用户: Aripriner/Higher Algebra/infinty monoids

定义 0.1. We denote by ${\Gamma }^{\mathrm{op}}$ the category whose objects are finite sets ans whose morphisms are partially defined maps(only defined for some elements, not all). Write $⟨n⟩=\{1\cdots ,n\}$. There is a functor $Cut:{\Delta }^{\mathrm{op}}\to {\Gamma }^{\mathrm{op}}$ defined by $Cut([n])=⟨n⟩$ and for a morphism $\alpha :[m]\to [n]$ in $\Delta$, we define $Cut({\alpha }^{\mathrm{op}})$ via $$\begin{array}{r}Cut\left({\alpha }^{op}\right)=\{\begin{array}{ll}undefined& ifi⩽\alpha \left(0\right)\\ j& if\alpha \left(j-1\right)<i⩽\alpha \left(j\right)\\ undefined& if\alpha \left(n\right)<i\end{array}\end{array}$$

定义 0.2. If $\mathcal{C}$ is an $\infty$-category with finite products, a cartesian commutative monoid in $\mathcal{C}$ is a functor $X:{\Gamma }^{\mathrm{op}}\to \mathcal{C}$ such that $X\circ Cut:{\Delta }^{\mathrm{op}}\to \mathcal{C}$ is a cartesian monoid. If it is even a cartesian group, $X$ is called a cartesian commutative group in $\mathcal{C}$. We let $\mathrm{CMon}(\mathcal{C}),\mathrm{CGrp}(\mathcal{C})$ denote the full category spanned by them.

定义 0.3. An $\infty$-category is called semi-additive if it admits finite products and finite coproducts, its initial and final objects agree, and for any $x,y\in \mathcal{C}$ the natural map $$\begin{array}{r}\left(\begin{array}{cc}i{d}_x& 0\\ 0& i{d}_y\end{array}\right):x\bigsqcup y\to x\times y\end{array}$$is an equivalence, and we denote their product and coporduct as $x\oplus y$.
We say $C$ is additive iff it is semi-additive and the map $$\begin{array}{r}\left(\begin{array}{cc}i{d}_x& i{d}_x\\ 0& i{d}_x\end{array}\right):x\oplus x\to x\oplus x\end{array}$$is an equivalence, for example $\mathcal{K}(\mathbb{Z})$ and $\mathcal{D}(\mathbb{Z})$.

命题 0.4. If $\mathcal{C}$ is semi-additive, then the natural forgetful functors $$\begin{array}{r}\mathrm{CMon}(\mathcal{C})\to \mathrm{Mon}(\mathcal{C})\to \mathcal{C}\end{array}$$are equivalences. If $C$ is additive, so are $$\begin{array}{r}\mathrm{CGrp}(\mathcal{C})\to \mathrm{Grp}(\mathcal{C})\to \mathcal{C}\end{array}$$

推论 0.5. If $\mathcal{C}$ is additive, then there is a canonical lift $$\begin{array}{r}{\mathrm{Hom}}_{\mathcal{C}}:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathrm{CGrp}(\mathrm{Ani})\end{array}$$If $\mathcal{C}$ is only semi-additive, we get a lift to $\mathrm{CMon}(\mathrm{Ani})$.

证明. In general, if $F:\mathcal{C}\to \mathcal{D}$ is a finite-preserving functor, then it induces a functor $F_{\ast }:\mathrm{CGrp}(\mathcal{C})\to \mathrm{CGrp}(\mathcal{D})$ and $F_{\ast }:\mathrm{CMon}(\mathcal{C})\to \mathrm{CMon}(\mathcal{D})$ because it perserves the Segal conditions.
Notice that ${\mathrm{Hom}}_{\mathcal{C}}:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathrm{Ani}$ does not preserve finite products, but the Yoneda embedding ${\mathrm{Hom}}_{\mathcal{C}}(c,-):\mathcal{C}\to \mathrm{Ani}$ does. Thus we can lift it to the functor ${\mathrm{Hom}}_{\mathcal{C}}(c,-):\mathcal{C}\simeq \mathrm{CGrp}(\mathcal{C})\to \mathrm{CGrp}(\mathrm{Ani})$.
To show then functorality in $c$, consider the Yoneda embedding ${\mathrm{Hom}}_{\mathcal{C}}:{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Fun}(\mathcal{C},\mathrm{Ani})$. Define the full subcategory ${\mathrm{Fun}}^{lex}(\mathcal{C},\mathrm{Ani})\subseteq \mathrm{Fun}(\mathcal{C},\mathrm{Ani})$ spanned by the finite product preserving functors. We have the map $$\begin{array}{r}{\mathrm{Fun}}^{\times }(\mathcal{C},\mathrm{Ani})\to \mathrm{Fun}(\mathrm{CGrp}(\mathcal{C}),\mathrm{CGrp}(\mathrm{Ani}))\simeq \mathrm{Fun}(\mathcal{C},\mathrm{CGrp}(\mathrm{Ani}))\end{array}$$by construction. And the image of the Yoneda embedding lies in ${\mathrm{Fun}}^{\times }(\mathcal{C},\mathrm{Ani})$. Thus we have ${\mathrm{Hom}}_{\mathcal{C}}:{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Fun}(\mathcal{C},\mathrm{CGrp}(\mathrm{Ani}))$. Similarly for the semi-additive case.

