Let $V$ be a vector space , $T:V \to V$ be a linear operator , then is it true that
$\ker (T) \cap R(T) \cong R(T)/R(T^2) $ ( where $R(T)$ denotes the range of $T$ ) ?
I know that the statement is true when $V$ is finite dimensional as I can show that if $V$ is finite
dimensional , then $\dim (\ker (T) \cap R(T))=rank (T) - rank (T^2)=\dim R(T)/R(T^2)$ ,
but I don't know what happens in general . Please help . Thanks in advance