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?



How do we handle the symmetry condition in nCob and TQFTs?
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.


