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Let $(A_n, \phi_n)$ be an inductive system of $C^*$ algebras and let $B$ be an arbitary $C^*$ algebra.
I want to prove $(\varinjlim A_n)\otimes_{max} B \cong \varinjlim (A_n \otimes_{max} B)$.
This should be a well known fact (for $C^*$-algebraists), however I can't find a proof in the literature, but there are similar questions: See here (1) When do tensor products of C*-algebras commute with colimits? or here (2) https://math.stackexchange.com/questions/1972645/tensor-product-commutes-with-direct-limits. The proof is nearly the same as in (2), except the proof that $\lambda$ as in (2) is injective and here injectivity of $\phi_n$ is not required.

However, using the notations as in (2), I need help to verify that $\lambda$ is injective. Injectivity of $\lambda$ follows if you can check that $\ker(\varphi^n\otimes_{max} \mathrm{id}_B)=\overline{\bigcup\limits_{j>i}\ker(\varphi_{j,i}\otimes_{max} \mathrm{id}_B)}$, and the last claim follwos from $\ker(\varphi^n\otimes_{max} \mathrm{id}_B)=\ker(\varphi^n)\otimes_{max} B$ (see (1)). However, how to prove $$\ker(\varphi^n\otimes_{max} \mathrm{id}_B)=\ker(\varphi^n)\otimes_{max} B$$ in this general case, if it's true? It seems that the exactness of the tensor product-functor $-\otimes_{max}B$ is important for it. Or do you know another proof for injectivity of $\lambda:\varinjlim (A_n \otimes_{max} B)\to (\varinjlim A_n)\otimes_{max} B$?

Another approach to proof $(\varinjlim A_n)\otimes_{max} B \cong \varinjlim (A_n \otimes_{max} B)$ might be using the universal property of maximal tensor products. The idea is:

Starting with $\hom (\varinjlim\limits_n (A_n \otimes_{max} B), C)\cong \varprojlim\limits_n\hom(A_n \otimes_{max} B,C)$, the universal property of maximal TP's is that $\hom(A_n \otimes_{max} B,C)$ can be identified with $\{(\alpha_n,\beta_n)| \alpha_n: A_n\to C, \beta_n: B\to C, [\alpha_n(A_n),\beta_n(B)]=0\}$. Now, it looks like $\beta_n$ does not really depend on $n$ (for exact arguments, I'm not sure if a case distinction for unital and non-unital c*algebras is needed), and then use the universal property of colimits to obtain $\hom (\varinjlim\limits_n (A_n) \otimes_{max} B, C)$. Or is there anything wrong with it?

However, my main question remains: How to prove that $\lambda:\varinjlim (A_n \otimes_{max} B)\to (\varinjlim A_n)\otimes_{max} B$ injective?

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  • $\begingroup$ Have you tried defining an inverse for your map $\lambda$ by using the universal property of $\otimes_{max}$? In case all algebras are unital and the connecting maps preserve units, this should be easy. $\endgroup$
    – Ruy
    Commented Jun 25, 2017 at 21:49
  • $\begingroup$ Can you show that $(-) \otimes_{max} B$ preserves all short exact sequences? If so, consider$0\rightarrow \ker(\phi^n)\rightarrow A \rightarrow \im(\phi^n)\rightarrow 0$ $\endgroup$
    – m07kl
    Commented Jun 26, 2017 at 23:54
  • $\begingroup$ @Ruy thank you. yes, I have tried this (under this unital - assumptions). However, I was curious how to prove injectivity in a way as stated in the thread. $\endgroup$
    – Sabrina Gemsa
    Commented Jun 27, 2017 at 6:28
  • $\begingroup$ @m07kl thank you. I am already aware of this fact and that you have to use this fact in this way (even though I just mentioned that you have to use the exactness of $\otimes_{max}$ in some way), however, I got stuck. $\endgroup$
    – Sabrina Gemsa
    Commented Jun 27, 2017 at 6:32
  • $\begingroup$ $\ker(\varphi^n\otimes_{max} \mathrm{id}_B)=\ker(\varphi^n)\otimes_{max} B$ seems to be wrong. I'm going to edit my question later. $\endgroup$
    – Sabrina Gemsa
    Commented Jun 29, 2017 at 10:26

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