$\square$

Recall that we have the adjunction $$\begin{array}{r}B:\mathrm{Mon}(\mathrm{Ani})\leftrightarrow \ast /\mathrm{Ani}:\Omega \end{array}$$Now $B=||$ and $\Omega$ both preseve finite products($B$ by explicit computations and $\Omega$ by right adjointness). Thus we can lift this adjunction:$$\begin{array}{r}B:\mathrm{CMon}(\mathrm{Mon}(\mathrm{Ani}))\leftrightarrow \mathrm{CMon}(\ast /\mathrm{Ani}):\Omega \end{array}$$It is easy to see the equivalence $\mathrm{CMon}(\ast /\mathcal{C})\simeq \mathrm{CMon}(\mathcal{C})$. We want to show the diagram

commutes and all arrows are equivalences.
To prove this statement, we let ${\mathrm{Cat}}_{\infty }^{\times }\subseteq {\mathrm{Cat}}_{\infty }$ be the subcategory spanned by $\infty$-categories admitting finite products and functors preserving finite products. Also let ${\mathrm{Cat}}_{\infty }^{add}\subseteq {\mathrm{Cat}}_{\infty }^{semi-add}\subseteq {\mathrm{Cat}}_{\infty }^{\times }$ denote the full subcategory spanned by additive and semi-additive $\infty$-categories.

定理 0.6. If an $\infty$-category $\mathcal{C}$ has products, then $\mathrm{CMon}(\mathcal{C})$ and $\mathrm{CGrp}(\mathcal{C})$ are semi-additive and additive respectively. Furthermore, the functors $$\begin{array}{rl}\mathrm{CMon}& :{\mathrm{Cat}}_{\infty }^{\times }\to {\mathrm{Cat}}_{\infty }^{semi-add}\\ \mathrm{CGrp}& :{\mathrm{Cat}}_{\infty }^{\times }\to {\mathrm{Cat}}_{\infty }^{add}\end{array}$$are right adjoints to the inclusion functors.

Using this theorem, we can get a proof of the claim. First we prove that the maps $forget:\mathrm{CMon}(\mathrm{CMon}(\mathcal{C}))\to \mathrm{CMon}(\mathcal{C})$ and $forge{t}_{\ast }:\mathrm{CMon}(\mathrm{CMon}(\mathcal{C}))\to \mathrm{CMon}(\mathcal{C})$ agree and are equivalences. In fact $forget:\mathrm{CMon}(-)\to -$ is the counit of the adjunction in theroem 0.6(see the proof). Hence $forge{t}_{\ast }$ is an equivalence. Because the inclusion map is full fatihful, we know that the other is an equivalence.
Moreover, by writing down the triangle identies we find that the maps are inverses to the unit $\mathrm{CMon}(\mathcal{C})\to \mathrm{CMon}(\mathrm{CMon}(\mathcal{C}))$(one regards $\mathrm{CMon}(\mathcal{C})\in {\mathrm{Cat}}_{\infty }^{\times }$ and the other regard $\mathrm{CMon}(\mathcal{C})\in {\mathrm{Cat}}_{\infty }^{semi-add}$ then using the identities respectively).
We want to replace one $\mathrm{CMon}$ by $\mathrm{Mon}$. To do this, note that if we have $\mathrm{CMon}(\mathcal{C})$ is semi-additive, we have $\mathrm{CMon}(\mathrm{CMon}(\mathcal{C}))\simeq \mathrm{Mon}(\mathrm{CMon}(\mathcal{C}))$. Also $\mathrm{CMon}(\mathrm{Mon}(\mathcal{C}))\simeq \mathrm{Mon}(\mathrm{CMon}(\mathcal{C}))$ by direct inspection as the full subcategory of $\mathrm{Fun}({\Delta }^{\mathrm{op}}\times {\Gamma }^{\mathrm{op}},\mathcal{C})$.
To prove theorem 0.6, we need a lemma

引理 0.7. Let $\mathcal{C}$ be a category with finite products. Then $\mathcal{C}$ is semi-additive if the following conditions hold:

1.

The terminal objects $\ast \in \mathcal{C}$ is also initial.

2.

We have a natural transformation ${\mu }_x:(\Delta :x\mapsto x\times x)\mapsto i{d}_{\mathcal{C}}$ such that both compositions $$\begin{array}{rl}x& \simeq x\times \ast \stackrel{i{d}_x\times 0}{⟶}x\times x\stackrel{{\mu }_x}{\to }x\\ x& \simeq \ast \times x\stackrel{0\times i{d}_x}{⟶}x\times x\stackrel{{\mu }_x}{\to }x\end{array}$$are homotopic to $i{d}_x$ for all $x\in \mathcal{C}$, and the diagram commutes up to homotopy.

