I know if $a,b$ are elements of a unitral algebra $A$, then $1-ab$ is invertible if and only if $1-ba$ is invertible. And it follows from the observation that if $1-ab$ has inverse $c$ then $1-ba$ has inverse $1+bca$. But I can't prove it.
Inverse of $1-ba$ when $a,b$ are the elements of unital algebra $A$
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spectral-theory
spectral-sequences
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0I don't know much about unitral algebras, but problems that look like this are usually solved by writing down $(1-ba)(1+bca)$ and simplifying. – 2017-01-12
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So we're given that $(1-ab)c=1$. That is, $c-abc=1$. Then: \begin{align} (1-ba)(1+bca)&=1+bca-ba-babca\\ &=1-ba+b(c-abc)a\\ &=1-ba+b(1)a\\ &=1-ba+ba\\ &=1 \end{align}