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I am reading Blackadar's book on Operator algebras. In $\Pi 9.6.5$ Blackader says that

Maximal Tensor products commute with arbitrary limits.

In the same book the proof of this fact is not given.In the literature also i could not find a proof of this. Can someone give a sketch or reference for the proof of the same result?

Thanks in advance

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  • $\begingroup$ Can you give a reference for what Blackadar means by "limits" here? Is this "limits" in the sense of category theory? Naively I would expect them to work better with "colimits" $\endgroup$ – Yemon Choi Feb 8 at 13:07
  • $\begingroup$ @YemonChoi: By limits he meant the directed limits. Probably the same holds for colimits also but I am not sure. $\endgroup$ – Math Lover Feb 9 at 7:01
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I think it goes like this (I am only dealing with the unital case, I am not exactly sure how this works in the non-unital case).

Let $(A_{i})_{i\in I}$ be an inductive system of unital $C^{\ast}$-algebras with unital connecting homomorphisms $f_{ij}: A_{i} \to A_{j}$ and let's denote the limit by $A$. If $B$ is another unital $C^{\ast}$-algebra, we would like to show that the inductive limit of $(A_{i}\otimes_{\rm{max}} B)_{i\in I}$ is isomorphic to $A\otimes_{\rm{max}} B$. In order to do so, we will show that $A\otimes_{\rm{max}}B$ has the required universal property, i.e. any compatible family of $\ast$-homomorphisms from $A_{i}\otimes_{\rm{max}} B$ to a unital $C^{\ast}$-algebra $C$ gives rise to a $\ast$-homomorphism from $A\otimes_{\rm{max}} B$. Note now that, by the universal property of the maximal tensor product, a $\ast$-homomorphism $\varphi_{i}: A_{i}\otimes_{\rm{max}} B \to C$ is given by $\varphi_{i} = \theta_{i}\cdot \psi_{i}$, where $\theta_{i}:A_{i} \to C$ and $\psi_{i}: B \to C$ are $\ast$-homomorphisms with commuting ranges. As connecting maps are of the form $f_{ij}\otimes \rm{Id}$, we can check that $\psi_{i}=\psi_{j}$, which we call $\psi$ from now on, and $\theta_{i} = \theta_{j}\circ f_{ij}$. Since $A$ is the limit of $A_i$'s, we get a $\ast$-homomorphism $\theta:A\to C$. As the union of ranges of $A_i$'s inside $A$ is dense, this $\ast$-homomorphism has range commuting with the range of $\psi$, so we get a $\ast$-homomorphism $\varphi: A\otimes_{\rm{max}} B \to C$ given by $\varphi(x\otimes y):= \theta(x)\psi(y)$.

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  • $\begingroup$ Sorry can you give a reference for universal property of maximal tensor product? Thank you very much! $\endgroup$ – Math Lover Feb 9 at 7:03
  • $\begingroup$ 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 $1\otimes \mathrm{B}$. $\endgroup$ – Mateusz Wasilewski Feb 11 at 8:43
  • $\begingroup$ Thank you. For non unital case do we need to proceed by adjoining identity or some other way? $\endgroup$ – Math Lover Feb 11 at 10:09
  • $\begingroup$ 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). $\endgroup$ – Mateusz Wasilewski Feb 11 at 10:35

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