Let $H$ be a separable infinite dimensional Hilbert space, $M \subset B(H)$ a von Neumann algebra and $A \subset M$ a separable $C^\*$-algebra such that $A''=M$. 


Let $p,q \in M\_{\infty}(A)$ be (Murray–von Neumann) equivalent projections.  

Let $U\_{p,q} = \{u \in M \otimes B(H) \text{ partial isometry } \vert uu^\*=p$, $u^\*u=q \}$


*Question*: Is is true that $U\_{p,q} \cap M\_{\infty}(A) \neq \emptyset$ ?


*Application*: let $\Gamma$ a countable torsion-free ICC group, $H=l^2(H)$, $A=C^\*\_r(\Gamma)$, $M=L\Gamma = A''$.  

$M \otimes B(H)$ is a ${\rm II}\_\infty$-factor with a trace $Tr$. Two projections $p,q$ are equivalent iff $Tr(p)=Tr(q)$.  

If the question above admits a positive answer for $A$: could we deduce that $K\_0(A)$ is a subgroup of $\mathbb{R}$?  

The problem is that $K\_0$ is defined ([here](http://www.univ-orleans.fr/mapmo/membres/chatterji/Bernst.pdf) p8) for idempotents (more general that projections). 



Murray–von Neumann equivalence on C$^*$-algebra and von Neumann algebra
In general the answer is no; here is one example. Let $A=\{\lambda I +K:\lambda\in \mathbb C, \ K \text{ compact }\}$ (the unitization of the compacts in $B(H)$). Then $M=B(H)$. Let $q$ be the identity and $p$ a projection of codimension one. Then every member of $U\_{p,q}$ is an isometry which is unitarily equivalent to the unilateral shift, and hence has Fredholm index $-1$. But every Fredholm operator in $A$ has index $0$. 


