Proof. In the listed order, these results are [Reference Haugseng, Hebestreit, Linskens and NuitenHHLN23b, Lemma 2.4, Theorem 2.18, Lemma 2.14, Proposition 2.15, and Proposition 4.9].

Notation 2.6. We will denote maps in $\mathcal {N}$ as .

Warning 2.8. Beware that the condition that $\mathcal {X} \to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ admits cocartesian lifts of morphisms in $\mathcal {F}^{{\mathrm{op}}}$ is strictly stronger than demanding that the pullback $\mathcal {X} \times _{\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})} \mathcal {F}^{{\mathrm{op}}}\to \mathcal {F}^{\mathrm{op}}$ be a cocartesian fibration.

Proof. We note that $\mathcal {C}^{{\mathcal {N}\text{-}\amalg }}$ is Barwick’s unfurling construction, and all claims about it follow immediately from [Reference Haugseng, Hebestreit, Linskens and NuitenHHLN23b, Theorem 3.1, Example 3.4, Corollary 3.18].

It remains to see that $\operatorname {\mathrm {Triv}}(\mathcal {C})$ is a cocartesian fibration and that ${{\mathrm{incl}}}_{\mathcal {C}}$ preserves cocartesian lifts of backwards maps. The former follows immediately from the fact that $\mathcal {F}^{{\mathrm{op}}}\to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ is itself an $(\mathcal {F},\mathcal {N})$ -operad, which in turn follows immediately from the fact that the subcategory $\mathcal {F}^{{\mathrm{op}}}\subset \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ is right cancelable.

Finally, to see that the map ${{\mathrm{incl}}}_{\mathcal {C}}$ preserves cocartesian edges over $\mathcal {F}^{{\mathrm{op}}}$ we may appeal to the characterization of cocartesian edges in $\mathcal {C}^{{\mathcal {N}\text{-}\amalg }}$ from Theorem 3.1 of op. cit.

Notation 2.18. We denote arrows in M by and arrows in E by .

Proof. The first item is the content of [Reference Barkan, Haugseng and SteinebrunnerBHS25, Proposition 2.2.2]. The adjunction in (2) is proven by combining Proposition 2.1.4 and Proposition 2.2.4 of op. cit, while $\mathrm{Cat}$ -linearity of $\operatorname {\mathrm {Env}}(-)$ is proven in Proposition 5.3.4. Finally, (3) is Proposition 2.3.2 of op. cit.

Proof of Proposition 2.26.

We begin by showing the first addendum, that $(-)^{{\mathcal {N}\text{-}\amalg }}$ preserves all limits. Note that the forgetful functor $\mathrm {Op}^\ast _{\mathcal {N}}(\mathcal {F}) \to \mathrm{Cat}_{/\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})}$ is a conservative right adjoint. The composite of $(-)^{{\mathcal {N}\text{-}\amalg }}$ with it factors as

where $\Phi $ sends a cartesian fibration $\mathcal {X} \to \mathcal {F}$ to the adequate triple $(\mathcal {X}, \mathcal {X}_{\mathrm{ct}}, \mathcal {X} \times _{\mathcal {F}} \mathcal {N})$ over $(\mathcal {F},\mathcal {F},\mathcal {N})$ (the former is indeed adequate by [Reference Haugseng, Hebestreit, Linskens and NuitenHHLN23b, Proposition 2.6]). By Theorem 2.5 limits in $\operatorname {\mathrm {AdTrip}}$ are computed pointwise and $\operatorname {\mathrm {Span}}$ is a right adjoint, so that it remains to see that the functors

$$\begin{align*}\mathrm{Cat}_{/\mathcal{F}}^{\mathrm{ct}} \to \mathrm{Cat}_{/\mathcal{F}},\ \mathcal{X} \mapsto \mathcal{X}^{\mathrm{ct}} \quad\mathrm{and}\quad \mathrm{Cat}_{/\mathcal{F}}^{\mathrm{ct}} \to \mathrm{Cat}_{/\mathcal{N}},\ \mathcal{X} \mapsto \mathcal{X} \times_{\mathcal{F}} \mathcal{N} \end{align*}$$

