ScalingStacks

Given a semistrict braided monoidal 2-category $\mathcal{C}$, we would like to embed it in its center by a braided monoidal 2-functor $\mathcal{F}:\mathcal{C}\to\mathcal{Z}(\mathcal{C})$. Developing a general definition of ‘braided monoidal 2-functor’ would require a fair amount of work. Luckily, in our case we can restrict ourselves to a very strict sort of braided monoidal 2-functor which is easy to define. The following definition should not be taken as fundamental; it is simply designed to be the strictest one for which our embedding theorem holds.

Proof - First let us show that $\mathcal{F}$ is a monoidal 2-functor. For this, we must define a pseudonatural 1-morphism $\xi_{A,B}:\mathcal{F}(A)\otimes\mathcal{F}(B)\to\mathcal{F}(A\otimes B)$, where $A,B\in\mathcal{C}$. We let

To be a morphism in $\mathcal{Z}(\mathcal{C})$, $\xi_{A,B}$ has to satisfy $({\to}\otimes(\bullet\otimes\bullet))$. This is equivalent to the axiom $((\bullet\otimes\bullet)\otimes(\bullet\otimes\bullet))$ in $\mathcal{C}$. We show that $\xi$ is natural, not merely pseudonatural. To this end we first show that for any morphism $f\colon A\to A^{\prime}$ in $\mathcal{C}$ the following diagram commutes ‘on the nose’. (Remember our shorthand symbol for tensor products in $\mathcal{Z}(\mathcal{C})$.)

The morphism “first right, then down” equals

The morphism “first down, then right” equals

These two $\mathcal{Z}(\mathcal{C})$-morphisms are equal, since by $(({\to}\otimes\bullet)\otimes\bullet)$, the underlying 2-morphisms are equal:

Then we must show naturality with respect to morphisms of the form $g\colon B\to B^{\prime}$, which is similar. Finally, it is easy to show that $\xi$ is also compatible with 2-morphisms. Using the axiom $((\bullet\otimes\bullet\otimes\bullet)\otimes\bullet)$ we see that $\xi$ fulfills the associativity condition on the nose, so we can define $\alpha$ to be the identity.

Next, we show that $\mathcal{F}$ is braided and, in addition, $\mathcal{F}_{R}={\rm id}$.

This is done by the following calculation.

Here $\xi^{-1}=(1_{A\otimes B},\tilde{T}_{(A,B|-)})$ is the inverse of $\xi$, as can be easily verified using the composition law for 1-morphisms in $\mathcal{Z}(\mathcal{C})$. The third equation holds by our assumption that $S^{+}=S^{-}$. The fifth equation holds according to our definition of the braiding in $\mathcal{Z}(\mathcal{C})$.

Finally, we must check that both diagrams in the definition of a strong braided monoidal 2-functor commute. In the first diagram all the 2-morphisms except $3$ and $4$ are identities. Note that the 2-morphism labeled $1$, namely $T_{(\mathcal{F}(X),\xi)}$, is the identity, since the 1-morphism part of $\xi$ is the identity, and that by an application of axiom $(\bullet\otimes{\Downarrow})$, face $1$ commutes on the nose. The remaining 2-morphisms $3$ and $4$ are equal.

In the second diagram the 2-morphism $3$ is the identity. Here, the 2-morphism $1$ is defined to be $R_{(X,Y|Z)}^{-1}$ and hence agrees with $4$. The remaining 2-morphisms are identities, so this diagram also commutes. ∎