Let $R$ be a noncommutative ring with $1$. I need to show that for any $r\in R$ either $r$ or $1-r$ is left-invertible if and only if either $r$ or $1-r$ is a unit.
Proof. Indeed, if either $r$ or $1-r$ is a unit, then either $r$ or $1-r$ is left-invertible. Conversely, it suffices to show that every left-invertible element $r\in R$ is a unit. Let $r\in R$ be left-invertible. There exists $x\in R$ such that $xr=1$.
Case 1 . If $x$ is left-invertible, there exists $y\in R$ such that $yx=1$. Now, \begin{align} xr = 1 \implies yxr=y \implies r=y \implies rx=1. \end{align} Thus, $r$ is a unit.
Case 2. If $x$ is not left-invertible, it follows by the hypothesis that $1-x$ is left-invertible. Let $z\in R$ be such that $z(1-x)=1 \iff 1-z=-zx$. Notice that if $1-z$ is left-invertible, then so is $-zx$ which implies that $x$ is left-invertible, a contradiction. By the hypothesis, $z$ is left-invertible with left-inverse $z'$ (say). Hence, \begin{align} z(1-x) = 1 \implies z'z(1-x)=z' \implies 1-x=z' \implies zz'=1. \end{align} That is, $z$ is a unit with inverse $1-x$. Now, \begin{align} r=1r=z(1-x)r=zr-zxr=zr-z \implies z=(z-1)r. \end{align} Observe that if $r$ is a unit, then $z-1$ turns to be a unit and thus $1-z$ is left-invertible, a contradiction.
How is that ?!. This procedure shows that I can never get I need to prove ?!. Where is the trick in my proof ?!.
Thanks in advance.