Observation 2.2 All four types of functors above preserve pairwise products and moreover any group-like additive functor splits extensions.

Proof. For the preservation of products for additive functors, simply note that

is a split Verdier square. To see that extension splitting functors preserve products, note that the map from their definition factors as

\[ F(\operatorname{Seq}(\mathcal{C})) \longrightarrow F(\mathcal{C}^2) \longrightarrow F(\mathcal{C})^2. \]

The composite being an equivalence implies that $F(\operatorname {Seq}(\mathcal {C})) \longrightarrow \mathcal {F}(\mathcal {C}^2)$ admits a retraction, and it also admits a section induced by the functor $\mathcal {C}^2 \rightarrow \operatorname {Seq}(\mathcal {C})$ taking $(x,y)$ to the split fibre sequence $x \rightarrow x \oplus y \rightarrow y$. Thus, the first map in the above composition is an equivalence, and so is the second. But generally, the map $F(\mathcal {C} \oplus D) \rightarrow F(\mathcal {C}) \times F(\mathcal {D})$ is a retract of $F((\mathcal {C} \oplus \mathcal {D}) \oplus (\mathcal {C} \oplus \mathcal {D})) \rightarrow F(\mathcal {C} \oplus \mathcal {D}) \times F(\mathcal {C} \oplus \mathcal {D})$, which we have just shown is an equivalence.

The second claim follows from

being a split Verdier square together with the splitting lemma (in the category of ${\mathrm {E}_\infty }$-groups in $\mathcal {E}$); the splitting lemma itself is a direct consequence for example of [Reference Calmès, Dotto, Harpaz, Hebestreit, Land, Moi, Nardin, Nikolaus and Steimle5, Lemma 1.5.12].

Proof. This exact version is proven in [Reference Haugseng, Hebestreit, Linskens and Nuiten10, § 2].

Proof. Given 4.1 and 2.5 note only that $\operatorname {\mathcal {K}}$ evidently preserves products, and so lifts uniquely to $\operatorname {Mon}_{\mathrm {E}_\infty }(\mathrm {An})$ (in addition to $\operatorname {Mon}_{\mathrm {E}_1}(\mathrm {An})$) since $\mathrm {Cat}_\infty ^\mathrm {st}$ is semi-additive. Thus, the ${\mathrm {E}_1}$-structure underlying the canonical ${\mathrm {E}_\infty }$-structure is also induced by the loop multiplication and this is group-like.

Proof of proposition 4.3 We first review the (well-known) equivalence $\operatorname {Span}(\mathcal {C})\to \operatorname {Span}(\mathcal {C}^\mathrm {op})$ for stable $\mathcal {C}$. It is the identity on objects, and on morphisms it is given by first completing a span to a bicartesian square and then forgetting the initial vertex:

(Here $F_{10}=F_{00}\cup _{F_{01}} F_{11}$.) To define this equivalence on higher cells, denote by $\hat{{\rm Q}}_n(\mathcal {C})\subset \operatorname {Fun}([n]\times [n]^\mathrm {op}, \mathcal {C})$ the full subcategory of diagrams $F$ such that each square

is bicartesian. Restriction along the inclusion $\operatorname {TwAr}[n]\to [n]\times [n]^\mathrm {op}$ defines an equivalence

(*)\begin{equation} \hat{{\rm Q}}_n(\mathcal{C})\xrightarrow \simeq \operatorname{Q}_n(\mathcal{C}) ; \end{equation}

(to see that this is indeed an equivalence, we note that diagrams on $[n]\times [n]^\mathrm {op}$ or on $\operatorname {TwAr}[n]$ lie in $\hat{{\rm Q}}_n(\mathcal {C})$ or $\operatorname {Q}_n(\mathcal {C})$ if and only if they are left Kan extended from the full subcategory $\mathcal {J}_n$ of pairs $(i,j)$ where $i=0$ or $j=n$, so both categories restrict to $\operatorname {Fun}(\mathcal {J}_n, \mathcal {C})$ by an equivalence.)

The duality of $[n]\times [n]^\mathrm {op}$ that switches the entries induces an equivalence

\[ \operatorname{Fun}([n]\times [n]^\mathrm{op}, \mathcal{C})\xrightarrow{\operatorname{op}} \operatorname{Fun}(([n]\times [n]^\mathrm{op})^\mathrm{op}, \mathcal{C}^\mathrm{op})^\mathrm{op} \to \operatorname{Fun}([n]\times [n]^\mathrm{op}, \mathcal{C}^\mathrm{op})^\mathrm{op}, \]

by pre-composition. This in turn restricts to an equivalence

\[ \hat{{\rm Q}}_n(\mathcal{C})\xrightarrow\simeq \hat{{\rm Q}}_n(\mathcal{C}^\mathrm{op})^\mathrm{op}. \]

From this we obtain the desired equivalence $\operatorname {Span}(\mathcal {C})\to \operatorname {Span}(\mathcal {C}^\mathrm {op})$ by applying (*) and taking groupoid cores in each simplicial degree, and then passing to the associated categories.

