Operator K-theory and the unitary group

Question

Suppose that $A$ is a C*-algebra whose unitary group is contractible (e.g. $B(H)$ or more generally the stable multiplier algebra of any C*-algebra). It is clear from the definition that $K_1(A) = 0$, but I think it is also true that $K_0(A)$ is necessarily $0$. Is this correct? Is there an elementary proof?

Answer 1

Here is an idea: To show that $K_1(A)=0$ you actually only use that $U(A)$ is path connected. Now it suffices to show that $U(\Sigma A)$ (unitary group of suspension) is path connected if $U(A)$ is contractible.