Now we give the proof of theorem 0.6.

证明. We first prove that $\mathrm{CMon}(\mathcal{C})$ is semi-additive by lemma 0.7. The constant functor $const\ast$ is the final object. To see it is initial, let $X\in \mathrm{CMon}(\mathcal{C})$ and compute $$\begin{array}{r}\mathrm{Nat}(const\ast ,X)\simeq {\mathrm{Hom}}_{\mathcal{C}}(\ast ,\underset{<n>\in {\Gamma }^{\mathrm{op}}}{\lim }{X}_n)\simeq {\mathrm{Hom}}_{\mathcal{C}}(\ast ,X_0)\simeq {\mathrm{Hom}}_{\mathcal{C}}(\ast ,\ast )\simeq \ast \end{array}$$Now we need to construct functorial maps ${\mu }_M:M\times M\to M$ for all $M\in \mathrm{CMon}(\mathcal{C})$. Consider the map $\times :{\Gamma }^{\mathrm{op}}\times {\Gamma }^{\mathrm{op}}\to {\Gamma }^{\mathrm{op}}$ by taking the product of sets and morphisms. It induces a functor $\mathrm{Fun}({\Gamma }^{\mathrm{op}},\mathcal{C})\to \mathrm{Fun}({\Gamma }^{\mathrm{op}},\mathrm{Fun}({\Gamma }^{\mathrm{op}},\mathcal{C}))$. We can see that the Segal conditions are preserved. So we get a functor $$\begin{array}{r}Double:\mathrm{CMon}(\mathcal{C})\to \mathrm{CMon}(\mathrm{CMon}(\mathcal{C}))\end{array}$$By construction, $Doubl{e}_1$ is just the identity and $Doubl{e}_2$ is naturally equivalent to the diagonal $\Delta :M\to M\times M$. Hence we define ${\mu }_M$ to be the multiplication $m:Doubl{e}_2\to Doubl{e}_1$ induced by $m:<2>\to <1>,1,2\mapsto 1$. This transformation is functoral and these conditons also hold.
Thus $\mathrm{CMon}(\mathcal{C})$ is semi-additive by the lemma 0.7. To show $\mathrm{CGrp}(\mathcal{C})$ is additive, we can know that for $G\in \mathrm{CGrp}(\mathcal{C})$, the codiagonal map $G\times G\to G$ is induced by ${\mu }_G$, hence by the multiplication on $Double(G)$. Thus we only need to show that $Double(G)$ is a cartesian commutative group it self. This is easy.
Finally, to show the adjunctions, just notice that using $\mathrm{CMon}(\mathcal{C})$ is semi-additive, then for $R=\mathrm{CMon}(-)$ and $\eta :R\Rightarrow id$, we have $\eta R:R\circ R\Rightarrow R$ and $R\eta :R\circ R\Rightarrow R$ are both equivaleces. So by some nonsenses, we have the adjunctions.

$\square$

推论 0.8. There is a commutative diagram of horizontal adjunction

We also have:

1.	Restricting $B$ to $\mathrm{CGrp}(\mathrm{Ani})$ gives a fully faithful functor $B:\mathrm{CGrp}(\mathrm{Ani})\to \mathrm{CMon}(\mathrm{Ani})$.
2.	Both functors $B,\Omega$ actually take values in $\mathrm{CGrp}(\mathrm{Ani})$.
3.	$\Omega B$ is left adjoint to the inclusion $\mathrm{CGrp}(\mathcal{C})\subseteq \mathrm{CMon}(\mathcal{C})$.

In particular, the top row adjunction restricts to the adjunction

推论 0.9. We get a commutative solid diagram

Finally we talk about free ${\mathbb{E}}_{\infty }$-monoids. Similarly, the forgetful functor $\mathrm{CMon}(\mathrm{Ani})\to \mathrm{Ani}$ has a left adjoint ${\mathrm{Free}}^{\mathrm{CMon}}:\mathrm{Ani}\to \mathrm{CMon}(\mathrm{Ani})$. For $X\in \mathrm{Ani}$, it is given by $$\begin{array}{r}{\mathrm{Free}}^{\mathrm{CMon}}(X{)}_1\simeq \coprod _{n⩾0}{X}_{h{\mathfrak{S}}_n}^n\end{array}$$We need to define homotopy-quotient under the action of a group. In general, given a functor $F:BG\to \mathcal{C}$, where $\mathcal{C}$ is an $\infty$-cateogry with colimits and $G$ is a discrete group, we put $$\begin{array}{r}{F}_{hG}:={\mathrm{colim}}_{BG}F\end{array}$$On each $X^n$, to get this homotopy-quotient, we give a functor $B\mathfrak{S}\to core(\mathrm{Ani})$ sending the $0$-complex of $B{\mathfrak{S}}_n$ to $X^n$. This is the same as giving a functor $\mathfrak{S}\to {\Omega }_{X^n}(core(\mathrm{Ani}))\simeq {\mathrm{Hom}}_{core(\mathrm{Ani})}(X^n,X^n)$. Then we just conder the permutation actions.