$K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras 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?
 A: There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map.
(My first example took $A = C[0,3]$ and let $A_0$ be the set of all polynomials in $x$ with rational coefficients and $p: A \mapsto \mathbb{C}$ the point evaluation at $x = e$. Since $e$ is transcendental, $p$ is injective on $A_0$. But $A_0$ is not a $*$-subalgebra over $\mathbb{C}$.)
A: 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 )
