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. Can someone give a sketch or reference for the proof of the same result?
Thanks in advance
"Arbitrary limits" means inductive limits here. Given an inductive system $(A_i,\phi_{ij})$ with limit $(A,\phi_i)$, the system $(A_i\otimes_\max B,\phi_{ij}\otimes \mathrm{id}_B)$ is an inductive system. The precise statement of Blackadar's claim is that $(A\otimes_\max B,\phi_i\otimes \mathrm{id}_B)$ is an inductive limit of this system.
In my opinion the easiest way to check this uses the universal properties of inductive limits and maximal tensor products. For the sake of simplicity I assume that everything is unital.
We want to prove that $(A\otimes_\max B,\phi_i\otimes \mathrm{id_B})$ is the inductive limit of $(A_i\otimes_\max B,\phi_{ij}\otimes \mathrm{id}_B)$. Note that we already used the universal property for the definition of the maps $\phi_i\otimes \mathrm{id}_B$ and $\phi_{ij}\otimes\mathrm{id}_B$. We have to show that for every $C^\ast$-algebra $C$ and $\ast$-homomorphisms $\chi_i\colon A_i\otimes_\max B\to C$ such that $\chi_j\circ (\phi_{ij}\otimes\mathrm{id}_B)=\chi_i$ there exists a unique $\ast$-homomorphism $f\colon A\otimes_\max B\to C$ such that $f\circ(\phi_i\otimes\mathrm{id}_B)=\chi_i$.
Let $\alpha_i\colon A_i\to A_i\otimes_\max B,\,\alpha_i(a_i)=a_i\otimes 1$ and $\beta_i\colon B\to A_i\otimes_\max B,\,\beta_i(b)=1\otimes b$ (if you think of $A_i\otimes_\max B$ in terms of the universal property, these maps are part of the definition). Then $\chi_j\circ \alpha_j\circ \phi_{ij}=\chi_i\circ\alpha_i$. by the universal property of $(A,\phi_i)$ there exists a unique $\ast$-homomorphism $g\colon A\to C$ such that $g\circ\phi_i=\chi_i\circ\alpha_i$.
Moreover, $\chi_i\circ \beta_i(b)=\chi_i(1\otimes b)=\chi_j(1\otimes b)=\chi_j\circ\beta_j(b)$. Let $h=\chi_i\circ\beta_i$ (which is now independent of $i$). By the universal property of $A\otimes_\max B$ there exists a unique $\ast$-homomorphism $f\colon A\otimes_\max B\to C$ such that $f(a\otimes b)=g(a)h(b)$. In particular, $$ f(\phi_i(a_i)\otimes\mathrm{id}_B(b))=g(\phi_i(a_i))h(b)=\chi_i(\alpha_i(a))\chi_i(\beta_i(b))=\chi_i(a_i\otimes b). $$ Thus $f\circ(\phi_i\otimes\mathrm{id}_B)=\chi_i$.
To prove uniqueness of $f$, you basically have to go through the whole proof backwards. I leave this as an exercise together with the non-unital case. And there are probably more effective ways to write this proof down.