ScalingStacks

Proof. We denote the $\infty$-category $\mathcal{D}\times_{\mathcal{C}}\mathcal{C}_{X/}$ by $\mathcal{D}_{X/}$. Let $\eta\colon X\to RL\left(X\right)$ be the $X$-component of the unit of the adjunction $L\dashv R$. By T.2.1.2.1, the projections $p_{0}\colon\mathcal{C}_{\eta/}\to\mathcal{C}_{X/}$ and $p_{1}\colon\mathcal{C}_{\eta/}\to\mathcal{C}_{RL\left(X\right)/}$ are left fibrations. Moreover, since $\Delta^{\left\{1\right\}}\hookrightarrow\Delta^{1}$ is right anodyne, the map $p_{1}$ is an equivalence of $\infty$-categories. By T.2.2.3.3, we can choose an inverse $p_{1}^{-1}\colon\mathcal{C}_{RL\left(X\right)/}\to\mathcal{C}_{\eta/}$ to $p_{1}$ that strictly commutes with the projections to $\mathcal{C}$. We obtain a commutative diagram of simplicial sets

There is an induced map from the upper left corner to the pullback of the outer rectangle without the upper left corner, which is another commutative diagram of simplicial sets

Since left fibrations are closed under base change (T.2.1.2.1), the vertical maps are left fibrations over $\mathcal{D}$. Hence, to show that the top map is an equivalence it is enough to show that the induced map on fibers is a homotopy equivalence (T.2.2.3.3). For every $Y\in\mathcal{D}$ we get a map

$$\operatorname{Map}_{\mathcal{D}}^{R}\left(L\left(X\right),Y\right)\to\operatorname{Map}_{\mathcal{C}}^{R}\left(X,R\left(Y\right)\right),$$

which is by construction obtained by applying the functor $R$ and pre-composing with the unit $\eta\colon X\to RL\left(X\right)$. By the universal property of the unit map this is a homotopy equivalence for all $Y\in\mathcal{D}$ and therefore the map $\mathcal{D}_{L\left(X\right)/}\to\mathcal{D}_{X/}$ is an equivalence of $\infty$-categories. ∎

Proof. Let $p\colon\mathcal{M}\to\Delta^{1}$ be the coCartesian fibration associated with the functor $L$ (which is also Cartesian, since $L$ has a right adjoint). We can assume that we have a commutative diagram

such that $s|_{\Delta^{\left\{0\right\}}\times\mathcal{C}}=\operatorname{Id}$, $s|_{\Delta^{\left\{1\right\}}\times\mathcal{C}}=L$ and $s|_{\Delta^{1}\times\left\{X\right\}}$ is a coCartesian edge of $\mathcal{M}$ for every $X\in\mathcal{C}$ (combine T.5.2.1.1 and T.5.2.1.3). It is clear from T.1.2.9.2 that for any pair of $\infty$-categories with objects $X\in\mathcal{C}$ and $Y\in\mathcal{D}$ there is a canonical isomorphism

$$\left(\mathcal{C}\times\mathcal{D}\right)_{\left(X,Y\right)/}\simeq\mathcal{C}_{X/}\times\mathcal{D}_{Y/}.$$

Hence, we get an induced commutative diagram

The functor $p_{X}$ is a Cartesian and coCartesian fibration by the duals of T.2.4.3.1(1) and T.2.4.3.2(1). Moreover, an edge in $\mathcal{M}_{X/}$ is (co)Cartesian if and only if its projection to $\mathcal{M}$ is (co)Cartesian by the duals of T.2.4.3.1(2) and T.2.4.3.2(2), which shows that the functor $L_{X}$ is associated with $p_{X}$. It follows that $L_{X}$ has a right adjoint $R_{X}$.