preserve limits. The second functor factors as the right adjoint $\mathrm{Cat}_{/\mathcal {F}}^{\mathrm{ct}} \to \mathrm{Cat}_{/\mathcal {N}}^{\mathrm{ct}}$ followed by the forgetful $\mathrm{Cat}_{/\mathcal {N}}^{\mathrm{ct}} \to \mathrm{Cat}_{/\mathcal {N}}$ , which preserves limits by [Reference Gepner, Haugseng and NikolausGHN17, Theorem 1.2]. Analogously, the first functor actually also lands in $\mathrm{Cat}^{\mathrm{ct}}_{/\mathcal {F}}$ , where under the equivalence with $\operatorname {\mathrm {Fun}}(\mathcal {F}^{\mathrm{op}},\mathrm{Cat})$ it corresponds to postcomposing with the groupoid core functor, which clearly preserves limits.

Next, we will employ the adjoint functor theorem of [Reference LurieLur09, Corollary 5.5.2.9] to see that $(-)^{{\mathcal {N}\text{-}\amalg }}$ admits a right adjoint. The domain is clearly presentable and the codomain is by [Reference Barkan, Haugseng and SteinebrunnerBHS25, Proposition 2.3.7] and so it remains to see that the functor preserves colimits. This can be tested after postcomposing with the conservative left adjoint $\operatorname {\mathrm {Env}}(-)$ and then evaluating at every $A \in \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ . We may consider the following pullbacks, where $\pi _A$ denotes the composite $\mathcal {N}_{/A} \to \mathcal {N} \to \mathcal {F}$ :

The bottom left square is a pullback by definition, while the bottom right is by the fact that forward maps in a span category are left cancelable.Footnote ⁷ The top right square is a pullback by definition of the envelope. The rectangle consisting of the top middle and right squares is a pullback by the definition of $\mathcal {C}^{{\mathcal {N}\text{-}\amalg }}$ and the fact that $\operatorname {\mathrm {Ar}}(\mathcal {N}) \to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ factors as . Finally, to see that the rectangle consisting of the entire top row is a pullback, recall that $\operatorname {\mathrm {Span}}$ is a right adjoint from the category of adequate triples. Viewing $\mathcal {N}_{/A} \simeq \operatorname {\mathrm {Span}}_{\simeq ,\mathrm{all}}(\mathcal {N}_{/A})$ as span category with backwards maps only the equivalences, we compute that the pullback is the span category with forwards maps $\mathrm{Un}^{\mathrm{ct}}(\pi _A^*{\mathcal {C}})$ , and backwards morphisms the subcategory on cartesian morphisms lying over $\operatorname {\mathrm {\iota }}(\mathcal {N}_{/A})$ . However a cartesian map lying over an equivalence is itself an equivalence, and thus the pullback has no nontrivial backwards maps.

We conclude that the following diagram commutes:

Each functor in the bottom composite commutes with colimits: for the first two and the last one this is clear and for $\mathrm{fgt}$ this was shown in [Reference RamziRam26, Corollary A.5]. Thus, the top composite preserves colimits for all $A\in \mathcal {F}$ , and hence also $(-)^{{\mathcal {N}\text{-}\amalg }}$ does.

Next, since $(-)^{{\mathcal {N}\text{-}\amalg }}$ preserves limits, it is symmetric monoidal for the cartesian symmetric monoidal structures. By Remark 2.13 the $\mathrm{Cat}$ -modules structures on both source and target are similarly defined by symmetric monoidal functors from $\mathrm{Cat}$ , and so it will suffice for $\mathrm{Cat}$ -linearity to show that the diagram

commutes as a diagram of symmetric monoidal functors or, equivalently, as a diagram of ordinary functors. This follows at once from the equivalences

