# The Unapologetic Mathematician

## Braidings on Span 2-Categories

Now that we can add a monoidal structure to our 2-category of spans, we want to add something like a braiding.

So, what’s a braided monoidal 2-category? Again, we need some data:

• For any objects $A$ and $B$ an equivalence $R_{A,B}:A\otimes B\rightarrow B\otimes A$ called the “braiding” of $A$ and $B$.
• For any 1-morphism $f:A\rightarrow A'$ and object $B$, a 2-isomorphism $R_{f,B}:R_{A',B}\circ(f\otimes B)\Rightarrow(B\otimes f)\circ R_{A,B}$ called the braiding of $f$ and $B$.
• For any object $A$ and morphism $g:B\rightarrow B'$, a 2-isomorphism $R_{A,g}:R_{A,B'}\circ(A\otimes g)\Rightarrow(g\otimes A)\circ R_{A,B}$ called the braiding of $A$ and $g$.
• For any objects $A$, $B$, and $C$, 2-isomorphisms $\tilde{R}_{(A|B,C)}:(B\otimes R_{A,C})\circ(R_{A,B}\otimes C)\Rightarrow R_{A,B\otimes C}$ and $\tilde{R}_{(A,B|C)}:(R_{A,C}\otimes B)\circ(A\otimes R_{B,C})\Rightarrow R_{A\otimes B,C}$ called the “braiding coherence 2-morphisms”.

We define the braiding to be an equivalence rather than an isomorphism because we don’t want to ask it to be exactly invertible. That is, there will be another 1-morphism $R^*_{A,B}:B\otimes A\rightarrow A\otimes B$ with 2-isomorphisms $R^*_{A,B}\circ R_{A,B}\cong1_{A\otimes B}$ and $R_{A,B}\circ R^*_{A,B}\cong1_{B\otimes A}$. The 2-isomorphisms braiding objects with 1-morphisms act in the place of naturality relations, allowing us to pull 1-morphisms back and forth through the object braiding. And then the hexagon identities from the definition of a braiding in a 1-category are now weakened to the braiding coherence 2-morphisms.

So, where are we going to find this structure for our bicategory $\mathbf{Span}(\mathcal{C})$? Well, let’s assume we have a braiding $\beta_{A,B}$ on $\mathcal{C}$. Then we’ll just define $R_{A,B}$ to be the span $(A\otimes B)\stackrel{1_{A\otimes B}}{\leftarrow}(A\otimes B)\stackrel{\beta_{A,B}}{\rightarrow}(B\otimes A)$. This is actually a 1-isomorphism, using the obvious inverse.

As we saw before with the tensorator, it turns out that we can set all of the $R_{f,B}$ and the $R_{A,g}$ to be the appropriate identities, as well as all the braiding coherence relations. Requiring the the monoidal product on $\mathcal{C}$ preserve pullbacks turns out to be an extremely powerful condition!

And now what are the conditions that make this data into a braided monoidal 2-category?

1. For 1-morphisms $f:A\rightarrow A'$ and $g:B\rightarrow B'$, we have
$(\bigotimes_{g,f}\circ R_{A,B})\bullet((g\otimes A')\circ R_{f,B})\bullet(R_{A',g}\circ(f\otimes B))=$
$((B'\otimes f)\circ R_{A,g})\bullet(R_{f,B'}\circ(A\otimes g))\bullet(R_{A',B'}\circ\bigotimes_{f,g}^{-1})$
2. For any 1-morphisms $f:A\rightarrow A'$ and $f':A\rightarrow A'$, 2-morphism $\alpha:f\Rightarrow f'$, and object $B$, we have
$((B\otimes\alpha)\circ R_{A,B})\bullet R_{f,B}=R_{f',B}\bullet(R_{A',B}\circ(\alpha\otimes B))$
3. For any 1-morphisms $g:B\rightarrow B'$ and $g':B\rightarrow B'$, 2-morphism $\beta:g\Rightarrow g'$, and object $A$, we have
$((A\otimes\beta)\circ R_{A,B})\bullet R_{A,g}=R_{A,g'}\bullet(R_{A,B'}\circ(\beta\otimes A))$
4. For any 1-morphisms $f:A\rightarrow A'$ and $f':A'\rightarrow A''$ and object $B$, we have
$((B\otimes f')\circ R_{f,B})\bullet(R_{f',B}\circ(f\otimes B))=R_{f'\circ f,B}$
5. For any 1-morphisms $g:B\rightarrow B'$ and $g':B'\rightarrow B''$ and object $A$, we have
$((A\otimes g')\circ R_{A,g})\bullet(R_{A,g'}\circ(A\otimes g))=R_{A,g'\circ g}$
6. For any objects $A$, $B$, and $C$, and 1-morphism $f:C\rightarrow C'$, we have
$R_{A,B\otimes f}\bullet(\tilde{R}_{(A|B,C')}\circ(A\otimes B\otimes f))=$
$((B\otimes f\otimes A)\circ\tilde{R}_{(A|B,C)})\bullet((B\otimes R_{A,f})\circ(R_{A,B}\otimes C))\bullet((B\otimes R_{A,C'})\circ\bigotimes_{R_{A,B},f})$
7. There are five more conditions like the last one, with the 1-morphism in other slots and different associations of the three terms.
8. For any objects $A$, $B$, $C$, $D$, we have
$\tilde{R}_{(A|B,C\otimes D)}\bullet((B\otimes\tilde{R}_{(A|C,D)})\circ(R_{A,B}\otimes C\otimes D))=$
$\tilde{R}_{(A|B\otimes C,D)}\bullet((B\otimes C\otimes R_{A,D})\circ(\tilde{R}_{(A|B,C)}\otimes D))$
9. There are two more conditions like the last one, corresponding to different ways of associating the four terms.
10. For any objects $A$, $B$, and $C$, we have
$(\tilde{R}_{(A|B,C)}\circ(A\otimes R_{B,C}))\bullet R_{A,R_{B,C}}^{-1}\bullet((R_{B,C}\otimes A)\circ\tilde{R}_{(A|B,C)}^{-1})=$
$((C\otimes R_{A,B})\circ\tilde{R}_{(A,B|C)}^{-1})\bullet R_{R_{A,B},C}\bullet(\tilde{R}_{(A,B|C)}\circ(R_{A,B}\otimes C))$.
11. $R_{\cdot,\cdot}$, $\tilde{R}_{(\cdot|\cdot,\cdot)}$, and $\tilde{R}_{(\cdot,\cdot|\cdot)}$ are each the identity whenever one of their slots is filled with the unit object $\mathbf{1}$.

Now in our currrent situation, almost all the structure here is trivial, and we have proofs to match! Specifically, all of the 2-morphisms that show up here are just the identities for various 1-morphisms. That is, they move between different ways of writing the same 1-morphism. For example, the first condition just reduces to saying that the composite of three applications of the identity 2-morphism on the span $R_{A',B'}\circ g\circ f$ are the same as three other applications of the identity on the same span. And on and on they go, identities on identities, and there’s ultimately nothing to do here.

So the upshot is that if we have a braiding on $\mathcal{C}$, then $\mathbf{Span}(\mathcal{C})$ is a braided monoidal 2-category. Dually, $\mathbf{CoSpan}(\mathcal{C})$ gets this structure if $\mathcal{C}$ is a braided monoidal category with pushouts preserved by the monoidal product.

October 12, 2007 Posted by | Category theory | 10 Comments