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Let $A$ be a commutative unital C*-Algebra and let $X= Spec(A) $ be the corresponding compact Hausdorff space of characters. By Gelfand-Naimark duality we know that $$ X \times X = Spec(A \coprod A) $$ or in other words $$ A \coprod A = \mathcal{C}(X \times X).$$

Is there an algebraic way to construct the coproduct, as for example some form of topological tensor product or similar?

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  • $\begingroup$ The theorem stated in the question here might be a good rule of thumb. My guess is the tensor product (which is unambiguous since commutative C$^*$-algebras are nuclear). Maybe check whether the tensor product is a coproduct. $\endgroup$ – Josh Keneda Sep 5 '16 at 12:03
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The maximal tensor product satisfies the following universal property:
Let $A, B,$ and $C$ be C$^*$-algebras. If $\phi: A \rightarrow C$ and $\psi: B \rightarrow C$ are $*$-homomorphisms whose images commute, then there's a unique $*$-homomorphism $\phi \otimes \psi : A\otimes_{max} B \rightarrow C$ such that $(\phi\otimes \psi) (a \otimes b) = \phi(a)\psi(b)$. See page 193 of Murphy's book for a proof here.

If we take $A \otimes_{max} B$ with inclusions $i_A (a) = a \otimes 1$ and $i_B(b) = 1 \otimes b$, then the universal property above guarantees that we have found our coproduct in C$^*$-alg$_{com}^1$. Note that the "max" in the previous sentence wasn't necessary, since $A$ and $B$ are nuclear anyway. So you could just as easily take the spatial tensor product, if you like that construction better.

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  • $\begingroup$ The Murphy book I linked to also describes the maximal and spatial tensor products, in case you hadn't seen them. $\endgroup$ – Josh Keneda Sep 6 '16 at 8:39

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