$$ \begin{align*} (\mathrm{const} K)^{{\mathcal{N}\text{-}\amalg}} &= \operatorname{\mathrm{Span}}_{\mathrm{ct},\mathcal{N}}(\mathrm{Un}^{\mathrm{ct}}(\mathrm{const} K)) \simeq \operatorname{\mathrm{Span}}(K_{\simeq,\mathrm{all}} \times ({\mathcal{F}},{\mathcal{F}},{\mathcal{N}}))\\ &\simeq \operatorname{\mathrm{Span}}_{\simeq,\mathrm{all}}(K)\times \operatorname{\mathrm{Span}}_{\mathcal{N}}({\mathcal{F}}) \simeq K \times \operatorname{\mathrm{Span}}_{\mathcal{N}}({\mathcal{F}})=\mathrm{const} K, \end{align*} $$

where $K_{\simeq ,\mathrm{all}}$ denotes K viewed as adequate triple with only forwards maps.

On the other hand, $\operatorname {\mathrm {Triv}}(-)$ is clearly a $\mathrm{Cat}$ -linear left adjoint, since we may simply restrict the $\mathrm{Cat}$ -linear adjunction

induced by the inclusion $i\colon \mathcal {F}^{{\mathrm{op}}} \to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ to an adjunction between $\mathrm {Cat}^\ast (\mathcal {F})$ and $\mathrm {Op}^\ast _{\mathcal {N}}(\mathcal {F})$ . This also shows that the right adjoint of $\operatorname {\mathrm {Triv}}$ is $\mathrm{fgt} = i^*$ , with the evident unit. That $\operatorname {\mathrm {Triv}}(-)$ is fully faithful follows immediately from the fact that $\mathcal {F}^{\mathrm{op}} \times _{\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})}\mathcal {F}^{\mathrm{op}}$ is equivalent to $\mathcal {F}^{\mathrm{op}}$ . Finally, in view of the pullback in Lemma 2.16 defining the natural map ${{\mathrm{incl}}}\colon \operatorname {\mathrm {Triv}} \Rightarrow (-)^{{\mathcal {N}\text{-}\amalg }}$ , it follows immediately that ${{\mathrm{incl}}}$ is also a $\mathrm{Cat}$ -linear transformation.

Proof. We want to show that the unique diagram

is a colimiting cocone in $\mathrm {Op}^\ast _{{\mathcal {N}}}({\mathcal {F}})$ , where $1$ denotes the final object of $\mathrm {Cat}^\ast ({\mathcal {F}})$ , so that $1^{\mathcal {N}\text{-}\amalg }=\operatorname {\mathrm {Span}}_{\mathcal {N}}({\mathcal {F}})$ is the final $({\mathcal {F}},{\mathcal {N}})$ -preoperad. As $(-)^{\mathcal {N}\text{-}\amalg }$ is a left adjoint, it suffices for this that the colimit of the ’s in $\mathrm {Cat}^\ast ({\mathcal {F}})$ is terminal. As $\operatorname {\mathrm {Fun}}({\mathcal {F}}^{\mathrm{op}},\operatorname {\mathrm {Spc}})\hookrightarrow \operatorname {\mathrm {Fun}}({\mathcal {F}}^{\mathrm{op}},\mathrm{Cat})=\mathrm {Cat}^\ast ({\mathcal {F}})$ is both a left and a right adjoint, this just amounts to the well known fact that the colimit of the Yoneda embedding is terminal.

Construction 2.35. The target projection $t \colon \operatorname {\mathrm {Ar}}_{\mathcal {N}}(\mathcal {F}) \to \mathcal {F}$ is a cartesian fibration classifying the functoriality of the slices $\mathcal {N}_{/B}$ in pullbacks along arbitrary maps $A \to B$ in $\mathcal {F}$ . We denote this cartesian straightening by $\mathcal {N}_{/-} \colon \mathcal {F}^{\mathrm{op}} \to \mathrm{Cat}$ . Now let $A \in \mathcal {F}$ and consider the pullback on the left:

Clearly $tp$ is a cartesian fibration, and a general morphism in it can be depicted as the diagram on the right. It is $tp$ -cartesian precisely if the square is a pullback. We denote the cartesian straightening of $tp$ by