The equivalence $\operatorname {Span}(\operatorname {Ar}(\mathcal {C}))\to \operatorname {Span}(\operatorname {TwAr}(\mathcal {C}))$ is essentially obtained by performing the above procedure in the target of the arrows; thus, on morphisms it is given by the rule

On higher cells, it is obtained from a map of simplicial animae $\hat{{\rm Q}}_n(\operatorname {Ar}(\mathcal {C}))\to \operatorname {Q}_n(\operatorname {TwAr}(\mathcal {C}))$ given by (restriction of) the following composite:

\begin{align*} \operatorname{Hom}_{\mathrm{Cat}_\infty}([n]\times [n]^\mathrm{op}\times [1], \mathcal{C}) \xrightarrow{\operatorname{TwAr}} & \operatorname{Hom}_{\mathrm{Cat}_\infty}(\operatorname{TwAr}([n]\times [n]^\mathrm{op}\times[1]), \operatorname{TwAr}(\mathcal{C})) \\ & \longrightarrow \operatorname{Hom}_{\mathrm{Cat}_\infty}(\operatorname{TwAr}[n], \operatorname{TwAr}(\mathcal{C})) \end{align*}

where the last map is restriction along the embedding

\begin{align*} \operatorname{TwAr}[n] & \longrightarrow \operatorname{TwAr}([n]\times [n]^\mathrm{op}\times [1]) \\ (i\to j) & \longmapsto ((i,j,0)\to (j,i,1)) . \end{align*}

To see that this map of simplicial animae is indeed an equivalence, it suffices to consider the cases $n=0$ and $n=1$ by the Segal condition: But for $n=0$, it is the identity and for $n=1$ it is given by the rule explained above.

The commutativity of the diagram follows directly from the definition of the equivalences.

Proof. Unwinding definitions we have to show that

\begin{align*} \operatorname{Hom}_{x/\operatorname{Span}(f)}& ((w,x \xleftarrow{\varphi} f(w) \xrightarrow{\mathrm{id}} f(w)),(v,x \xleftarrow{\chi} y \xrightarrow{\psi} f(v))) \\ & \simeq \operatorname{Hom}_{x/f}((\bar y, f(\bar y) \simeq y \xrightarrow{\chi} x), (w,f(w) \xrightarrow{\varphi} x)) \end{align*}

naturally in $(w,f(w) \xrightarrow {\varphi } x)$. The left-hand side, call it $L_\varphi$, unwinds to be the pullback

where the right vertical map is

\begin{align*} & \operatorname{Hom}_{\operatorname{Span}(\mathcal{C})}(w,v) \xrightarrow{\operatorname{Span}(f)} \operatorname{Hom}_{\operatorname{Span}(\mathcal{D})}(f(w),f(v)) \xrightarrow{-{\circ} (x \xleftarrow{\phi}\ f(w)\ \xrightarrow{id} f(w))}\nonumber\\ & \quad\operatorname{Hom}_{\operatorname{Span}(\mathcal{D}}(x,f(v)) . \end{align*}

But per definition

\begin{align*} & \operatorname{Hom}_{\operatorname{Span}(\mathcal{C})}(w,v) \simeq \mathrm{Cr} \operatorname{Fun}(\operatorname{TwAr^{r}}([1]),\mathcal{C}) \times_{\mathrm{Cr}(\mathcal{C} \times \mathcal{C})} \{(w,v)\}\nonumber\\ & \quad \simeq \mathrm{Cr}(\mathcal{C}/w) \times_{\mathrm{Cr} \mathcal{C}} \mathrm{Cr}(\mathcal{C}/v) \end{align*}

and similarly for the lower right-hand term, allowing us to rewrite this pullback as

where the right vertical map identifies component-wise as

\[ \mathcal{C}/w \xrightarrow{f} \mathcal{D}/f(w) \xrightarrow{ \varphi \circ{-}} \mathcal{D}/x \quad \text{and} \quad \mathcal{C}/v \xrightarrow{f} \mathcal{D}/f(v). \]

Switching the order of pullbacks, the fibre of the right-hand map is contractible (since $f$ is a right fibration), so this pullback rewrites as

where $\bar y \rightarrow v$ is the lift of $\psi$ to $\mathcal {C}$. Switching the order of pullbacks back this gives

where the right-hand vertical map is

\[ \operatorname{Hom}_{\mathcal{C}}(\bar y,w) \xrightarrow{f} \operatorname{Hom}_{\mathcal{D}}(y,f(w)) \xrightarrow{\phi \circ{-}} \operatorname{Hom}_{\mathcal{D}}(y,x). \]

But this pullback also describes the right-hand term in the equivalence we have to produce, and the whole procedure above is readily checked to be natural in $(w,\phi \colon f(w) \to x)$.

Proof of proposition 4.4 For the cofinality claim, we have to check that $|(c,d)/\operatorname {Span}(s,t)| \simeq \, \ast$ for all $(c,d) \in \mathcal {C} \times \mathcal {C}^\mathrm {op}$. But from the lemma we find $|(c,d)/\operatorname {Span}(s,t)| \simeq |(s,t)/(c,d)|$ and the category $(s,t)/(c,d)$ has an initial object: One easily checks that $\mathrm {id} \colon 0 \rightarrow 0$ is initial in $\operatorname {TwAr^{r}}(\mathcal {C})$, and whenever a functor preserves the initial objects, all its slices inherit these.

Proof. We have to show that restriction along the natural transformation $\mathrm {Cr} \Rightarrow \mathcal {K}$ gives an equivalence

\[ \operatorname{Nat}(\mathcal{K},F) \longrightarrow \operatorname{Nat}(\mathrm{Cr},F) \]

for every grouplike additive $F \colon \mathrm {Cat}_\infty ^\mathrm {st} \rightarrow \mathrm {An}$. For a general functor $F \colon \mathrm {Cat}_\infty ^\mathrm {st} \rightarrow \mathrm {An}$ set $G(F) = \Omega |F \operatorname {Q}-|$, so that $G(\mathrm {Cr}) \simeq \mathcal {K}$. The inclusion

\[ \mathcal{C} \rightarrow \operatorname{Q}_1(\mathcal{C}), \quad x \longmapsto (0 \leftarrow x \rightarrow 0) \]

induces a natural transformation $F \rightarrow G(F)$ which in turn extends to a natural transformation $\eta \colon \mathrm {id} \Rightarrow G$ (where $\eta _{\mathrm {Cr}}\colon \mathrm {Cr} \Rightarrow \mathcal {K}$ is of course the transformation considered above). Now consider the commutative diagram

the upper square commutes simply because $G$ is a functor, and the other three parts are consequences of the naturality of $\eta$. We now claim that the arrows labelled by $\simeq$ are equivalences: This is an immediate consequence of the following propositions. A diagram chase then implies that the entire diagram consists of equivalences.

