Let $A$ be a nonunital C*-algebra and let $I \subset A$ be a dense, $*$-closed, 2-sided ideal. I was under the impression that there existed some "obvious" argument proving that $I$ carries all the $K$-theory of $A$, but it does not seem so clear to me... For example, it's not hard to prove things like this:

Claim:For any projection $e \in M_n(A)$, there are projections in $e' \in M_n(I)$ arbitrarily near to $e$.

Sketch:Since $M_n(I)$ is dense in $M_n(A)$, we can find $x \in M_n(I)$ as near as we like to $e$. Without loss of generality, $x$ is self-adjoint or replace it by $\frac{x+x^*}{2}$ which is also close to $e$ and belongs to $M_n(I)$ since $I$ is self-adjoint. Define \begin{align*} f(t) = \begin{cases} 0 & \text{ if } t< 1/2 \\ 1/t & \text{ if } t > 1/2 \\ \end{cases} && && \chi(t) = \begin{cases} 0 & \text{ if } t < 1/2 \\ 1 & \text{ if } t > 1/2 \\ \end{cases} \end{align*} and notice that $\chi(t) = t f(t)$. Now, since $\operatorname{spec}(x) \subset \mathbb{R}$ is concentrated near $\{0,1\}$, we have that the projection $e' = \chi(x)$ is very close to $x$. Moreover, $e' = x f(x)$ belongs to $M_n(I)$ since the latter is an ideal in $M_n(A)$.

Now, at first glance, the above claim would seem to imply the map $K_0(I) \to K_0(A)$ induced by the containment $I \subset A$ is surjective, but there is a gap. By the usual definition, $K_0(A)$ is a subgroup of $K_0( \tilde A)$, where $\tilde A$ is the unitization of $A$. So, it would seem we actually need to prove the containment $\tilde I \subset \tilde A$ induces an isomorphism on $K$-theory (note that $\tilde I$ is not an ideal in $\tilde A$!). So, given any projection $e = a + e_0 \in M_n(\tilde A)$ where $a \in M_n(A)$ and $e_0$ is a projection in $M_n(\mathbb{C}$), how does one adapt the argument above to produce a nearby projection $e' \in M_n(\tilde I)$? One can go ahead and replace $a$ by $x \in M_n(I)$ close to $a$, and then show that $e' = \chi(x+e_0)$ is a projection close to $e$ with scalar part $e_0$, but I am uncertain how to show $e' - e_0$ belongs to $M_n(I)$, or indeed if this should hold in general.

local $C^\ast$-algebras. And you don't need it to be an ideal. – AlexE May 1 at 21:22