In fact, $\mathcal {F}_{/A} \times _{\mathcal {F}} \mathcal {N}_{/-} \in \mathrm {Cat}^\ast (\mathcal {F})$ is covariantly functorial in A via postcomposition; this follows immediately from the fact that given $f \colon A \to B$ in $\mathcal {F}$ , the functor $f_! \colon \mathcal {F}_{/A} \to \mathcal {F}_{/B}$ over $\mathcal {F}$ is a map of cartesian fibrations. We therefore obtain a functor

$$\begin{align*}\mathcal{F}_{/\bullet} \times_{\mathcal{F}} \mathcal{N}_{/-} \colon \mathcal{F} \to \mathrm{Cat}^\ast(\mathcal{F}),\ A \mapsto \mathcal{F}_{/A} \times_{\mathcal{F}} \mathcal{N}_{/-}. \end{align*}$$

Proof. Let $X \in \mathcal {F}$ , and consider a map in $\mathcal {N}$ . Clearly the pullback $n^* \colon \mathcal {N}_{/B} \to \mathcal {N}_{/A}$ admits a left adjoint given by postcomposition with n, and similarly one checks that we obtain an adjunction

For example, the unit map at some is given by the map $\varphi $ in the following commutative diagram

Note that $\varphi $ is indeed a map in $\mathcal {N}$ by left cancellability, and this is one of the crucial points where this assumption is needed.

The fact that the pullback square (1) induces a left-adjointable square, that is, that the induced square

(2)

is horizontally left adjointable, is a simple but tedious exercise in pullback pasting.

For the second claim, we have to show that is right $\mathcal {N}$ -coadjointable. Since 2-functors preserve adjointable squares (see, e.g., [Reference Heyer and MannHM24, Lemma D.2.7]), it suffices to show that $\mathcal {F}_{/\bullet } \times _{\mathcal {F}} \mathcal {N}_{/-} \colon \mathcal {F} \to \mathrm {Cat}^\ast (\mathcal {F})$ is right $\mathcal {N}$ -coadjointable.

So fix in $\mathcal {N}$ . We need to show that the $\mathcal {F}$ -functor $n_! \colon \mathcal {F}_{/A} \times _{\mathcal {F}} \mathcal {N}_{/-} \to \mathcal {F}_{/B} \times _{\mathcal {F}} \mathcal {N}_{/-}$ induced by postcomposing with n admits a right adjoint. For some fixed $X \in \mathcal {F}$ , one checks that pullback along induces a right adjoint

$$\begin{align*}n^*(X) \colon \mathcal{F}_{/B} \times_{\mathcal{F}} \mathcal{N}_{/X} \to \mathcal{F}_{/A} \times_{\mathcal{F}} \mathcal{N}_{/X} \end{align*}$$

to $n_!(X)$ . By Remark 2.14 the condition for these to assemble into a right adjoint to the $\mathcal {F}$ -functor $n_!$ is that for each $f \colon X \to Y$ the square

is vertically right adjointable, which is again a simple exercise in pullback pasting.

Finally, it remains to see that a pullback square as in (1)

which is horizontally right adjointable, that is, that the Beck–Chevalley transformation $f^{\prime }_!n^{\prime *} \Rightarrow n^*f_!$ of $\mathcal {F}$ -functors is an equivalence. This can be checked pointwise at some $X \in \mathcal {F}$ , where the argument is entirely analogous to the horizontal left adjointability of (2) considered above.

Proof. Recall from Example 2.17 that is naturally equivalent to $\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F}_{/A})\to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ , so that its envelope is then given by

Identifying $\operatorname {\mathrm {Ar}}_{\mathcal {N}}(\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F}))$ as in Remark 2.25 and appealing to Theorem 2.5, we see that is naturally equivalent to the category

(3) $$ \begin{align} \operatorname{\mathrm{Span}}_{\mathrm{pb},\mathcal{N}}(\operatorname{\mathrm{Ar}}_{\mathcal{N}}(\mathcal{F})\times_{\mathcal{F}}\mathcal{F}_{/A}), \end{align} $$

whose morphisms are spans of the form

(4)

On the other hand, (3) also precisely agrees with by definition, see Definition 2.15 and Construction 2.35.