Proof of proposition 5.2 The simplicial object $\operatorname {Null}(\mathcal {C})$ is split (over $0$) in the sense of [Reference Lurie12, § 6.1.3], so in particular $|F\operatorname {Null}(\mathcal {C})| \simeq \, \ast$ for every (not necessarily additive) $F \colon \mathrm {Cat}_\infty ^\mathrm {st} \rightarrow \mathrm {An}$. Further it follows that the natural map

\[ \eta\colon F(\mathcal{C}) \rightarrow \Omega|F\operatorname{Q}(C)| \]

is essentially by definition induced by applying $F$ and realisation to the square

where the top horizontal map is induced by $\mathcal {C} \rightarrow \operatorname {Null}(\mathcal {C})_0, x \mapsto (0 \leftarrow x \rightarrow 0)$. Thus, we need to show that the square remains cartesian after applying $F$ and realisation. Using the equifibrancy lemma (compare again the proof of 2.7), we can do this by showing that the map of simplicial animae $F\operatorname {Null}(\mathcal {C})\Rightarrow F\operatorname {Q}(\mathcal {C})$ is equifibred for every group-like additive functor $F \colon \mathrm {Cat}_\infty ^\mathrm {st} \rightarrow \mathrm {An}$.

It is easy to check from the Segal condition that it suffices to treat the squares

where the vertical maps are one of $d_0,d_1$ and $d_2$. For $d_1$ and $d_2$, these squares are split Verdier squares (before applying $F$). For the remaining case, we first note that $d_0$ is a split Verdier projection in both vertical maps, so that it suffices to compare vertical fibres over $0$ (since the map $\pi _0F\operatorname {Q}_2(\mathcal {D}) \rightarrow \pi _0F\operatorname {Q}_1(\mathcal {D})$ is surjective since it is split by the degeneracy $s_0$). But on the left, this fibre identifies with $\mathcal {D} \times \mathcal {D}$ and on the right with $\operatorname {Ar}(\mathcal {D})$, and the functor between them identifies with $(d,d') \mapsto (d' \rightarrow d\oplus d')$. But this map is clearly a right inverse to $(s,\operatorname {cof}) \colon \operatorname {Ar}(\mathcal {D}) \rightarrow \mathcal {D}^2$, so an equivalence after applying $F$ by 2.4.

Proof of proposition 5.3 Unwinding definitions, one finds that the two maps in question

\[ \Omega\lvert F\operatorname{Q} (-)\rvert \Longrightarrow \Omega\lvert \Omega\lvert F \operatorname{Q}^2(-)\rvert \rvert \]

are induced by the maps into the different $\Omega$- and $\operatorname {Q}$-terms. In particular, the composites

\[ \Omega\lvert F\operatorname{Q} (-)\rvert \Longrightarrow \Omega\lvert \Omega\lvert F \operatorname{Q}\operatorname{Q}(-)\rvert \rvert \Longrightarrow \Omega^2 \lvert F\operatorname{Q}^2(-)\rvert \]

are exchanged by flipping both the $\Omega$ and the $\operatorname {Q}$-terms; here the second map is the canonical limit–colimit interchange pulling the right-hand $\Omega$ through the outer realisation. We now claim that this is an equivalence, finishing the proof.

To this end, note that for each $k \in \Delta$ the sequence

\[ \Omega\lvert F \operatorname{Q}\operatorname{Q}_k(\mathcal{C})\rvert \longrightarrow \, \ast \, \longrightarrow \lvert F \operatorname{Q}\operatorname{Q}_k(\mathcal{C})\rvert \]

(with the appropriate homotopy) is not just a fibre, but also a cofibre sequence of $\mathcal {E}_\infty$-groups: This is equivalent to the assertion that $\pi _0\lvert F \operatorname {Q}\operatorname {Q}_k(\mathcal {C})\rvert = 0$ and this holds for any stable category $\mathcal {D}$ in place of $\operatorname {Q}_k(\mathcal {C})$: The functor

\begin{align*} \operatorname{Q}_0\mathcal{D} = \mathcal{D} & \longrightarrow \operatorname{Q}_1(\mathcal{D}) \\ x & \longmapsto (0 \leftarrow 0 \rightarrow x) \end{align*}

composes to the identity with $d_1 \colon \operatorname {Q}_1(\mathcal {D}) \rightarrow \mathcal {D}$ and to $0$ with $d_0$, so the claim is a consequence of $F$ being reduced.

But then it follows that also

\[ \lvert \Omega\lvert F \operatorname{Q}^2(\mathcal{C})\rvert \rvert \rightarrow \, \ast \, \rightarrow \lvert F \operatorname{Q}^2(\mathcal{C})\rvert \]

is a cofibre sequence of ${\mathrm {E}_\infty }$-groups, so in particular a fibre sequence of underlying animae. Looping the resulting equivalence $\lvert \Omega \lvert F \operatorname {Q}^2(\mathcal {C})\rvert \rvert \simeq \Omega \lvert F \operatorname {Q}^2(\mathcal {C})\rvert$ once gives the claim.

Proof. Since the cotensor $(-)^{\mathcal {I}}$ has a left adjoint, given by the tensor $(-)_{\mathcal {I}}$, we deduce that $(-)^{\mathcal {I}}$ preserves limits (for an arbitrary category ${\mathcal {I}}$). On the other hand, for a finite poset ${\mathcal {I}}$, the functor $(-)^{\mathcal {I}}$ also has a right adjoint equally given by $(-)_{\mathcal {I}}$, see [Reference Calmès, Dotto, Harpaz, Hebestreit, Land, Moi, Nardin, Nikolaus and Steimle4, Definition 6.5.1 and Lemma 6.5.6], so $(-)^{\mathcal {I}}$ also preserves colimits.

