The maximal tensor product is a continuous functor

I am trying to prove continuity of the maximal tensor product functor. I have a problem in the proof that I cannot see how to handle; If anyone could give me a clue on how to go on from here, I would really appreciate it. So here it goes:

Let $$B$$ be a $$C^*$$-algebra. I am trying to show that if $$A_1\xrightarrow{\varphi_1} A_2\xrightarrow{\varphi_2}\dots$$ is an inductive limit of $$C^*$$-algebras with inductive limit $$(A,\{\mu_n\})$$, then $$$$\varinjlim(A_1\otimes_{\max}B\xrightarrow{\varphi_1\otimes_{\max}\text{id}_B}A_2\otimes_{\max}B\xrightarrow{\varphi_2\otimes_{\max}\text{id}_B}\dots)=(A\otimes_{\max}B,\{\mu_n\otimes_{\max}\text{id}_B\}).\;\;\;(\star)$$$$

Since $$-\otimes_{\max}B$$ acts as a functor, we easily have that $$(\mu_{n+1}\otimes_{\max}\text{id}_B)\circ(\varphi_n\otimes_{\max}\text{id}_B)=\mu_n\otimes_{\max}\text{id}_B$$ for all $$n\geq1$$. By the universal property of the inductive limit, in order to verify $$(\star)$$, we need to check that:

1. $$A\otimes_{\max}B=\overline{\bigcup_{n=1}^\infty(\mu_n\otimes_{\max}\text{id}_B)(A_n\otimes_{\max}B)}$$
1. For any $$n\geq1$$: $$\ker(\mu_n\otimes_{\max}\text{id}_B)=\{x\in A_n\otimes_{\max}B: \lim_{m\to\infty}\|\varphi_{m,n}\otimes_{\max}\text{id}_B(x)\|=0\}.$$

Condition (1) is easily verified with some density arguments. Condition (2) is the tricky one; of course the RHS is easily seen to be a subset of LHS, so we have boiled it down to the other inclusion. From what I've seen around here (see this post e.g.) in relevant posts, one needs to use the exactness of the maximal tensor product functor. So we consider the short exact sequence $$0\to\ker(\mu_n)\xrightarrow{\iota}A_n\xrightarrow{\pi}\mu_n(A_n)\to0$$ where $$\pi$$ is the co-restriction of $$\mu_n$$, i.e. $$\pi:A_n\to\mu_n(A_n)$$, $$\pi(a)=\mu_n(a)$$ for all $$a\in A_n$$. We apply the maximal tensor product to conclude that $$0\to\ker(\mu_n)\otimes_{\max}B\xrightarrow{\iota\otimes_{\max}\text{id}_B}A_n\otimes_{\max}B\xrightarrow{\pi\otimes_{\max}\text{id}_B}\mu_n(A_n)\otimes_{\max}B\to0$$ is a short-exact sequence, in particular $$\ker(\pi\otimes_{\max}\text{id}_B)=\text{Im}(\iota\otimes_{\max}\text{id}_B)$$. Now this last equation would make things really easy for us to continue if $$\pi=\mu_n$$ (i.e. if $$\mu_n:A_n\to A$$ was surjective, but of course this is not necessarily the case). My question is, how does one handle the general case?

Sure, if $$j:\mu_n(A_n)\to A$$ is the inclusion map, then $$\mu_n=j\circ\pi$$. If the map $$j\otimes_{\max}\text{id}_B$$ was injective (which, since $$B$$ is ANY C*-algebra, happens if-f every non-degenerate rep. of $$\mu_n(A_n)$$ extends uniquely to a c.c.p. map on $$A$$, see Brown-Ozawa, Chapter 3), then sure, $$\ker(\mu_n\otimes_{\max}\text{id}_B)=\ker(\pi\otimes_{\max}\text{id}_B)$$ and things go on smoothly. But again, this is not always the case..

Edit: By Narutaka Ozawa's hint, I figured it out:

First one proves the result for the special case that $$B$$ is unital, each $$A_n$$ is unital and all the connecting maps preserve the unit. This is done by using the universality of the maximal tensor product to construct an inverse to $$C\to A\otimes_{\max}B$$, where $$C=\varinjlim(A_n\otimes_{\max}B)$$.

Second, one shows the following lemma: Given two inductive systems $$(A_n,\varphi_n)$$ and $$(B_n,\psi_n)$$ with inductive limits $$(A,\{\mu_n\})$$ and $$(B,\{\nu_n\})$$ respectively such that there exist injective *-homomorphisms $$j_n:A_n\to B_n$$ such that $$j_n\circ\varphi_n=\psi_n\circ j_{n+1}$$, then there exists a unique *-homomorphism $$\omega:A\to B$$ such that $$\omega\circ\mu_n=\nu_n\circ j_n$$ and $$\omega$$ is injective (the non-trivial part is injectivity of $$\omega$$).

Finally, the general case follows from the above lemma, the fact that the unitization functor is continuous and the fact that the maximal tensor product preserves inclusions when applied to ideals (more specifically when applied to $$A\subset\tilde{A}$$).

$$\otimes_{\max}$$ is continuous by universality: $$\varinjlim (A_n\otimes_{\max}B)$$ is a C*-completion of the algebraic tensor product $$A\otimes_{\mathrm{alg}} B$$ and one is left to check maximality (which is easy).
On the other hand, $$A_n\otimes_{\max}B \to A\otimes_{\max}B$$ may not be faithful. For example, consider a non-nuclear C-subalgebra $$B$$ of a nuclear C-algebra $$D$$ and put $$A_n:=D\otimes\cdots\otimes D\otimes \bigotimes_{k>n}B \subset \bigotimes_{k\in{\mathbf N}} D=:A.$$ Let's say the tensor product is the spatial one here. Since $$B\otimes_{\max}B^{\mathrm{op}}\neq B\otimes_{\min}B^{\mathrm{op}}$$, the map $$B\otimes_{\max} B^{\mathrm{op}}\to D\otimes_{\max}B^{\mathrm{op}}=D\otimes_{\min}B^{\mathrm{op}}$$ is not faithful. This implies that for every $$n, the map $$A_n\otimes_{\max} B^{\mathrm{op}}\to A_m\otimes_{\max}B^{\mathrm{op}}$$ is not faithful.
• Hello, thank you very much for your reply. Right now I am trying to figure out why $\varinjlim(A_n\otimes_{\max}B)$ must be a C*-tensor product of $A,B$. If this becomes clear to me, then I can see that universality is verified, so the C*-tensor product is the maximal one. The example is also very nice, thanks! Jan 29, 2021 at 17:23