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Feb 20, 2019 at 10:27 vote accept Math Lover
Feb 11, 2019 at 10:35 comment added Mateusz Wasilewski I haven't checked the details, but I suppose so. One point which might be important, is that a non-unital algebra is an ideal in its unitisation and inclusions of ideals interact nicely with the maximal tensor product (which is not the case for arbitrary inclusions).
Feb 11, 2019 at 10:09 comment added Math Lover Thank you. For non unital case do we need to proceed by adjoining identity or some other way?
Feb 11, 2019 at 8:43 comment added Mateusz Wasilewski It really follows from the definition: the norm in the maximal tensor product is computed as the supremum of the norms under pairs of $\ast$-homomorphisms with commuting ranges. On the other hand, if we are given a $\ast$-homomorphism from the tensor product $A\otimes B$ (of unital algebras), we get a pair of homomorphisms with commuting ranges by restricting to $A\otimes \mathrm{1}$ and $1\otimes \mathrm{B}$.
Feb 9, 2019 at 7:03 comment added Math Lover Sorry can you give a reference for universal property of maximal tensor product? Thank you very much!
Feb 8, 2019 at 10:32 history answered Mateusz Wasilewski CC BY-SA 4.0