Relative K-theory and split exact sequences of C* algebras

Although this question was posted and answered a very long time ago, I just stumbled upon it and I thought it might be worthwhile if an answer is provided near the post (it actually seems like a full answer is not yet available).

First I will assume $A$ is untital. Second I will use the fact that given $[(p',q',x')]in K_0(A,A/J)$ I can write it as $[(ell_k,q,ell_k)]$ where
$pi(q)=pi(ell_k)=ell_k$ where $ell_k$ is the standard (bigger than $ktimes k$) matrix over the complex numbers with ones on the first $k$ entries of the diagonal and zeros elsewhere. A sketch of the proof is:

-note that
$$[(p',q',x')]=[(p'oplus 1-p',q'oplus 1-p', x'oplus 1-p')]$$
-note that
$$[(p'oplus 1-p',q'oplus 1-p', x'oplus 1-p')]=[(ell_k,q'',x'')]$$
- Use the fact that given a partial isometry $y$ we can find a path of unitaries in matrices of quadruple the size from the identity to $V$ such that
$V^*yy^*V=y^*y$ and $yy^*Vy^*y=y$ (where we include partial isometries and projections in the upper left hand corner in bigger matrices).

-So let $u_t$ this path of unitaries for the partial isometry $pi(x)$ and let $v_t$ a lift starting at the identity.

-Note that
$$[(ell_k,q'',x'')]=[(ell_k, q, x)],$$
where $ell_k=pi(ell_k)=pi(q)=pi(x)$ (use $(ell_k,v_t^*q''v_t,v_t^*x'')$).

-Note that
$$[(ell_k, q, x)]=[(ell_k,q,ell_k)]$$ by a linear path $tell_k+(1-t)x$ since
$pi(x)=pi(ell_k)$.

Denote by $sigma:A/Jrightarrow A$ the section which exists by assumption. Now suppose $r=[(ell_k, q, ell_k)]$ is a general element in $K_0(A,A/J)$ such that $[ell_k]-[q]=0$ in $K_0(A)$. Then there must exist a path of unitaries $u_t$ starting at the identity such that $u_1^*qu_1=ell_k$. So we have
$r=[(ell_k,ell_k, u_1^*ell_k)]$. Now denote $U:=sigma(pi(u_1))$ then
$pi(U)=pi(u_1)$ and thus by a linear path $r=[(ell_k,ell_k, U^*ell_k)]$. Now note that $U^*ell_k U=sigmapi(u_1^*ell_ku_1)=ell_k$ which shows that $(ell_k,ell_k,U^*ell_k)$ is degenerate thus $r=0$. So we find that the kernel of $K_0(A,A/J)$ is trivial. Which was to be shown.