Proof of proposition 6.2 It suffices to show that the relevant map is an equivalence in each simplicial degree of the $\operatorname {Q}$-construction, i.e. that the canonical map

\[ \vert F(\operatorname{Fun}^{\operatorname{Q}_k\mathcal{A}}([-], \operatorname{Q}_k\mathcal{B}))\vert \to \vert F(\operatorname{Q}_k\mathcal{C})\vert \]

of spaces is an equivalence. By the lemma,

\[ \operatorname{Q}_k \mathcal{A} \to \operatorname{Q}_k \mathcal{B} \to \operatorname{Q}_k \mathcal{C} \]

is also a Verdier projection, in view of the equivalence

\[ \operatorname{Q}_k\mathcal{C} \simeq \operatorname{Fun}(\mathcal{J}_k, \mathcal{C}), \]

where $\mathcal {J}_k\subset \operatorname {TwAr}[k]$ is the sub-poset

Thus, we are reduced to the case $k=0$ which holds by assumption.

Proof. Denote by $t_y \colon S(y) \rightarrow \mathcal {B}$ the functor taking targets. The proof has three steps, the first two of which in fact work for an arbitrary $S$ containing the identities of $\mathcal {C}$.

1. Step 1: Whenever the functor

   \[ \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} \operatorname{Hom}_{\mathcal{B}}(-,y') \colon \mathcal{B}^\mathrm{op} \rightarrow \mathrm{An} \]
   inverts $S$, it agrees with $\operatorname {Hom}_{\mathcal {B}[S^{-1}]}(-,y) \colon \mathcal {B}^\mathrm {op} \rightarrow \mathrm {An}$.

2. Step 2: We have

   \[ \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} \operatorname{Hom}_{\mathcal{B}}(-,y') \simeq (t_y)_! \, \mathrm{const}_\ast{\simeq} \lvert -{/}t_y \rvert \]
   as functors $\mathcal {B}^\mathrm {op} \rightarrow \mathrm {An}$.

3. Step 3: If $S$ satisfies the assumptions from the statement and $f\colon x \rightarrow x'$ is in $S$, then $f^* \colon x'/t_y \rightarrow x/t_y$ admits a left adjoint.

Since adjunctions give equivalences on realisations, the theorem follows.

Proof of step 1 Denoting by $p \colon \mathcal {B}^\mathrm {op} \rightarrow \mathcal {B}[S^{-1}]^\mathrm {op}$ the localisation functor we compute

\begin{align*} p_! \left( \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} \operatorname{Hom}_{\mathcal{B}}(-,y')\right) & \simeq \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} p_! \operatorname{Hom}_{\mathcal{B}}(-,y') \\ & \simeq \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} \operatorname{Hom}_{\mathcal{B}[S^{{-}1}]}(-,y') \\ & \simeq \operatorname{Hom}_{\mathcal{B}[S^{{-}1}]}(-,y) \end{align*}

as follows. The first equivalence holds because left Kan extension is a left adjoint and thus preserves colimits, and the second one holds because representable functors left Kan extend to representable functors by Yoneda's lemma. For the last equivalence, observe that the functor

\begin{align*} S(y) & \longrightarrow \operatorname{Fun}(\mathcal{B}^\mathrm{op},\mathrm{An}) \\ (y \to y') & \longmapsto \operatorname{Hom}_{\mathcal{B}[S^{{-}1}]}(-,y') \end{align*}

inverts all morphisms in $S(y)$ by 2-out-of-3 for equivalences in $\mathcal {B}[S^{-1}]$, and that $\lvert S(y) \rvert$ is contractible since it has $\mathrm {id}_y$ as an initial object. But if a functor $\mathcal {B}^\mathrm {op} \rightarrow \mathrm {An}$ inverts $S$, then its left Kan extension along $p$ is simply the induced functor $\mathcal {B}[S^{-1}]^\mathrm {op} \rightarrow \mathrm {An}$ by inspection of universal properties.

Proof of step 2 The right-hand equivalence is a direct consequence of the pointwise formula for Kan extensions:

\[ ((t_y)_! \, \mathrm{const}_\ast)(x) \simeq \mathop{\mathrm{colim}}_{x/t_y} \mathrm{const}_\ast{\simeq} \lvert x/t_y \rvert \]

For the left-hand one, we use

\[ \mathop{\mathrm{colim}}_{f \in S(y)} \operatorname{Hom}_{S(y)}(g,f) \simeq \lvert g/S(y) \rvert \simeq \; \ast \]

to compute

\begin{align*} (t_y)_! \, \mathrm{const}_* & \simeq (t_y)_! \left( \mathop{\mathrm{colim}}_{f \in S(y)} \operatorname{Hom}_{S(y)}(-,f) \right)\simeq \mathop{\mathrm{colim}}_{f \in S(y)} (t_y)_! \operatorname{Hom}_{S(y)}(-,f)\nonumber\\ & \simeq \mathop{\mathrm{colim}}_{(y\to y')\in S(y)} \operatorname{Hom}_{\mathcal{B}}(-,y') . \end{align*}

Proof of step 3 The left adjoint is easily checked to be given by taking an object $x \rightarrow z \leftarrow y$ (with left pointing arrow in $S$) to $x' \rightarrow p \leftarrow y$, where

is a pushout; this pushout exists by assumption since $f \colon x \rightarrow x'$ is in $S$ and similarly the composite $y \rightarrow z \rightarrow p$ is in $S$ since $S$ is closed under pushouts and composition. We leave it to the reader to check the adjunction property.

