I'm reading a text on linear preservers and I'm having difficulty understaning one proposition.
Let $A$ and $B$ be unital algebras and let $\phi:A\rightarrow B$ be an injective linear map. And let's add another postulate: $c\in A$ and $d\in B$ are such that $\phi(a)\phi(b)=d$ whenever $a,b\in A$ and $ab=c$. The text is trying to figure out some possible relations between fixed points $c$ and $d$. More precisely, it's trying to prove that if $d$ is left (right) invertible, then $c$ is also left (right) invertible.
Here is how the proof goes:
- $\phi(ann_l(c))\subseteq ann_l(d)$ and $\phi(ann_l(c))\subseteq ann_l(d)$.
$ann_l(x)$ denotes all elements $y$ for which $yx=0$ and $ann_r(x)$ denotes all elements $y$ for which $xy=0$.
- Since $d$ is left invertible, we have $ann_r(d)=\{0\}$. Then from 1) we know that $\phi(ann_r(c))=\{0\}$. Since $\phi$ is assumed to be injective, we can deduce that $ann_r(c)=\{0\}$.
However, $ann_r(c)=\{0\}$ does not imply $c$ is left invertible in general. So I'm wondering how we can prove the fact that $c$ is left invertible.
The statement that $c$ must be left invertible is false for arbitrary algebras over fields. As a counterexample, take $F$ to be any field, $A=F[x]$, $B=F[x,x^{-1}]$, $\phi$ the inclusion map, and $c=d=x$.
The paper you referenced is mostly about Banach algebras, so perhaps the author meant to assume the algebras in Proposition 3.1 had additional properties, such as being Banach algebras. But Proposition 3.1(2) and (3) are false as stated for general algebras, and your original comment is exactly the reason why the proof is not valid: having zero annihilator does not usually imply invertibility. If you are really interested, you could write to the author and ask him whether he meant to include additional assumptions.