Is there such a mapping that is open mapping.

Question

Let X be the banach space and Y be the normed space.$T\in \mathscr{B}(X,Y)$.The operator T is surjective.If Y is a banach space, then T is an open mapping from the open mapping theorem. My question is if, on the contrary, we know that T is an open mapping, is Y necessarily a banach space?In other words, is there a certain operator T s.t. Y is not a Banach space, but T is an open mapping.I'm really clueless about this issue, please help me.

Answer 1

Consider the quotient space $Z=X/\operatorname{ker}(T)$. Since $\operatorname{ker}(T)$ is closed, $Z$ is a Banach space. Take the quotient map $\tilde T\colon Z\to Y$. This map is continuous and bijective. Since $T$ is bounded and open, then $\tilde T$ is bounded and open. Since $\tilde T$ is bijective, bounded and open means bicontinuous. So, you have an isomorphism of two spaces $Z$ and $Y$. Since $Z$ is a Banach space, $Y$ has to be a Banach space.

… I think it works.