Let $M$ be a von Neumann algebra and $a\in A$. Then for the range projections $[a]\sim [a^\ast]$, since when $a=v|a|$ is the polar decomposition $v^\ast v$ is the projection onto $ran(v^\ast)=ker(v)^\bot=ker(a)^\bot=ran(a^\ast)$.
Apparently $vv^\ast$ is supposed to be the range projection of $a$, but how do I see this?
Thanks in advance!
Soultion: Now, in the view of daylight, I think my admittedly ridiculous problem was the following:
It is known (without having any further knowledge on the polar decomposition of $a^*$ that $vv^*$ is the orthogonal projection onto $ran\ v$ and that this is in fact: $ran\ v = (ker\ v^*)^\bot \stackrel{!}{=} (ker\ a^*)^\bot = ran\ a^*$, beacause we also know from the polar decomposition of $a$, that $ran\ a =ran\ v$, which in turn implies : $(ran\ a)^\bot=(ran\ v)^\bot$ and $ker\ a^* = ker\ v^*$.
My brain was too stubborn to think that the indicated equation might be as easy as it is and is not due to knowing anything about the polar decomposition of $a^*$.
Thank you @Martin Argerami for your patience nevertheless!
You have $$ \operatorname{ran} vv^*=(\ker vv^*)^\perp=(\ker v^*)^\perp=\overline{\operatorname{ran} a}. $$ For the middle equality, it is true in general that $\ker x^*x=\ker x$. The inclusion $\ker x\subset\ker x^*x$ is trivial. The converse is easily seen if $x\in B(H)$: in that case, if $x^*x\xi=0$, then $$ 0=\langle x^*x\xi,\xi\rangle=\langle x\xi,x\xi\rangle=\|x\xi\|^2. $$