Let $A$ and $B$ be two $C^\*$-algebras, and let $p:A \to B$ be a surjective norm-decreasing $\*$-homomorphism which is injective on a dense $\*$-sub-algebra of $A$. Can such a map have non-trivial kernel, and if so, is it possible that the $K$-theory groups of $A$ and $B$ can be non-isomorphic?



$K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras
Yes to both.$\newcommand{\Cst}{{\rm C}^\*}$
 The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any non-amenable discrete group will provides examples for your first question.


There are non-amenable groups $G$ for which $q$ induces an isomorphism on $K$-theory — I think the standard name for such groups is $K$-amenable. Lance proved that free groups have this property. But infinite groups with Kazhdan's Property (T) do not have this property because the so-called Kazhdan projection in the full group $\Cst$-algebra lies in the kernel of $q$. (Thanks to Jamie Gabe in comments for clarifying/sharpening my original statement.)


(See also this MO question [$\*$-algebras, completions, and $K$-theory](https://mathoverflow.net/questions/315907/algebras-completions-and-k-theory) )


