A $(n+1)$-topological quantum field theory $\mathcal{T}$ is a rigid symmetric monoidal functor from the category $(n+1)$-Cob of $n$-manifolds and $(n+1)$-cobordisms to FdVect. My question is about the symmetric monoidal structure of $(n+1)$-Cob and how it plays well with functoriality:

We consider our manifolds abstractly (as opposed to being embedded in some ambient space). The monoidal structure in $(n+1)$-Cob is given by disjoint union, thus $$M\otimes N = M \sqcup N$$ for manifolds $M,N$, and likewise for cobordisms. But since our manifolds are just abstract, surely this is exactly the same object as $$N\otimes M=N\sqcup M.$$ But when we pass to FdVect our functor gives $$ \mathcal{T}(M\otimes N)=\mathcal{T}M \otimes \mathcal{T}N $$ while $$ \mathcal{T}(N\otimes M) = \mathcal{T}N \otimes \mathcal{T}M $$ which are certainly isomorphic but distinct objects.

So clearly something is broken in my understanding of this, but what? Do we somehow distinguish between $M\otimes N$ and $N\otimes M$ in $(n+1)$-Cob? If so, how, given that these are just abstract manifolds?

  • $\begingroup$ You should think about what, really, you mean by the collection of symbols $M \sqcup N$. If you want this construction to give you an actual abstract manifold, it should also give you an actual abstract underlying set: which set is it? $\endgroup$ Jul 1, 2014 at 14:23
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    $\begingroup$ Yes, that's exactly what I mean: there are then canonical isomorphisms $M \sqcup N \to N \sqcup M$, which are not the identity maps, and which give the cobordism category a symmetric monoidal structure. $\endgroup$ Jul 1, 2014 at 14:36
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    $\begingroup$ Not quite, labelling by the set's "name" is not a good idea: what would $M \sqcup M$ be in that case? $\endgroup$ Jul 1, 2014 at 14:57
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    $\begingroup$ Any symmetric tensor category can be strictified, so there's certainly some version of the bordism category where $M \sqcup N = N \sqcup M$. Scrictification is not a very natural thing to do though, and Oscar's right that with standard definitions $M \sqcup N \neq N \sqcup M$. $\endgroup$ Jul 1, 2014 at 15:04
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    $\begingroup$ While it is possible to arrange for $M \amalg N = N \amalg M$ as objects, it is not in general possible to make the "swap" isomorphism $M \amalg N \to N \amalg M$ equal to the identity morphism. So there's really no point in forcing $M \amalg N = N \amalg M$ anyway. $\endgroup$
    – Zhen Lin
    Jul 1, 2014 at 17:13

1 Answer 1


As Oscar has explained in comments, with the most common definitions it's just not true that $M \sqcup N$ is exactly the same as $N \sqcup M$. But even if you were working with some version of the category of sets where they were equal, this wouldn't be a problem because tensor functors don't have to be strict. In other words, you don't know that $\mathscr{F}(M \sqcup N) = \mathscr{F}(M)\otimes \mathscr{F}(N)$, but instead only that you have isomorphisms $\mathscr{F}(M \sqcup N) \cong \mathscr{F}(M)\otimes \mathscr{F}(N)$ satisfying the natural coherence conditions. In particular, $\mathscr{F}(M \sqcup N) = \mathscr{F}(N \sqcup M)$ does not imply that $\mathscr{F}(M)\otimes \mathscr{F}(N) = \mathscr{F}(N)\otimes \mathscr{F}(M)$, but only that the latter are isomorphic in a coherent way.


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