TITLE: Can the monoidal structure on manifolds be strictified?
QUESTION [6 upvotes]: I'm asking this question purely out of curiosity.
Let $\{M_\alpha\}_{\alpha\in A}$ be a collection of closed smooth manifolds, with exactly one in every diffeomorphism class of closed smooth manifold.  Hence for each $\alpha,\beta\in A$, there exists a unique $\alpha\ast\beta\in A$ for which $M_{\alpha\ast\beta}$ is diffeomorphic to $M_\alpha\times M_\beta$.  Hence we get an binary operation $\ast:A\times A\to A$ which is associative and commutative.

Can we choose diffeomorphisms $\phi_{\alpha\beta}:M_\alpha\times M_\beta\to M_{\alpha\ast\beta}$ which are associative? what about associative and commutative?

To be precise, associativity means the following diagram commutes:
$$\begin{matrix}M_\alpha\times M_\beta\times M_\gamma&\rightarrow&M_{\alpha\ast\beta}\times M_\gamma\cr\downarrow&&\downarrow\cr M_\alpha\times M_{\beta\ast\gamma}&\rightarrow&M_{\alpha\ast\beta\ast\gamma}\end{matrix}$$
and commutativity means the following diagram commutes:
$$\begin{matrix}M_\alpha\times M_\beta&\rightarrow&M_{\alpha\ast\beta}\cr\downarrow&&\downarrow\cr M_\beta\times M_\alpha&\rightarrow&M_{\beta\ast\alpha}\end{matrix}$$

REPLY [4 votes]: You cannot have commutativity even for zero-dimensional manifolds, as you will see by considering the case where $\alpha=\beta$ and $M_\alpha=M_\beta=\{0,1\}$.
For associativity, my guess is that the monoid of isomorphism classes is probably a free abelian monoid (so we have "unique factorization").  If that is true then you can certainly arrange associativity.  It would be a bit fiddly to write down a precise proof and I do not have time just now.  However, you will get the idea if you first consider a category in which all objects are isomorphic to $A^i$ for some fixed object $A$ and some $i\in\mathbb{N}$, and then consider a category in which every object is of the form $A^i\times B^j$.