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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?

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    $\begingroup$ All star-HMs between Cstar algebras are automatically norm-decreasing (assuming you mean what I would call contractive) $\endgroup$
    – Yemon Choi
    Mar 4 '20 at 23:54
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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 )

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    $\begingroup$ It might be worth adding that the Kazhdan projection $p$ has non-trivial $K_0$-class in the kernel of $q$ since $p$ is mapped to a generating projection in $K_0(\mathbb C) = \mathbb Z$ by the trivial representation. $\endgroup$
    – Jamie Gabe
    Mar 5 '20 at 0:14
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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 $\mathbb{C}$.)

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