Proof of proposition 6.8 Applying theorem 6.9 to $S\subset \mathcal {B}$ and $S\subset S$ (for the latter, noting that pushouts in $S$ are computed in the ambient category $\mathcal {B}$), we see that

\[ \operatorname{Hom}_{S[S^{{-}1}]}(x,y)\to\operatorname{Hom}_{\mathcal{B}[S^{{-}1}]}(x,y) \]

is computed by the formula

\[ \mathop{\mathrm{colim}}_{(y\to y')\in y/S} \operatorname{Hom}_{S}(x,y')\to \mathop{\mathrm{colim}}_{(y\to y')\in y/S} \operatorname{Hom}_{\mathcal{B}}(x,y'). \]

Since $S\subset \mathcal {B}$ is a subcategory, this is a directed colimit of full inclusions of subspaces, and therefore a full inclusion itself.

This shows the first part. For the second part, we first note that all three conditions are conditions on the respective homotopy categories, and that the homotopy categories of the localisations admit a (classical) calculus of fractions as a consequence of theorem 6.9 (cf. [Reference Cisinski8, Corollary 7.2.12]).

We first show the implication (i)$\Rightarrow$(iii): If $f$ has source and target in $S$ and becomes invertible in $\mathcal {B}[S^{-1}]$, then under (i) it is represented by a zig-zag in $\operatorname {Ho}(S)$ so that, by calculus of fractions and $2$-out-of-$3$, $f$ belongs itself to $S$. The implication (iii)$\Rightarrow$(ii) is trivial, since equivalences satisfy $2$-out-of-$6$. Finally, assume (ii) holds, and let $f$ be an invertible morphism in $\mathcal {B}[S^{-1}]$ between objects of $S$; we need to show that it is represented by a zig-zag in $S$.

For this, we may clearly assume that $f$ is a morphism in $\mathcal {B}$. Applying calculus of fractions again, we see that a morphism $f$ of $\mathcal {B}$ is split mono in the localisation if and only if, after post-composition with a morphism $g$ of $\mathcal {B}$, it lies in $S$. If $f$ is even an equivalence in the localisation, then so is $g$, and applying the same argument to $g$, we find another morphism $h$ such that $h\circ g\in S$; then $f\in S$ by $2$-out-of-$6$.

Proof of theorem 6.1 The groupoid-core functor satisfies $(\ast )$ by 6.6 and the discussion preceding it. From 6.2 we conclude that the functor $|\mathrm {Cr} \operatorname {Q}|$ also does, so by corollary 2.9, $|\mathrm {Cr} \operatorname {Q}|\colon \mathrm {Cat}_\infty ^\mathrm {st}\to \mathrm {An}$ is Verdier-localising and therefore also $\operatorname {\mathcal {K}}=\Omega |\mathrm {Cr}\operatorname {Q}|$. For the second claim, we recall that a fibre sequence of ${\mathrm {E}_\infty }$-groups gives rise to a fibre sequence of spectra if (and only if) it is surjective on $\pi _0$, but any Verdier projection induces an epimorphism on $\operatorname K_0$ by the formula for $\operatorname K_0$ (or by noting that $\pi _0 |\mathrm {Cr} \operatorname {Q}(\mathcal {C})|=0$ for any $\mathcal {C}$ and using that $|\mathrm {Cr} \operatorname {Q}|$ is Verdier-localising).

First proof The first proof we give is based on the construction $\operatorname {\mathcal {K}}(\mathcal {C}) \simeq \Omega |\mathrm {Cr}\operatorname {Q}(\mathcal {C})|$ and the analysis made for the localisation theorem. From 2.7 applied to $|\mathrm {Cr} \operatorname {Q}-|$, we obtain a fibre sequence

\[ |\mathrm{Cr} \operatorname{Q}\mathcal{A}| \longrightarrow |\mathrm{Cr} \operatorname{Q}\mathcal{B}| \longrightarrow ||\mathrm{Cr}\operatorname{Q}\operatorname{Fun}^\mathcal{A}([-],\mathcal{B})|| \]

and by inspection

\[ \mathrm{Cr}\operatorname{Q}\operatorname{Fun}^\mathcal{A}([-],\mathcal{B}) \simeq \mathrm{Cr} \operatorname{Fun}^{\operatorname{Q}\mathcal{A}}([-],\operatorname{Q}\mathcal{B}) \]

as bisimplicial animae. As in § 6, we can identify

\[ |\mathrm{Cr} \operatorname{Fun}^{\operatorname{Q}_n\mathcal{A}}([-],\operatorname{Q}_n\mathcal{B})| \simeq |\operatorname N \operatorname{Q}_n(\mathcal{B})_{\operatorname{Q}_n(\mathcal{A})}| \simeq |\operatorname{Q}_n(\mathcal{B})_{\operatorname{Q}_n(\mathcal{A})}| . \]

But if $\mathcal {A} \rightarrow \mathcal {B}$ is dense, so is $\operatorname {Q}_n(\mathcal {A}) \rightarrow \operatorname {Q}_n(\mathcal {B})$. Thus, $\operatorname {Q}_n(\mathcal {B})/\operatorname {Q}_n(\mathcal {A}) \simeq 0$. As explained in 6.7, this gives that $|\operatorname {Q}_n(\mathcal {B})_{\operatorname {Q}_n(\mathcal {A})}|$ is discrete with components $\operatorname K_0(\operatorname {Q}_n(\mathcal {B}))/\operatorname K_0(\operatorname {Q}_n(\mathcal {A}))$. By direct inspection, one finally finds that $\operatorname K_0(\operatorname {Q}(\mathcal {C}))$ is the edgewise subdivision of $\mathrm {Bar}(\operatorname K_0(\mathcal {C}))$, so that in total

\[ ||\mathrm{Cr}\operatorname{Q}\operatorname{Fun}^\mathcal{A}([-],\mathcal{B})|| \simeq |\mathrm{Bar}(\operatorname K_0(\mathcal{B})/\operatorname K_0(\mathcal{A}))| \]

is an Eilenberg–Mac Lane space in degree $1$. Looping the fibre sequence from the start of this proof now gives the claim.