By Lemma 2.16(2) and the definition of the envelope, we obtain pullback squares

Note that a general morphism (4) in is cocartesian precisely if the map is an equivalence (cf. the description of cocartesian morphisms in $\operatorname {\mathrm {Env}}(1)= \operatorname {\mathrm {Ar}}_{\mathcal {N}}(\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F}))$ from Remark 2.25). Thus, is precisely the subcategory on the cocartesian morphisms. The identification of $\operatorname {\mathrm {Env}}({{\mathrm{incl}}})$ therefore follows from the general fact that for any functor $F \colon \mathcal {C} \to \mathrm{Cat}$ with cocartesian straightening $p \colon \mathcal {X} \to \mathcal {C}$ , the inclusion $\mathcal {X}_{\mathrm{cc}} \hookrightarrow \mathcal {X}$ of the subcategory of cocartesian morphisms corresponds under cocartesian straightening to the natural transformation $\operatorname {\mathrm {\iota }} F \Rightarrow F$ .

Proof. After Proposition 2.37, the only nontrivial fact is the identification of the bottom arrow. However, combining the fact that $\operatorname {\mathrm {Triv}}$ is fully faithful and that the natural Yoneda equivalence is given by evaluating at , we see that the bottom horizontal equivalence in the square is given by the composite

so the claim follows from the description of the unit $\eta $ of Theorem 2.23.

Proof. The claimed equivalence follows immediately from the identification of the equivalence

with $\operatorname {\mathrm {ev}}_{\mathrm{id}_A,\mathrm{id}_A}$ from Corollary 2.38. In the case that A is terminal, we see that it does not have any automorphisms in $\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ , hence by the Yoneda lemma also $\operatorname {\mathrm {hom}}_{\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})}(A,-)$ does not have any automorphisms, which yields the uniqueness of the equivalence.

Proof. The first claim is immediate from the fact that is essentially surjective, and $\mathbb {U}$ is given by restricting along this. For the second claim, recall that is the right adjoint in a $\mathrm{Cat}$ -linear adjunction. In particular, it preserves the $\mathrm{Cat}$ -cotensoring. Since it is also a 2-functor, it therefore sends the adjunction

witnessing that $\mathcal {C}$ has fiberwise K-colimits for $K \in \mathcal {K}$ to the analogous adjunction

which then shows that also admits fiberwise K-colimits.

Proof. This follows immediately from the fact that $\operatorname {\mathrm {Fun}}_{\mathcal {F}}^{\mathcal {N}}(-,\mathcal {C}) \colon \mathrm {NCat}^\ast _{\mathcal {N}}(\mathcal {F})^{\mathrm{op}} \to \mathrm{Cat}$ is a 2-functor (the mapping-functor of the 2-category $\mathrm {NCat}^\ast _{\mathcal {N}}(\mathcal {F})$ ), the identification from Corollary 2.38, and the fact that is right $\mathcal {N}$ -coadjointable by Lemma 2.36(3).

Proof. It is easy to see from the assumptions on p that the supposed inverse factors through $\mathcal {X}_{A \times B}$ and defines a section of the cocartesian transport $\Phi \colon \mathcal {X}_{A \times B} \to \mathcal {X}_A \times \mathcal {X}_B$ . In particular, $\Phi $ is essentially surjective. To see that $\Phi $ is also fully faithful, consider $X,Y \in \mathcal {X}_{A \times B}$ . We have the following two commutative diagrams

The bottom square on the right is a pullback since $X \to \mathrm{pr}_{A!}X$ is cocartesian. The vertical maps are fiber sequences, which give the top equivalence, which is also the left factor in the vertical equivalence in the left diagram, the other factor coming from the analogous diagram for B. Also the vertical maps in the left square are fiber inclusions. By assumption on p, we know that , so that the bottom and hence also the top map in that square is an equivalence. By 2-out-of-3, we then conclude that $\Phi $ is fully faithful.

