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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: $$\begin{matrix}M_\alpha\times M_\beta&\rightarrow&M_{\alpha\ast\beta}\cr\downarrow&&\downarrow\cr M_\beta\times M_\alpha&\rightarrow&M_{\beta\ast\alpha}\end{matrix}$$

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  • $\begingroup$ For the last diagram you also need to choose maps $\psi_{\alpha\beta} \colon M_{\alpha*\beta} \to M_{\beta*\alpha}$. (They are not yet part of your data.) On the other hand, you can define them via the other three isomorphisms in that diagram. Commutativity of the commutativity diagram is then a direct consequence. $\endgroup$
    – jmc
    Commented Feb 22, 2014 at 7:46
  • $\begingroup$ For the associativity, I don't think that you can get a commutative diagram in general. I would fancy that you also need maps $\chi_{\alpha\beta\gamma} \colon M_{(\alpha*\beta)*\gamma} \to M_{\alpha*(\beta*\gamma)}$. (Of course the domain and codomain are both just $M_{\alpha*\beta*\gamma}$, but I don't think you can arrange for $\chi_{\alpha\beta\gamma}$ to be the identity morphism in general. $\endgroup$
    – jmc
    Commented Feb 22, 2014 at 7:50
  • $\begingroup$ @jmc the maps $\Psi$ and $\chi$ your are talking about, in my understanding of the problem are assumed to be the identity, isn't? $\endgroup$
    – Daniele Zuddas
    Commented Feb 22, 2014 at 10:31
  • $\begingroup$ @jmc: Daniele is right, I meant for these maps to be the identity. But certainly we could also pick an associator map $\chi_{\alpha\beta\gamma}:M_{\alpha\ast\beta\ast\gamma}\to M_{\alpha\ast\beta\ast\gamma}$ and a commutator map $\psi_{\alpha\beta}:M_{\alpha\ast\beta}\to M_{\alpha\ast\beta}$ to force the diagrams to commute. Then the question of interest would be: can we choose the associator and commutator to satisfy the natural compatibility relations? (e.g. the pentagon relation for the associator). $\endgroup$
    – John Pardon
    Commented Feb 22, 2014 at 18:17
  • $\begingroup$ @JohnPardon – I guessed that is what you meant. However, in the commutativity diagram you actually used an unnamed map $M_{\alpha*\beta} \to M_{\beta*\alpha}$ so it was not clear to me whether you assumed it to be the identity or not. $\endgroup$
    – jmc
    Commented Feb 22, 2014 at 20:19

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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 $A^i\times B^j$.

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    $\begingroup$ The monoid is not free abelian. For example, Donaldson showed that there exist simply connected $4$-manifolds $M_1$ and $M_2$ which are $h$-cobordant but not diffeomorphic. But then $M_1\times S^2$ and $M_2\times S^2$ are $h$-cobordant (multiply the original $h$-cobordism by $S^2$), and thus diffeomorphic. $\endgroup$
    – John Pardon
    Commented Feb 22, 2014 at 18:28

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