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In the answer to question Localization of symmetric monoidal category, it was mentioned that '' Assuming that the tensor product of two morphisms in $S$ is again in $S$, the localised category should inherit a symmetric monoidal structure, just by the universal property.''

So I want to know by which universal property we can show that $\mathcal{M}[S^{-1}]$ inherit a symmetric monoidal structure?

Since I cannot comment on the original answer, I posted this as a new (stupid) question.

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  • $\begingroup$ I would appreciate it a lot if you could favour to answer this question..@Neil Strickland $\endgroup$ – kousaka Nov 30 '16 at 10:23
  • $\begingroup$ You may would like to ask this question on math.stackexchange instead. Anyway, the assumption that $S$ is closed with respect to the monoidal product $\otimes$ implies that the composition of the localisation functor $L_S:\mathcal{M}\to \mathcal{M}[S^{-1}]$ with $\otimes$ sends a pair morphisms $(s,t)$ in $S\times S$ to an isomorphisms in $\mathcal{M}[S^{-1}]$. Thus, the composition $L_S\circ \otimes$ factorises through $L_S\times L_S$, defining the desired monoidal product on $\mathcal{M}[S^{-1}]\times\mathcal{M}[S^{-1}]$. $\endgroup$ – user337830 Nov 30 '16 at 12:45
  • $\begingroup$ The universal property used here is that of the functor $L_S\times L_S:\mathcal{M}\times\mathcal{M}\to \mathcal{M}[S^{-1}]\times\mathcal{M}[S^{-1}]$. $\endgroup$ – user337830 Nov 30 '16 at 12:47
  • $\begingroup$ @user337830 Your explanation is very understandable, thanks a lot!!! $\endgroup$ – kousaka Nov 30 '16 at 14:01
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    $\begingroup$ Note that using the "@" here will not send a notification to Neil Strickland, since he has nothing to do with this post. $\endgroup$ – David White Nov 30 '16 at 16:10
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A reference regarding the monoidal structure on $M[S^{-1}]$ is Brian Day's Note on monoidal localisations. It also talks about the enriched setting, and about monoidal completion. It includes a proof of Neil Strickland's point about requiring $S$ to be closed under $\otimes$, from the other MO thread you linked to.

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When $S=\mathcal{C}$, it is fairly straightforward to define a symmetric monoidal structure on the groupoid completion $\mathcal{G}\mathcal{C}:=\mathcal{C}\left[ S^{-1} \right]$. I couldn't find a reference for this fact, so I'll write it out in this answer. The goal is to prove the following claim: the groupoid completion of a symmetric monoidal category is a symmetric monoidal category.

Let $\mathcal{C}$ be a small symmetric monoidal category and let $\mathcal{G}\mathcal{C}$ denote the groupoid completion of $\mathcal{C}$. By the universal property of the groupoid completion, this defines a functor $\mathcal{G}(-): \mathbf{Cat}\to \mathbf{Gpd}$, to the category of small groupoids. This functor is left-adjoint: for any small category $\mathcal{D}$ and any small groupoid $\mathcal{E}$, there is a natural bijection of sets: \begin{equation}\label{eq adj} \mathbf{Gpd}(\mathcal{G}\mathcal{D},\mathcal{E}) \cong \mathbf{Cat}(\mathcal{D}, \mathcal{E}), \end{equation} (This is just the universal property.) In fact, this bijection can be enriched to an isomorphism of categories:

Equation 1 \begin{equation} \underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D},\mathcal{E}) \cong \underline{ \mathbf{Cat} }(\mathcal{D}, \mathcal{E}). \end{equation}

By virtue of being a left-adjoint, the functor $\mathcal{G}(-)$ preserves arbitrary colimits. It turns out that it also preserves finite products:

Lemma 1: For any pair of small categories $\mathcal{C}$ and $\mathcal{D}$, there is a natural isomorphism $\mathcal{G}(\mathcal{C}\times\mathcal{D})\cong\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}$.

