Proving operator image is closed

Question

Let $T:X\to Y$ be a linear bounded and surjective operator between Banach spaces $X$ and $Y$. I want to prove that if $A+\ker T$ is closed then $T(A)$ is closed. I tried using the open mapping theorem but it didn't work. I tried saying that $X\setminus (A+\ker T)$ is open so $T(X\setminus (A+\ker T))$ is open but I don't think that $T(X \setminus (A+\ker T)) = Y \setminus T(A)$ which would imply that $T(A)$ is closed. Thanks.

Answer 1

Since $T$ is surjective, it is open. Let $U$ be the complementary space of $A+Ker(T)$ it is open. This implies that $T(U)$ is open, and the complementary of $T(A)=T(A+Ker(T))$ is $T(U)$ since $T$ is surjective. To show this, remark that $y\in T(U)\cap T(A+Ker(T))$ implies $y=T(x_1)=T(x_2), x_1\in A+ker(T), x_2\in U$. This implies that $T(x_2-x_1)=0$ and $x_2-x_2\in Ker(T)$, we deduce that $x_2\in Ker(T)+A$ contradiction. Since $T$ is surjective, $T(U)\cup T(A+Ker(T))=Y$. We deduce that $T(A)$ is closed.