Algebraic $K_1$ group for a $C^*$-algebra Let $A$ be a $C^*$-algebra: then one defines topological $K_1$ group as $GL_{\infty}(A^+)/\Big(GL_{\infty}(A^+)\Big)_0$ where $A^+$ denotes $A$ with the unit adjointed (even if $A$ already had a unit: in this case $A^+$ is isomorphic to $A \oplus \mathbb{C}$ but this definition turns out to be equivalent if we insert $A$ instead of $A^+$). The algebraic $K_1(A)$ is defined as $K_1(A)=GL_{\infty}(A)/[GL_{\infty}(A),GL_{\infty}(A)]$ where ${H,H}$ is a commutator subgroup and it is known that it is not the same as topological $K_1$. I read somewhere that $K_1^{top}(A)$ may be defined somehow similarly to the algebraic $K_1$ as $K_1^{top}(A)=GL_{\infty}(A)/ \overline{[GL_{\infty}(A),GL_{\infty}(A)]}$, at least for unital $C^*$-algebras. My question is the following:

How to show that these two aproaches are equivalent?  

 A: One must show that 
$$
\overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0.
$$
The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.
Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that
these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$.
In this case one has
$$
(e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}.
$$
The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can 
express the matrix
$$
\begin{pmatrix}
1 & & & \\
& -\frac 1 n  & &\\
&& -\frac 1 n &\\
&&&\ddots
\end{pmatrix}
$$
as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h\otimes 1_{n+1}$ to get diag$(h,-h/n,\ldots,-h/n)$ as a commutator.