Proof. The second statement follows from the first and lemma 7.5: The cofibre sequence of ${\mathrm {E}_\infty }$-groups

\[ F^\mathrm{grp}(\mathcal{A})\to F^\mathrm{grp}(\mathcal{B}) \to F(\mathcal{B})/F(\mathcal{A}) \]

(with last term discrete) is a fibre sequence of animae, with last map $\pi _0$-surjective, and the functor

\[ \pi_0\colon \operatorname{Mon}_{\mathrm{E}_\infty}(\mathrm{An})\to \mathrm{CMon} \]

commutes with colimits, since it admits the discrete inclusion as a right adjoint.

Let us prove the first statement. The chain of dense inclusions

\[ \mathcal{A}\to \mathcal{B}\to \mathcal{A}^\natural \; ( = \mathcal{B}^\natural) \]

induces a cofibre sequence of ${\mathrm {E}_\infty }$-groups

\[ F(\mathcal{B})/F(\mathcal{A})\to F(\mathcal{A}^\natural)/F(\mathcal{A})\to F(\mathcal{A}^\natural)/F(\mathcal{B}) \]

and similarly with $F^{\mathrm {grp}}$, so it suffices to consider the case $\mathcal {B}=\mathcal {A}^\natural$.

We claim that the functor $F'\colon \mathrm {Cat}_\infty ^\mathrm {st}\to \operatorname {Mon}_{\mathrm {E}_\infty }(\mathrm {An})$ given by the formula

\[ F'(\mathcal{A}):=F(\mathcal{A}^\natural)/F(\mathcal{A}) \]

(quotient of ${\mathrm {E}_\infty }$-monoids) represents the quotient $(F\circ (-)^\natural )/F$ in the category of additive functors $\mathrm {Cat}_\infty ^\mathrm {st}\to \operatorname {Mon}_{\mathrm {E}_\infty }(\mathrm {An})$. To see this, it will suffice to prove that $F'$ is additive. Since $F'$ is group-like, we can combine 2.4 and 2.5 with the splitting lemma to see that it suffices to prove that $F'$ sends the (split) Verdier sequence $\mathcal {A} \rightarrow \operatorname {Ar}(\mathcal {A}) \xrightarrow {t} \mathcal {A}$ to a (split) fibre sequence. In view of lemma 7.5, we only need to show that it induces a short exact sequences of abelian groups

\[ 0\to \pi_0 F'(\mathcal{A})\to \pi_0 F'(\operatorname{Ar}(\mathcal{A}))\to \pi_0 F'(\mathcal{A})\to 0. \]

For this, we note that short exact sequences of monoids are in particular cofibre sequences of monoids. Since $F$ and $F\circ (-)^\natural$ are additive, we see by commuting quotients that the sequence in question is a cofibre sequence (of abelian monoids or of abelian groups), and hence right exact. Also, the first map is (split) injective, so the sequence is indeed short exact.

Similarly, $(F^\mathrm {grp})'$ is additive and so represents the quotient of ($F^\mathrm {grp}\circ (-)^\natural )/F^\mathrm {grp}$ in the category of additive functors $\mathrm {Cat}_\infty ^\mathrm {st}\to \operatorname {Grp}_{\mathrm {E}_\infty }(\mathrm {An})$: The short exact sequence

\[ 0\to \pi_n (F^\mathrm{grp})'(\mathcal{A})\to \pi_n (F^\mathrm{grp})'(\mathcal{B})\to \pi_n (F^\mathrm{grp})'(\mathcal{C})\to 0 \]

is proven for $n=0$ as above, and for $n>0$ follows from the equivalence

\[ \Omega (G/H)\simeq \operatorname{fib}(H\to G) \]

valid for any map of ${\mathrm {E}_\infty }$-groups.

Next, we note that the canonical map $F\circ (-)^\natural \to F^\mathrm {grp}\circ (-)^\natural$ is a group completion: This follows from the fact that the endofunctor $(-)^\natural$ of $\mathrm {Cat}_\infty ^\mathrm {st}$ admits the minimalisation $(-)^\mathrm {min}$ as a left adjoint (a simple consequence of Thomason's theorem): This adjunction induces an adjunction on $\operatorname {Fun}(\mathrm {Cat}_\infty ^\mathrm {st}, \mathrm {An})$ so we have equivalences

\[ \operatorname{nat}(F\circ(-)^\natural, G) \simeq \operatorname{nat}(F, G\circ(-)^\mathrm{min})\!\simeq\! \operatorname{nat}(F^{\mathrm{grp}}, G\circ(-)^\mathrm{min}) \!\simeq \!\operatorname{nat}(F^{\mathrm{grp}}\circ(-)^\natural, G) \]

for every group-like additive functor $G$.

Thus, comparing universal properties, we see that the map $F'\to (F^\mathrm {grp})'$ is a group completion of $F'$: But $F'$ is already group-like, so it is an equivalence.

Proof. It follows from the cofinality theorem that $F^\mathrm {grp}$ sends dense functors to dense maps of ${\mathrm {E}_\infty }$-monoids (with the last two conditions being automatic for group-like functors). It remains to prove that $F^\mathrm {grp}$ preserves pullback squares

in $\mathrm {Cat}_\infty ^\mathrm {st}$ whose vertical maps are dense.

By assumption, this is true for $F$ and since $\pi _0 F(\mathcal {A}')\to \pi _0 F(\mathcal {B}')$ is injective, we see that also $\pi _0(F)$ sends the square to a pullback square. From density we then deduce that the map of quotient monoids

\[ \pi_0 F(\mathcal{B}')/\pi_0 F(\mathcal{A}')\to \pi_0 F(\mathcal{B})/\pi_0 F(\mathcal{A}) \]

has kernel zero. But by the cofinality theorem, this identifies with the corresponding map for $\pi _0 (F^\mathrm {grp})$, so we deduce that $\pi _0(F^\mathrm {grp})$ also sends the square to a pullback square. Applying the cofinality theorem again, we see that $F^\mathrm {grp}$ sends the square to a pullback square of animae.

