Here is a quotation of a book:
Proposition 3.7.1. If $0 \rightarrow J \rightarrow A \rightarrow (A/J)\rightarrow 0$ is an exact sequence, then for every $B$, the natural sequence $$0 \rightarrow J\otimes_{max} B \rightarrow A\otimes_{max} B \rightarrow(A/J)\otimes_{max}B\rightarrow 0$$ is also exact. (Here, $A$ is a C*-algebra and $J$ is a C*-subalgebra, I think.)
In order to prove the conclusion, of course, we need to show $J\otimes_{max} B \rightarrow A\otimes_{max} B$ is injective and $A\otimes_{max} B \rightarrow(A/J)\otimes_{max}B$ is surjective. However, I can understand the proof of this proposition.
In the proof, the author say "$J\otimes_{max} B \rightarrow A\otimes_{max} B$ is injective follows from Corollary 3.6.4, while $A\otimes_{max} B \rightarrow(A/J)\otimes_{max}B$ is surjective thanks to the fact that $^{*}$-homomorphisms of C$^{*}$-algebras always have closed range." I can not understand this explaination:
The $J$, here, is a hereditary subalgebra of $A$?
How to use "C*-algebra always have closed range" to illustrate the surjectivity?
Here is so called Corollary 3.6.4.
Corollary 3.6.4. If $A\subset B$ is hereditary subalgebra, then for every $C$ we have a natural inclusion $$A\otimes_{max} C\subset B\otimes_{max} C.$$
Here $J$ is a closed two-sided ideal of $A$. This is necessary in order for $A/J$ to be a $C^*$-algebra. Such an ideal is a hereditary subalgebra, so we can apply Corollary 3.6.4.
Working with the algebraic tensor products shows that the *-homomorphism from $A \otimes_{max} B$ to $(A/J) \otimes_{max} B$ has dense range. As the range is also closed, it must be all of $(A/J) \otimes_{max} B$.