Assuming that $R$ is fully faithful, we will show that $R_{X}$ is fully faithful by showing that the counit of the adjunction $L_{X}\dashv R_{X}$ is an equivalence. For every object, the counit map is an edge of $\mathcal{M}_{X/}$. Since the projection $\mathcal{M}_{X/}\to\mathcal{M}$ is conservative, it is enough to show that the counit map of $L_{X}\dashv R_{X}$ is mapped to the counit map of $L\dashv R$. Indeed, for an object $Y\in\mathcal{D}\simeq\mathcal{M}|_{\Delta^{\left\{1\right\}}}$, we choose a Cartesian edge $e\colon R\left(Y\right)\to Y$ and a coCartesian edge $d\colon R\left(Y\right)\to L\left(R\left(Y\right)\right)$, and combine them into a commutative diagram of the form:

where $f|_{\Delta^{\left\{0,1\right\}}}=d$ and $f|_{\Delta^{\left\{0,2\right\}}}=e$. Since $d$ is coCartesian, there exists a lift $\overline{f}\colon\Delta^{2}\to\mathcal{M}$ that gives an edge

$$\overline{f}|_{\Delta^{\left\{1,2\right\}}}=c\colon L\left(R\left(Y\right)\right)\to Y$$

that is isomorphic to the counit map of the adjunction $L\dashv R$ at $Y$ in the homotopy category $h\mathcal{D}$. We can similarly construct the counit map for an object of $\mathcal{M}_{X/}$. The assertion now follows from the above characterization of (co)Cartesian edges in $\mathcal{M}_{X/}$. ∎

Proof. Consider the commutative square

The vertical functors and the bottom horizontal functor preserve pullbacks. The right vertical functor is conservative. It follows that the top horizontal functor preserves pullbacks as well. ∎

Proof. First, we show that $Y^{\operatorname{\scriptsize{red}}}$ is in fact reduced. Applying $L$ to the defining diagram of $Y^{\operatorname{\scriptsize{red}}}$ and using the fact that $L$ preserves pullbacks, we see that the map $L\left(Y^{\operatorname{\scriptsize{red}}}\right)\to LR\left(\varnothing\right)$ is the pullback of the map $L\left(Y\right)\to LRL\left(Y\right)$, which is an equivalence (from the fact that the counit $LR\left(Y\right)\to Y$ is an equivalence, the zig-zag identities and the 2-out-of-3 property). It follows that the map $L\left(Y^{\operatorname{\scriptsize{red}}}\right)\to LR\left(\varnothing\right)$ is an equivalence, but $LR\left(\varnothing\right)\to\varnothing$ is an equivalence as well (since $R$ is fully faithful) and we are done.

Now, we show that $\rho$ is a co-localization. Let $Z$ be a reduced object. We have a homotopy pullback diagram of spaces

and we note that the space of maps from a reduced object to any object in the essential image of $R$ is contractible. ∎

Proof. Follows from 2.1.5 and the dual of T.5.2.7.8. ∎

Proof. The first claim follows from A.2.4.3.9 by passing to maximal $\infty$-subgroupoids. The second claim follows from A.2.4.3.11. ∎

Proof. The composition of forgetful functors

$$\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{un}}}\to\mathbf{Op}_{\infty}\to\mathbf{Cat}_{\infty}$$

has a left adjoint given as the composition of the corresponding left adjoints. The first one takes $\Delta^{0}$ to $\mathbf{Triv}$ (by A.2.1.4.8) and the second takes $\mathbf{Triv}$ to $\mathbf{Triv}\otimes\mathbb{E}_{0}\simeq\mathbb{E}_{0}$ (by A.2.3.1.9). Hence, for every reduced operad $\mathcal{P}$ (which is in particular unital), we get

$$\operatorname{Map}\left(\mathbb{E}_{0},\mathcal{P}\right)\simeq\operatorname{Map}\left(\Delta^{0},\underline{\mathcal{P}}\right)\simeq\underline{\mathcal{P}}^{\simeq}\simeq\Delta^{0}.$$

∎

Proof. By A.3.2.4.4, the $\infty$-operad $\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)^{\otimes}\to\mathbf{Fin}_{*}$ is also a symmetric monoidal $\infty$-category and so we only need to show that the unit object of $\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)$ is initial. Since $\mathcal{Q}$ is unital, the canonical map $\mathcal{Q}\to\mathbb{E}_{0}\otimes\mathcal{Q}$ is an equivalence of $\infty$-operads (by A.2.3.1.9) and therefore the forgetful functor

$$\underline{\operatorname{Alg}}_{\mathbb{E}_{0}\otimes\mathcal{Q}}\left(\mathcal{C}\right)\simeq\underline{\operatorname{Alg}}_{\mathbb{E}_{0}}\left(\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)\right)\to\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)$$

is an equivalence of $\infty$-categories. On the other hand, by A.2.1.3.10 we have

$$\underline{\operatorname{Alg}}_{\mathbb{E}_{0}}\left(\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)\right)\simeq\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)_{1/},$$

where $1\in\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)$ is the unit object and the projection $\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)_{1/}\to\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)$ is an equivalence of $\infty$-categories if and only if $1$ is initial (T.1.2.12.5). ∎