It remains to show that the fiber over the terminal object $1 \in S$ is contractible. By assumption, $\mathcal {X}$ has a terminal object $\mathbf{1}$ , which is contained in the fiber over 1, where it is in particular terminal again. It therefore suffices to show that all objects of $\mathcal {X}_1$ are equivalent (say, in $\mathcal {X}$ ). So let $X \in \mathcal {X}_1$ and pick a product $X \times \mathbf{1}$ in $\mathcal {X}$ . By assumption, p sends this to $1 \times 1 \simeq 1$ , hence both projections lie over equivalences, so are themselves equivalences, giving $X \simeq X \times \mathbf{1} \simeq \mathbf{1}$ .

For the second claim in this lemma, suppose $f \colon Y \to Z$ and $g \colon X \to Z$ are edges in $\mathcal {X}$ . If $(f,g) \colon X \to Y \times Z$ is cocartesian, then clearly f and g are too since by assumption on p projections are cocartesian. For the converse, we obtain the following commutative diagram, where maps labeled “ $\mathrm{cc}$ ” are cocartesian (and exist by assumption), and vertical maps are fiberwise

Since cocartesian maps compose, we see that the top right corner is actually $(pf)_!X$ , and hence the right vertical map is an equivalence since f is cocartesian. But this map is the image of the middle vertical map under $(\mathrm{pr}_{pY})_! \colon \mathcal {X}_{pY \times pZ} \to \mathcal {X}_{pY}$ . The same argument shows that its image under $(\mathrm{pr}_{pZ})_!$ is an equivalence, and since these functors induce an equivalence $\mathcal {X}_{pY \times pZ} \to \mathcal {X}_{pY} \times \mathcal {X}_{pZ}$ , it follows that the middle vertical map is already an equivalence, and hence $(f,g)$ is cocartesian.

Proof. Note that (2) implies that p is an $(S,S,\operatorname {\mathrm {\iota }} S)$ -operad. Hence the implications $(2) \Rightarrow (1),(3)$ follow from the above lemma. Moreover, the implication (3) $\Rightarrow $ (2) is clear since the identity on $X \times Y$ is always cocartesian, hence by (3) also the projections $X \leftarrow X \times Y \to Y$ are cocartesian.

It remains to show (1) $\Rightarrow $ (2). Let $p \colon \mathcal {X} \to S$ denote the cocartesian unstraightening of F, and let $A,B\in S$ , $X\in \mathcal {X}_A$ , and $Y\in \mathcal {X}_B$ be arbitrary. As F preserves products, there is a unique $Z\in \mathcal {X}_{A\times B}$ whose cocartesian pushforwards along the projections $A\gets A\times B\to B$ are given by X and Y, respectively. It therefore suffices to show that any $Y\in \mathcal {X}_{A\times B}$ is the product of its cocartesian pushforwards. This follows by running the argument for part (1) of the previous lemma backwards.

Namely, we let $X \in \mathcal {X}$ be arbitrary and consider the following commutative diagram

where we are interested in showing that $\Psi $ is an equivalence, and have taken vertical fibers over an arbitrary $(f,g) \colon pS \to A \times b$ to reduce to showing that $\psi $ is an equivalence. This in turn follows by 2-out-of-3, since the top vertical maps are equivalences by the universal property of cocartesian morphisms, and the top horizontal map is an equivalence since F preserves finite products. It remains to see that the unique object $\mathbf{1} \in \mathcal {X}_1 = F(1) = *$ is terminal in $\mathcal {X}$ . For $X \in \mathcal {X}$ , we have

$$\begin{align*}\mathcal{X}(X,\mathbf{1}) \simeq \mathcal{X}(X,1)_{pX \to 1} \simeq \mathcal{X}_1((pX \to 1)_!X,1) = *.\\[-33pt] \end{align*}$$

Proof.

1. 1. Suppose $p\colon \mathcal {O}\to \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ is an $(\mathcal {F},\mathcal {N})$ -operad. Then the pullback of p to $\mathcal {F}^{{\mathrm{op}}}$ is a cocartesian fibration which satisfies the conditions of Corollary 2.49(2). We conclude by said corollary that its straightening is a product preserving functor, that is, $\mathrm{fgt}(\mathcal {O})$ is an $\mathcal {F}$ -category.