Proof of lemma 7.5 We start by observing that $\pi _0(M/N) = \pi _0(M)/\pi _0(N)$ is a group by the cofinality assumption. We then need to show that

\[ \Omega(M/N)\simeq \Omega(M^\mathrm{grp}/N^\mathrm{grp}) \simeq \operatorname{fib}(N^\mathrm{grp}\to M^\mathrm{grp}) \]

is contractible. Since $\pi _0(N^\mathrm {grp})\to \pi _0(M^\mathrm {grp})$ identifies with the discrete group completion of $\pi _0(N)\to \pi _0(M)$, it is injective by cofinality, and we are left to show that the map $N^\mathrm {grp}\to M^\mathrm {grp}$ induces an equivalence on the base point components. We can prove this by showing it is a homology isomorphism because it is a map of $H$-spaces.

To show that the map on base point components is injective on homology, it suffices to prove the same for $N^\mathrm {grp}\to M^\mathrm {grp}$ itself. By the group completion theorem, this map is given by the composite

\[ H_*(N)[\pi_0(N)^{{-}1}]\to H_*(M)[\pi_0(N)^{{-}1}]\to H_*(M)[\pi_0(M)^{{-}1}] \]

where the first map is injective by the first assumption of cofinality (since localisation is exact), and the second map is an isomorphism by the second assumption.

To show surjectivity on the base point component, we first consider the chain of maps in the homotopy category of spaces

\[ M \to M^\mathrm{grp} \to M^\mathrm{grp}_0 \]

mapping an ${\mathrm {E}_\infty }$-monoid $M$ into the base point component of its group completion, where the first map is the canonical one and the second one subtracts $x$ in the component of $x$. On homology, these maps are given by base-change along the canonical ring homomorphisms

\[ \mathbb{Z}[\pi_0 M] \to \mathbb{Z}[\pi_0 M^\mathrm{grp}]\to \mathbb{Z} \]

in view of the group completion theorem and the commutativity of $M$ for the left map and the decomposition $M^\mathrm {grp} \simeq \pi _0 M^\mathrm {grp} \times M^\mathrm {grp}_0$ (natural in the homotopy category of $\mathrm {An}$) for the right map. We conclude the existence of a natural isomorphism

\[ H_*(M^\mathrm{grp}_0) \cong H_*(M)/\pi_0(M) \]

where in the target we take the quotient of the monoid action in the category of graded abelian groups.

Thus we need to show that the map $N\to M$ becomes surjective in homology after modding out the $\pi _0(M)$-action in the target. So let $a\in H_*(M)$ where we may assume that $a$ is defined in a single path component $M_x$ of $M$, for some $x\in \pi _0(M)$. By the second condition of cofinality, we may assume as well that $x\in \pi _0(N)$, in which case $a$ lifts to $H_*(N)$ by the first condition of cofinality.

Proof of proposition 8.2 By the equifibrancy lemma the square

is cartesian, since the map $F \operatorname {Null}(\mathcal {D}) \to F \operatorname {Q}(\mathcal {D})$ is equifibred as shown in the proof of 5.2. In other words,

\[ |F\operatorname{Q}(f)| \longrightarrow |F\operatorname{Q}(\mathcal{C})| \longrightarrow |F\operatorname{Q}(\mathcal{D})| \]

is a fibre sequence (the case $\mathcal {C}=0$ gave $F(\mathcal {D}) \simeq \Omega |F\operatorname {Q}(\mathcal {C})|$ in 5.2, since $\operatorname {Q}(0 \rightarrow \mathcal {D}) \simeq \mathrm {const}_\mathcal {D}$ by inspection). Rotating the fibre sequence above twice to the left therefore gives us the desired fibre sequence

\[ F(\mathcal{C}) \longrightarrow F(\mathcal{D}) \longrightarrow |F\operatorname{Q}(f)| . \]

To see that it is also a cofibre sequence, it suffices (by the long exact sequence) to check that $\pi _0|F\operatorname {Q}(\mathcal {C})| = 0$. But the functor

\begin{align*} \operatorname{Q}_0(\mathcal{C}) = \mathcal{C} & \longrightarrow \operatorname{Q}_1(\mathcal{C}) \\ c & \longmapsto (0 \leftarrow 0 \rightarrow c) \end{align*}

induces a homotopy between the $0$ and identity maps of $\pi _0|F\operatorname {Q}(\mathcal {C})|$, which gives the claim.

Proof of proposition 8.3 We shall realise both $\operatorname {Q}_n(i)$ and $\operatorname {Fun}^\mathcal {C}([n] * [n]^\mathrm {op},\mathcal {D})$ as the following full subcategory $P_n$ of $\operatorname {Fun}(\operatorname {TwAr^{r}}([0]*[n]*[n]^\mathrm {op}*[0]),\mathcal {D})$: A functor $F$ lies in $P_n$ if under the identification $[0]*[n]*[n]^\mathrm {op}*[0] \cong [1+n]*[1+n]^\mathrm {op}$,

1. (i) all squares in the ‘left half’ of $\operatorname {TwAr^{r}}([1+n]*[1+n]^\mathrm {op})$ go to cartesian squares, i.e.

   
   whenever $j \leq m$ (where $\epsilon \in \{l,r\}$ and $i_\epsilon$ denotes the $i$-th element of the left respectively right factor of $[1+n]*[1+n]^\mathrm {op}$),

2. (ii) $F$ vanishes on the ‘right half’ of $\operatorname {TwAr^{r}}([1+n]*[1+n]^\mathrm {op})$, i.e. we have $F(i_l \leq k_r) = 0 = F(i_r \leq k_r)$ whenever $i \geq k$, and

3. (iii) $F$ vanishes on the lower left corner of $\operatorname {TwAr^{r}}([1+n]*[1+n]^\mathrm {op})$, i.e. $F(0_l \leq 0_l) = 0$, and

4. (iv) $F$ takes values in $\mathcal {C}$ on all spots not of the form $(0_l \leq k_\epsilon )$.