Proof. We need to verify the hypothesis of 2.1.5. The underlying $\infty$-category functor $L\colon\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{un}}}\to\mathbf{Cat}_{\infty}$ is a composition of two functors $\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{un}}}\to\mathbf{Op}_{\infty}\to\mathbf{Cat}_{\infty}$. The first is a right adjoint by A.2.3.1.9 and the second is a right adjoint by A.2.1.4.10. Hence, the composition is a right adjoint as well and therefore preserves limits. Moreover, by 2.2.3, $L$ is also a left adjoint and its right adjoint is fully faithful. Hence, the functor $\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{un}}}\to\mathbf{Cat}_{\infty,*}$ also has a fully faithful right adjoint and we have $\left(\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{un}}}\right)^{\operatorname{\scriptsize{red}}}\simeq\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{red}}}$. Finally, by 2.1.5, the inclusion $\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{red}}}\simeq\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{red}}}\hookrightarrow\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{un}}}$ admits a right adjoint with the stated description. ∎

Proof. Recall the combinatorial simplicial model category $\mathbf{POp}_{\infty}$ of $\infty$-preoperads, whose underlying $\infty$-category is $\mathbf{Op}_{\infty}$ (see A.2.1.4). Let $\overline{\mathcal{Z}}_{0}\subseteq\overline{\mathcal{Z}}_{1}\subseteq\mathbf{Fin}_{*}$ be the following subcategories:

1. (1)

   The category $\overline{\mathcal{Z}}_{0}$ is discrete and contains only the objects $\left\langle 1\right\rangle$ and $\left\langle m\right\rangle$.

2. (2)

   The category $\overline{\mathcal{Z}}_{1}$ contains $\overline{\mathcal{Z}}_{0}$ together with a unique non-identity morphism, which is the active map $\alpha\colon\left\langle m\right\rangle\to\left\langle 1\right\rangle$.

We endow $\overline{\mathcal{Z}}_{0}$ and $\overline{\mathcal{Z}}_{1}$ with the induced (trivial) marking. Unwinding the definition, for any $\infty$-operad $\mathcal{P}$, the simplicial set $\operatorname{Map}_{\mathbf{POp}_{\infty}}\left(\overline{\mathcal{Z}}_{0},\mathcal{P}^{\natural}\right)$ is isomorphic to $\mathcal{P}_{\left\langle m\right\rangle}^{\simeq}\times\mathcal{P}_{\left\langle 1\right\rangle}^{\simeq}$. Moreover, given

$$\underline{X}=\left(X_{1}\oplus\cdots\oplus X_{m},Y\right)\in\mathcal{P}_{\left\langle m\right\rangle}^{\simeq}\times\mathcal{P}_{\left\langle 1\right\rangle}^{\simeq},$$

the fiber of the fibration (hence also the homotopy fiber)

$$\varphi_{\mathcal{P}}\colon\operatorname{Map}_{\mathbf{POp}_{\infty}}\left(\overline{\mathcal{Z}}_{1},\mathcal{P}^{\natural}\right)\to\operatorname{Map}_{\mathbf{POp}_{\infty}}\left(\overline{\mathcal{Z}}_{0},\mathcal{P}^{\natural}\right)\simeq\mathcal{P}_{\left\langle m\right\rangle}^{\simeq}\times\mathcal{P}_{\left\langle 1\right\rangle}^{\simeq}$$

