$X,Y$ be Banach , $T \in \mathcal B(X,Y)$ be onto ; then , for every sequence $y_n \to y \in Y$ , $\exists x_n \to x\in X$ s.t. $T(x_n)=y_n , T(x)=y$?

Question

Let $X,Y$ be Banach spaces , $T:X \to Y$ be a surjective continuous linear transformation , then is it true that for every convergent sequence $\{y_n\}$ in $Y$ , converging to $y \in Y$ , there exist a sequence $\{x_n\}$ in $X$ , converging to $x \in X$ , such that $T(x_n)=y_n , \forall n \in \mathbb N$ and $T(x)=y$ ?

Answer 1

If $T$ is onto, this is the open mapping theorem:

You first choose $x$ with $Tx=y$. Then you look at the images of small open balls around $x$. These images are open and contain $y$ and thus contain all but finitely many $y_n$. Hence, you can choose $x_n$ in the balls as needed.