Let $A$ be a unital $C^*$-algebra. Let $u$ be an isometry in $A$. Then we can write $1_A$, which is a projection, as $u^*u$. It is known that then $uu^*$ is also a projection. Can we produce any projection by such a decomposition?
In other words, given a projection $q \in A$, can we find an isometry $u \in A$ s.t. $q = uu^*$?
As a related question, is every projection in $A$ Murray-von-Neumann equivalent to $1_A$?
Definitely not.
Very simple example: Consider $X=[0,1]\cup[2,3]$ and the projection $p\in C(X)$ which is the indicator function of $[0,1]$, i.e. $p(t)=1$ for all $t\in[0,1]$ and $p(t)=0$ for all $t\in[2,3]$. If there existed an isometry $s\in C(X)$ such that $p=ss^*$, then since $C(X)$ is abelian we would have $p=1$, which is not the case.