Proof: The proof is purely formal. By the Yoneda lemma, it suffices to exhibit a natural isomorphism between the functors represented by $\mathcal{G}(\mathcal{C}\times\mathcal{D})$ and $\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}$. For any groupoid $\mathcal{E}$, we have natural isomorphisms \begin{equation*} \begin{aligned} \mathbf{Gpd}(\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}, \mathcal{E}) & \cong \mathbf{Gpd} (\mathcal{G}\mathcal{C}, \underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D}, \mathcal{E}))& \text{by the exponential law for groupoids}\\ & \cong \mathbf{Cat}(\mathcal{C}, \underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D}, \mathcal{E})) & \text{by the uni. prop. of $\mathcal{G}\mathcal{C}$}\\ & \cong \mathbf{Cat}(\mathcal{C}, \underline{ \mathbf{Cat} }(\mathcal{D}, \mathcal{E})) & \text{by eq. 1 }\\ & \cong \mathbf{Cat}(\mathcal{C}\times\mathcal{D}, \mathcal{E})& \text{by the exponential law for categories}\\ & \cong \mathbf{Gpd}(\mathcal{G}(\mathcal{C}\times\mathcal{D}), \mathcal{E})& \text{by the uni. prop. of $\mathcal{G}(\mathcal{C}\times\mathcal{D})$}. \end{aligned} \end{equation*}

Explicitly, the above isomorphism simply makes morphisms of the form $q_{\mathcal{C}\times\mathcal{D}}(f,g)$ correspond to morphisms of the form $(q_\mathcal{C}(f), q_{\mathcal{D}}(g))$.

Now given a symmetric monoidal category $\mathcal{C}$ with tensor product $\otimes$, we define a functor $\otimes'$ as the following composite:

Diagram 1

In order to show that $\otimes'$ is part of a symmetric monoidal structure on $\mathcal{G}\mathcal{C}$, first observe the following.

Remark: Consider the following diagram:

Diagram 2

Here the square commutes by definition of $\mathcal{G}(\otimes)$, and the triangle commutes by definition of $\otimes'$. Using the explicit description of the isomorphism, we see that commutativity of the above diagram indicates that we have \begin{equation} q_\mathcal{C}(f\otimes g) = q_\mathcal{C} (f) \otimes' q_\mathcal{C} (g), \end{equation} for any pair of morphisms $f$ and $g$ in $\mathcal{C}$.

Finally we have the following:

Proposition: The groupoid completion of a symmetric monoidal category is a symmetric monoidal category.

Proof: Let $(\mathcal{C}, \otimes, e, \alpha, \lambda, \rho, \gamma)$ be a symmetric monoidal category. The symmetric monoidal structure on the groupoid completion $\mathcal{G}\mathcal{C}$ is given by $(\otimes', e, q_\mathcal{C}(\alpha), q_\mathcal{C}(\lambda), q_\mathcal{C}(\rho), q_\mathcal{C}(\gamma))$.

The fact that this indeed defines a symmetric monoidal structure follows from the remark. To illustrate, we verify the triangle axiom. Let $a$ and $c$ be objects in $\mathcal{G}\mathcal{C}$ (\ie objects in $\mathcal{C}$). The triangle axiom for $\mathcal{C}$ states that the following diagram commutes.

Diagram 3

Applying the functor $q_\mathcal{C}$, we see that the following diagram commutes.

Diagram 4

Here we have used that $q_\mathcal{C}(1\otimes\lambda)=1\otimes q_\mathcal{C}(\lambda)$ and $q_\mathcal{C}(\rho\otimes1)=q_\mathcal{C}(\rho)\otimes 1$, which follows from the remark and functoriality of $q_\mathcal{C}$. This verifies the triangle axiom for $\mathcal{G}\mathcal{C}$, and the other axioms are verified similarly.

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