over $\underline{X}$ is homotopy equivalent to the multi-mapping space $\operatorname{Mul}_{\mathcal{P}}\left(\left\{X_{1},\dots,X_{m}\right\};Y\right)$. Let $\mathcal{Z}_{0}$ and $\mathcal{Z}_{1}$ be $\infty$-operads that are fibrant replacements of $\overline{\mathcal{Z}}_{0}$ and $\overline{\mathcal{Z}}_{1}$, respectively. Moreover, let $f\colon\mathcal{Z}_{0}\to\mathcal{Z}_{1}$ be a map corresponding to the inclusion $\overline{\mathcal{Z}}_{0}\hookrightarrow\overline{\mathcal{Z}}_{1}$. The functor $F\colon\left(\mathbf{Op}_{\infty}\right)_{\mathcal{Z}_{0}/}\to\mathcal{S}$, co-represented by $f\colon\mathcal{Z}_{0}\to\mathcal{Z}_{1}$, preserves limits. Furthermore, its value on $g\colon\mathcal{Z}_{0}\to\mathcal{P}$ fits by T.5.5.5.12 into a fiber sequence

which therefore identifies $F\left(\mathcal{P}\right)$ with $\operatorname{Mul}_{\mathcal{P}}\left(\left\{X_{1},\dots,X_{m}\right\};Y\right)$ for the objects $X_{1},\dots,X_{m},Y\in\mathcal{P}$ determined by $g$.

Let $U\colon\mathbf{Op}_{\infty,*}\to\left(\mathbf{Op}_{\infty}\right)_{\mathcal{Z}_{0}/}$ be the functor induced from the map $\mathcal{Z}_{0}\to\mathbf{Triv}$ corresponding to the inclusion $\overline{\mathcal{Z}}_{0}\hookrightarrow\mathbf{Triv}$. By T.1.2.13.8, the functor $U$ preserves limits. We define $G^{(m)}\colon\mathbf{Op}_{\infty,*}\to\mathcal{S}$ to be the composition of $F$ and $U$, which is limit-preserving as a composition of limit preserving-functors. Unwinding the definitions, $G^{(m)}$ indeed lifts $hG^{(m)}$. ∎

Proof. By 2.2.9 we have a pullback square of pointed unital $\infty$-operads

which, by 2.2.11, induces a pullback square of multi-mapping spaces

The bottom map is the map $\Delta^{0}\to\operatorname{Map}_{\mathcal{C}}\left(X^{\sqcup m},X\right)$ that chooses the fold map since it is induced from the map $\mathbb{E}_{\infty}=\left(\Delta^{0}\right)_{\sqcup}\to\underline{\mathcal{C}}_{\sqcup}$. The right vertical map is induced by pre-composition with the map $\sigma\colon X^{\sqcup m}\to X^{\otimes m}$, since it is induced by the adjunction

$$\left(-\right)_{\sqcup}\colon\mathbf{Cat}_{\infty}\leftrightarrows\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{un}}}\colon\underline{\left(-\right)}.$$

∎

Proof. Since $p$ is a pullback of the right fibration $\left(\mathcal{O}_{\operatorname{\scriptsize{act}}}^{\otimes}\right)_{/X}\to\mathcal{O}_{\operatorname{\scriptsize{act}}}^{\otimes}$ it is itself a right fibration. The simplicial set $\mathbf{Triv}_{\operatorname{\scriptsize{act}}}^{\otimes}$ is isomorphic to $\mathbf{Fin}^{\simeq}$ and is in particular a Kan complex. By T.2.1.3.3 the map $p$ is a Kan fibration. ∎

Proof. This follows directly from unwinding 2.3.3. ∎

Proof. Let $g\colon\mathcal{P}\to\mathcal{Q}$ be a map of reduced $\infty$-operads such that $g_{\mathbf{SSeq}}$ is an equivalence. The map $g$ is defined by a commutative triangle

To show that $g$ is an equivalence of $\infty$-operads, we need to show that $g^{\otimes}$ is an equivalence of $\infty$-categories. Since $\mathcal{P}$ and $\mathcal{Q}$ are reduced, it is clear that $g^{\otimes}$ is essentially surjective. To show that $g^{\otimes}$ is fully faithful, we can use the Segal conditions to reduce this to showing that the map

$$\mathcal{P}\left(n\right)=\operatorname{Mul}_{\mathcal{P}}(*^{(n)},*)\to\operatorname{Mul}_{\mathcal{Q}}(*^{(n)},*)=\mathcal{Q}\left(n\right)$$

is a homotopy equivalence for all $n$. By 2.3.5, those maps are induced by the equivalence $g_{\mathbf{SSeq}}$ and therefore are equivalences. ∎