2. 2. Suppose ${\mathcal {C}}$ is an $\mathcal {F}$ -category. Then its unstraightening satisfies the conditions of Corollary 2.49(2) as a functor to $\mathcal {F}^{{\mathrm{op}}}$ . However the inclusion $\mathcal {F}^{{\mathrm{op}}}\hookrightarrow \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})$ is product preserving, and so $\operatorname {\mathrm {Triv}}({\mathcal {C}})$ is an $(\mathcal {F},\mathcal {N})$ -operad.

3. 3. Consider the cartesian unstraightening $p\colon \mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})\to {\mathcal {F}}$ of an ${\mathcal {F}}$ -category ${\mathcal {C}}$ . By the dual of Corollary 2.49, $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ admits coproducts. We will verify the assumptions of [Reference Bachmann and HoyoisBH21, Corollary C.21(2)] to show that $\operatorname {\mathrm {Span}}_{\mathrm{ct},{\mathcal {N}}}(\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}}))$ admits products, computed as coproducts in $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ . To begin we show that

   $$\begin{align*}\hspace{4em}\amalg\colon \mathrm{Un}^{\mathrm{ct}}({\mathcal{C}})\times \mathrm{Un}^{\mathrm{ct}}({\mathcal{C}})\to \mathrm{Un}^{\mathrm{ct}}({\mathcal{C}}) \end{align*}$$
   is a functor of adequate triples. The fact that $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ preserves maps lying over maps in ${\mathcal {N}}$ follows from the fact that p preserves coproducts and Definition 2.51(2). That $\amalg $ preserves cartesian edges follows by Corollary 2.49 $^{{\mathrm{op}}}(3)$ . From this it follows immediately that $\amalg $ preserves ambigressive pullback squares, since any square in $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ over a pullback square in ${\mathcal {F}}$ such that two parallel sides are p-cartesian is a pullback square, see [Reference Haugseng, Hebestreit, Linskens and NuitenHHLN23b, proof of Proposition 2.6].

   Next, we show that the unit and counit of the $\amalg \dashv \Delta $ adjunction are cartesian edges. In the first case this follows from the fact that the diagonal is p-cartesian, once again using Corollary 2.49 $^{{\mathrm{op}}}(3)$ . In the second case this follows from the fact that the coprojections are p-cartesian by (2) of the above cited lemma. Finally, we have to show that the unit and counit naturality squares for maps in $\mathcal {N}$ are pullback squares. This follows immediately from the above characterization of pullback squares in $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ combined with the fact that the squares

   
   are pullbacks in $\mathcal {F}$ by [Reference Cnossen, Haugseng, Lenz and LinskensCHLL24, Remark 2.2.3 and Proposition 2.2.4]. We conclude that ${\mathcal {C}}^{{\mathcal {N}\text{-}\amalg }}= \operatorname {\mathrm {Span}}_{\mathrm{ct},\mathcal {N}}(\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}}))$ admits products computed as coproducts in $\mathrm{Un}^{\mathrm{ct}}({\mathcal {C}})$ . Using this, the fact that $p \colon \mathrm{Un}^{\mathrm{ct}}(\mathcal {C}) \to \mathcal {F}$ satisfies the dual of Corollary 2.49, and that backwards maps in $\operatorname {\mathrm {Span}}_{\mathrm{ct},\mathrm{all}}(\mathrm{Un}^{\mathrm{ct}}(\mathcal {C}))$ are $\operatorname {\mathrm {Span}}(p)$ -cocartesian, it follows that $\mathcal {C}^{{\mathcal {N}\text{-}\amalg }}$ is an $(\mathcal {F},\mathcal {N})$ -operad.

4. 4. The fact that $\operatorname {\mathrm {Env}}$ restricts as claimed follows immediately by applying Proposition 2.50 to $(S,E,M) = (\operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F}),\mathcal {F}^{\mathrm{op}},\mathcal {N})$ . The conditions of this cited proposition are satisfied by the definition of a weakly extensive span pair, see Definition 2.51.

