\{y_j\}_{j\in J}\) be objects of \(\Oo ^{\otimes }\). There is a commutative triangle

For a span \(\alpha \colon I\xleftarrow {u}K\xrightarrow {v}J\), the naturality in Lemma 14.1.6 identifies the induced map on fibers over \(\alpha \) with \[ \prod _{j\in J}\Oo (\{x_{u(k)}\}_{k\in v^{-1}(j)};y_j) \longrightarrow \prod _{j\in J}\Pp (\{f(x_{u(k)})\}_{k\in v^{-1}(j)};f(y_j)). \] This is an equivalence because \(f\) is fully faithful. Hence the top map in the triangle is an equivalence, since it is a map of animae over a fixed base which is an equivalence on every fiber. Thus \(f^{\otimes }\) is fully faithful and essentially surjective, and hence an equivalence over \(\Span (\Fin )\). □

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 \(\prod _{i \in I} \prod _{t \in T_i} X^t_i\). □

Proof. The maps \(Y \to Y_j\) exhibit \(Y\) as a product if and only if for every \(X \in \Oo ^{\otimes }_I\) the top map in the following square is an equivalence:

Since the bottom map is an equivalence, this is equivalent to the square being a pullback square, which is in turn equivalent to the condition that the top map induces isomorphisms on all fibers over \(\alpha \in \Hom _{\Span (\Fin )}(I,J)\). □

Proof. By assumption \(\widetilde {f}\) is inert for every \(f\). Conversely, we saw in the proof of (i) of the previous proposition that for every backwards span \(I \xleftarrow {\smash {f}} J \xrightarrow {=} J\) and every \(X \in \Oo ^{\otimes }_I\) the map \(\widetilde {f} \colon X \simeq \prod _{i \in I} X_i \to \prod _{j \in J} X_{f(j)}\) is a cocartesian lift of \(f\) starting in \(X\), so by uniqueness it follows that every cocartesian lift is of this form.

If \(\phi ^{\otimes }\) preserves finite products, it in particular preserves the maps \(\widetilde {f}\colon \prod _{i \in I} X_i \to \prod _{j \in J} X_{f(j)}\), hence all inert morphisms. Conversely, if \(\phi ^{\otimes }\) preserves inert morphisms, it preserves the projection maps \(Y \to Y_j\) for \(Y \in \Oo ^{\otimes }_J\) exhibiting \(Y\) as a product of the \(Y_j\), hence it preserves every nonempty finite product. It also preserves the terminal object: since \(\phi ^{\otimes }\) lies over \(\Span (\Fin )\), it carries the unique object of \(\Oo ^{\otimes }_{\emptyset }\simeq *\) to the unique object of \(\Pp ^{\otimes }_{\emptyset }\simeq *\). □

Proof. We claim that an inverse is given by left Kan extension.

Step 1: Given a functor \(F\colon C \to D\), we show that its left Kan extension \(i_!F\) along \(i\) exists. By the pointwise formula for left Kan extensions, it suffices to show that each of the colimits \[ (i_!F)(x_1, \dots , x_n) = \colim _{y \in C_{/(x_1, \dots , x_n)}} F(y) \] exists. Observe that the canonical map \[ \bigsqcup _{k=1}^n C_{/x_k} \to C_{/(x_1, \dots , x_n)} := C \times _{\Fin (C)} \Fin (C)_{/(x_1, \dots , x_n)} \] sending \((\phi \colon y \to x_k)\) to the map \((\{k\} \hookrightarrow \lra {n}, \phi )\) is an equivalence. Since \(D\) has finite coproducts, we see that colimits indexed by the slices \(C_{/(x_1, \dots , x_n)}\) exist in \(D\) if and only if colimits indexed by each \(C_{/x_k}\) exist. But the latter is clear: as \(C_{/x_k}\) admits a terminal object, colimits always exist and are given by evaluation at that terminal object. This shows that \(i_!F\) exists and that it is given on objects by \[ (i_!F)(x_1, \dots , x_n) = F(x_1) \sqcup \dots \sqcup F(x_n). \]

Step 2: It follows from the pointwise description of \(i_!F\) that it preserves finite coproducts. In particular, the restriction functor thus admits a left adjoint \[ i_!\colon \Fun (C,D) \to \Fun ^{\amalg }(\Fin (C),D). \]

Step 3: We now show that the unit \(\id \to i^*i_!\) and counit \(i_!i^* \to \id \) are natural isomorphisms. The unit is a natural isomorphism because \(i\) is fully faithful. If \(G\colon \Fin (C)\to D\) preserves finite coproducts, then the counit evaluated at \((x_1,\dots ,x_n)\) is the canonical map \[ G(x_1)\sqcup \dots \sqcup G(x_n) \longrightarrow G(x_1,\dots ,x_n), \] which is an isomorphism because \((x_1,\dots ,x_n)\) is the coproduct of the singleton families \((x_i)\) in \(\Fin (C)\). □

Proof. Let us first check that \(\Triv _C\) is an \(\infty \)-operad. By Construction 14.1.16, the \(\infty \)-category \(\Fin (C\catop )\) admits finite coproducts and \(q\) preserves them, so \(\Triv _C^{\otimes } = \Fin (C\catop )\catop \) admits finite products and \(q\catop \) preserves them; as the inclusion \(\Fin \catop \hookrightarrow \Span (\Fin )\) preserves finite products by part (1) of Lemma 13.3.8, the same is true for \(p_{\Triv _C}\). Since the products of \(\Triv _C^{\otimes }\) are the concatenations of tuples, the fiber over a finite set \(I\) is \(C^I\), whose decomposition \(C^I \simeq \prod _{i \in I}C\) is the one required in the definition of an \(\infty \)-operad.