In other words, $P_n$ consists of diagrams of the shape

where the lower left corner and the entire right half are zero, the $2n+2$ objects on the upper left diagonal marked by empty circles are in $\mathcal {D}$, the objects marked by filled circles are in the image of $\mathcal {C}$ and all squares in the left half are cartesian.

Now consider on the one hand the inclusion

\[ \alpha \colon \operatorname{TwAr^{r}}([0]*[n]) \longrightarrow \operatorname{TwAr^{r}}([0]*[n]*[n]^\mathrm{op}*[0]) \]

induced by the inclusion $[0] * [n] \subset ([0]*[n])*([0]*[n])^\mathrm {op}$. Since its image lies fully in the ‘left half’, the first and last two conditions guarantee that restriction along $\alpha$ yields a map

\[ \alpha^* \colon P_n \rightarrow \operatorname{Q}_n(i), \]

which is clearly natural in $n \in \Delta$.

Similarly, consider the map

\begin{align*} \beta \colon [n]*[n]^\mathrm{op} & \longrightarrow \operatorname{TwAr^{r}}([0]*[n]*[n]^\mathrm{op}*[0])\\ i_\epsilon & \longmapsto (0_l \leq i_\epsilon). \end{align*}

The first and last conditions imply that any map in the restriction of some $F \in P_n$ goes to the pullback of some map in $\mathcal {C}$, so in particular to an equivalence modulo $\mathcal {C}$, we therefore obtain a well-defined map

\begin{align*} \beta^* \colon P_n \rightarrow \operatorname{Fun}^\mathcal{C}([n]*[n]^\mathrm{op},\mathcal{D}), \end{align*}

which is again clearly natural in $n \in \Delta$.

As mentioned, the claim will now follow from both these maps $\alpha ^*$ and $\beta ^*$ being equivalences. We start with $\alpha ^*$. An inverse is constructed by the following two-step Kan extension. First, let $T_n \subseteq \operatorname {TwAr^{r}}([0]*[n]*[n]^\mathrm {op}*[0])$ denote the subposet given as the union of $\operatorname {TwAr^{r}}([0]*[n])$ and the ‘right half’ of $\operatorname {TwAr^{r}}([0]*[n]*[n]^\mathrm {op}*[0])$. Then consider

\begin{align*} \operatorname{Fun}(\operatorname{TwAr^{r}}([0]*[n]),\mathcal{D}) \!\xrightarrow{\mathrm{Lan}}\! \operatorname{Fun}(T_n,\mathcal{D}) \!\xrightarrow{\mathrm{Ran}}\! \operatorname{Fun}(\operatorname{TwAr^{r}}([0]*[n]*[n]^\mathrm{op}*[0]),\mathcal{D}). \end{align*}

As Kan extensions along fully faithful maps are fully faithful and right inverse to restriction, it remains only to check that the composite takes $\operatorname {Q}_n(i) \subseteq \operatorname {Fun}(\operatorname {TwAr^{r}}([0]*[n]),\mathcal {D})$ onto $P_n$. But from the pointwise formulae, it is easy to see that the first Kan extension is an extension by $0$, and that the pullback condition (i) is equivalent to being right Kan extended from $T_n$. This in turn implies immediately that for $F$ satisfying conditions (i) and (ii), condition (iv) is equivalent to $F\rvert _{\operatorname {TwAr^{r}}([n])}$ taking values in $\mathcal {C}$, which finishes the claim (since the vanishing condition in (iii) is also contained in the definition of $\operatorname {Q}_n(i)$).

We finally treat $\beta ^*$: Again the inverse is given by successive Kan extensions. First add a zero at $(0_l \leq 0_r)$ to a functor defined on the image of $\beta$ by left Kan extension, and then right Kan extend to add zeros at $(0_l \leq 0_l)$ and the ‘right half’, and finally left Kan extend once more to the whole of $\operatorname {TwAr^{r}}([0]*[n]*[n]^\mathrm {op}*[0])$. Again this process is right inverse to restriction along $\beta$ on the whole of $\operatorname {Fun}([n]*[n]^\mathrm {op},\mathcal {D})$, and it remains to check that it takes $\operatorname {Fun}^\mathcal {C}([n]*[n]^\mathrm {op},\mathcal {D})$ onto $P_n$.

But again it follows trivially from the pointwise formulae that the first two Kan extensions really are extensions by zero, precisely as required in conditions (ii) and (iii), and that being in the image of the second left Kan extension is equivalent to the squares in condition (i) being pushouts, so stability implies that these are equivalent conditions. To finally see that the image of $\operatorname {Fun}^\mathcal {C}([n]*[n]^\mathrm {op},\mathcal {D})$ also satisfies condition (iv), note that (as just discussed) the value at $(i_l \leq k_\epsilon )$, with $i < k$ or $i=k$ and $\epsilon = l$, of the extension $\operatorname {TwAr^{r}}([1+n]*[1+n]^\mathrm {op}) \rightarrow \mathcal {D}$ of some $F \colon \mathrm {im}(\beta ) \rightarrow \mathcal {D}$ sits in a cocartesian square

But its right most term is part of the ‘right half’ of $\operatorname {TwAr^{r}}([0]*[n]*[n]^\mathrm {op}*[0])$ (in fact it lies on the middle vertical line) and the upper left pointing map is an equivalence module $\mathcal {C}$ by assumption, so $F(i_l \leq k_\epsilon ) \in \mathcal {C}$.