5. 5. Let $X,Y\in {\mathcal {F}}$ arbitrary. Then the coproduct diagram $X\hookrightarrow X\amalg Y\hookleftarrow Y$ lifts through the forgetful functor $\pi \colon {\mathcal {F}}_{/X\amalg Y}\to {\mathcal {F}}$ to a coproduct diagram in ${\mathcal {F}}_{/X\amalg Y}$ . It will therefore suffice to show that is an ${\mathcal {F}}_{/X\amalg Y}$ -category. In view of Lemma 2.32 we may therefore assume, after replacing $({\mathcal {F}},{\mathcal {N}})$ by $({\mathcal {F}}_{/X\amalg Y},{\mathcal {F}}_{/X\amalg Y}\times _{\mathcal {F}}{\mathcal {N}})$ , that $X\amalg Y$ is final.

   In this case the extensivity of $\mathcal {F}$ implies that $\mathcal {F}\simeq \mathcal {F}_{/X\amalg Y} \simeq \mathcal {F}_{/X}\times \mathcal {F}_{/Y}$ . Under this equivalence the objects X and Y correspond to the objects $(\mathrm{id}_X,\emptyset )$ and $(\emptyset ,\mathrm{id}_Y)$ respectively. Now note that the equivalence $\mathcal {F}_{/\emptyset }\simeq \ast $ shows that $\emptyset $ is a strict initial object in any extensive category, that is, every map $X\to \emptyset $ is an equivalence. A simple computation shows that products by strict initial objects always exist (and are given by $\emptyset $ ), and so we conclude that for every $A\in \mathcal {F}$ , the products $X\times A$ and $Y\times A$ exists, and that

   
   is a coproduct diagram in $\mathcal {F}$ . In particular, we obtain an equivalence via the restrictions for any ${\mathcal {N}}$ -normed ${\mathcal {F}}$ -category ${\mathcal {C}}$ . Since is corepresented on $\mathrm {NCat}_{{\mathcal {N}}}({\mathcal {F}})$ , we conclude that
   
   via the restrictions. Lemma 2.33 then identifies this with the natural map , finishing the proof.

Proof. By Theorem 2.57(5) and Lemma 2.44 is an $\mathcal {F}$ -category admitting fiberwise sifted colimits. Moreover, Corollary 2.45 (which requires the assumption of left-cancellability) shows that it is left $\mathcal {N}$ -adjointable. Since $\mathcal {N}$ contains all fold maps by extensivity and is an $\mathcal {F}$ -category (and not just a precategory), this also gives fiberwise finite coproducts by [Reference Cnossen, Lenz and LinskensCLL25a, Proposition 4.2.14].

Notation 3.3. Recall that we denote maps in $\mathcal {N}$ by . We will further denote maps in $\mathcal {M}$ by and maps in $\mathcal {E}$ by .

Convention 3.21. Throughout, we fix a factorization system $(T,E,M)$ such that M and T have finite coproducts, and the inclusion $M\hookrightarrow T$ preserves them. The reader is welcome to think of $T = \operatorname {\mathrm {Span}}_{\mathcal {N}}(\mathcal {F})^{{\mathrm{op}}}$ for a weakly extensive span pair $(\mathcal {F},\mathcal {N})$ , even more so if it is part of a distributive context.

Warning 3.22. Compared to the rest of this paper, our basic object of study will be contravariant functors $T^{\mathrm{op}}\to \mathrm{Cat}$ instead of covariant functors $\operatorname {\mathrm {Span}}_{\mathcal {N}}({\mathcal {F}})\to \mathrm{Cat}$ . This results in various conventions being the opposite of the conventions used elsewhere in this paper: for example, when we specialize T as indicated above, then the norms will correspond to the left class E of the factorization system on T.

This is unfortunate, but unavoidable: we will heavily rely on the existing parametrized literature, and in particular on [Reference Martini and WolfMW24, Reference Cnossen, Lenz and LinskensCLL25b]Footnote ⁹ , which use the contravariant convention (for good reason).