For the final condition, consider objects \(X_i = \{x^i_a\}_{a \in A_i}\) of \(\Triv _C^{\otimes }\) for \(i \in I\) and a map \(f\colon J \to I\) of finite sets. Read in \(\Fin (C\catop )\), the morphism \(\widetilde {f}\colon \prod _{i \in I}X_i \to \prod _{j \in J}X_{f(j)}\) is the map \(\coprod _{j \in J}X_{f(j)} \to \coprod _{i \in I}X_i\) which restricts on the \(j\)-th summand to the inclusion of \(X_{f(j)}\); it lies over the map \(\bigsqcup _{j \in J}A_{f(j)} \to \bigsqcup _{i \in I}A_i\), \((j,a) \mapsto (f(j),a)\), and all of its components in \(C\catop \) are identities. It is therefore \(q\)-cartesian, and hence \(\widetilde {f}\) is \(q\catop \)-cocartesian. It remains \(p_{\Triv _C}\)-cocartesian after composing with the inclusion \(\Fin \catop \hookrightarrow \Span (\Fin )\): every morphism of \(\Triv _C^{\otimes }\) lies over a backwards span, and composing a non-backwards span with a backwards span is again non-backwards, so the mapping animae over the remaining spans are empty on both sides.

The description of the multimorphism animae follows: the active span \(I \xleftarrow {=} I \to \lra {1}\) lies in \(\Fin \catop \) only when \(I\) is a singleton, in which case it is the identity of \(\lra {1}\). Thus \(\Triv _C(\{x_i\}_{i \in I};y)\) is empty unless \(I = \{i_0\}\), in which case it is the anima of morphisms \(x_{i_0} \to y\) in the fiber \((\Triv _C)_{\lra {1}} \simeq C\).

We turn to the universal property, which will follow from the universal property of \(\Fin (C\catop )\). Indeed, applying Lemma 14.1.17 to \(C\catop \) and passing to opposite categories, we see that restriction along the inclusion \(j\colon C \hookrightarrow \Triv _C^{\otimes }\) of the fiber over \(\lra {1}\) induces an equivalence \[ j^*\colon \Fun ^{\times }(\Triv _C^{\otimes },E) \iso \Fun (C,E) \] for every \(\infty \)-category \(E\) that admits finite products. This equivalence is natural in \(E\), so applying it to \(E = \Oo ^{\otimes }\) and to \(E = \Span (\Fin )\) yields a commutative square

The composite \(p_{\Triv _C} \circ j\) is the constant functor at \(\lra {1}\), so passing to the fibers over \(p_{\Triv _C}\) and \(\const _{\lra {1}}\) gives equivalences \[ \Fun _{\Op _{\infty }}(\Triv _C,\Oo ) \iso \Fun (C,\Oo ^{\otimes })\times _{\Fun (C,\Span (\Fin ))}\{\const _{\lra {1}}\} \simeq \Fun (C,\Oo _{\lra {1}}), \] where the second equivalence holds because \(\Fun (C,-)\) preserves pullbacks. By construction, this equivalence is given by restriction to the underlying \(\infty \)-categories. □

Proof. The equivalence of Lemma 14.1.20 is given by passing to underlying \(\infty \)-categories and precomposing with the isomorphism \(C \iso (\Triv _C)_{\lra {1}}\), and is thus natural in both variables. Passing to groupoid cores yields the corresponding equivalence of mapping animae, so it exhibits \(\Triv _{(-)}\) as a left adjoint of \((-)_{\lra {1}}\). Taking \(\Oo = \Triv _D\) for an \(\infty \)-category \(D\), the equivalence \[ \Fun _{\Op _{\infty }}(\Triv _C,\Triv _D) \iso \Fun (C,(\Triv _D)_{\lra {1}}) \simeq \Fun (C,D) \] is inverse to the map induced by \(\Triv _{(-)}\), which is therefore fully faithful. □

Proof. Recall that \(\Triv ^{\otimes } = \Fin \catop \) and that \(p_{\Triv }\colon \Fin \catop \hookrightarrow \Span (\Fin )\) is the inclusion. The product operad \(\Triv \times \Oo \) is therefore represented by the pullback

By the defining properties of an \(\infty \)-operad, the left vertical map is a cocartesian fibration whose cocartesian straightening preserves finite products and sends \(\lra {1}\) to \(\Oo _{\lra {1}}\). It is therefore equivalent to \[ \Fin \catop \to \Cat _{\infty }, \qquad I \mapsto \Oo _{\lra {1}}^I. \] By Remark 14.1.19, its cocartesian unstraightening is precisely \[ q_{\Oo _{\lra {1}}}\catop \colon \Fin (\Oo _{\lra {1}}\catop )\catop \to \Fin \catop , \] which is the total category of \(\Triv _{\Oo _{\lra {1}}}\). □