Proof. Since all the steps in the construction are invariant, we may assume without loss of generality that $G$ is the identity functor and $U_{\mathcal{P}}=U_{\mathcal{Q}}$. In this case, the map $\alpha_{G}$ is given by applying $U_{\mathcal{Q}}$ to the composition

$$F_{\mathcal{Q}}\xrightarrow{F_{\mathcal{Q}}u}F_{\mathcal{Q}}U_{\mathcal{Q}}F_{\mathcal{Q}}\xrightarrow{cF_{\mathcal{Q}}}F_{\mathcal{Q}}$$

where $u$ and $c$ are the unit and counit of the adjunction $F_{\mathcal{Q}}\dashv U_{\mathcal{Q}}$. This composition is homotopic to the identity by the zig-zag identities. ∎

Proof. One only has to observe that the map $\tilde{f}\colon U_{\mathcal{P}}F_{\mathcal{P}}\left(X\right)\to U_{\mathcal{P}}\left(A\right)$ is a map of cones on $\mathcal{P}_{\mathbf{SSeq}}\left(X\right)$. Let $u_{X}\colon X\to U_{\mathcal{P}}F_{\mathcal{P}}\left(X\right)$ be the unit map of the free-forgetful adjunction at $X$. The adjunct map $F_{\mathcal{P}}\left(X\right)\to A$ induces a map $\tilde{f}^{\triangleright}\colon\left(\mathcal{\mathcal{C}}_{\operatorname{\scriptsize{act}}}^{\otimes}\right)_{/U_{\mathcal{P}}F_{\mathcal{P}}\left(X\right)}\to\left(\mathcal{\mathcal{C}}_{\operatorname{\scriptsize{act}}}^{\otimes}\right)_{/U_{\mathcal{P}}\left(A\right)}$. Inspecting Construction A.3.1.3.1, it can be seen that the cone diagram $\mathcal{P}_{\mathbf{SSeq}}\left(f\right)$ is equivalent to the composition of the universal cone diagram $\mathcal{P}_{\mathbf{SSeq}}\left(u_{X}\right)$ and $\tilde{f}^{\triangleright}$. ∎

Proof. We denote by $G\colon\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{C}\right)\to\underline{\operatorname{Alg}}_{\mathcal{P}}\left(\mathcal{C}\right)$ the forgetful functor induced by the map $g$. Let

$$f\colon X\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)\simeq U_{\mathcal{P}}GF_{\mathcal{Q}}\left(X\right)$$

be the unit map. It induces a cone diagram

$$\mathcal{P}_{\mathbf{SSeq}}\left(f\right)\colon\mathcal{P}_{\mathbf{SSeq}}\to\mathcal{\mathcal{C}}_{/U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)}$$

and, by 2.4.5, the associated map $\tilde{f}\colon U_{\mathcal{P}}F_{\mathcal{P}}\left(X\right)\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ is equivalent to the map $\operatorname*{colim}\mathcal{P}_{\mathbf{SSeq}}\left(X\right)\to U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ specified by the cone diagram $\mathcal{P}_{\mathbf{SSeq}}\left(f\right)$. On the other hand, inspecting Construction A.3.1.3.1, it can be seen that the diagram $\mathcal{P}_{\mathbf{SSeq}}\left(f\right)$ is obtained from the diagram

$$\mathcal{Q}_{\mathbf{SSeq}}\left(f\right)\colon\mathcal{Q}_{\mathbf{SSeq}}\to\mathcal{\mathcal{C}}_{/U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)}$$

by pre-composition with $g_{\mathbf{SSeq}}\colon\mathcal{P}_{\mathbf{SSeq}}\to\mathcal{Q}_{\mathbf{SSeq}}$ and that $\mathcal{Q}_{\mathbf{SSeq}}\left(f\right)$ exhibits $U_{\mathcal{Q}}F_{\mathcal{Q}}\left(X\right)$ as the colimit of $\mathcal{Q}_{\mathbf{SSeq}}\left(X\right)$. Thus, we get the desired equivalence. ∎