$Y$ be real NLS ; if there is a Banach space $X$ such that there is a continuous linear open mapping from $X$ to $Y$ then is $Y$ a Banach space?

Question

Let $Y$ be a real normed linear space ; if there exist a Banach space $X$ such that there exist a continuous linear open mapping from $X$ to $Y$ then is $Y$ a Banach space ?

Answer 1

Let $f: X \to Y$ be continuous, open linear map.

From openness follows surjectivity: let $y \in Y$, consider $B_1(0)_X \subset X$, then $f(B_1(0)_X)$ is open, thus contains some $B_\epsilon(0)_Y$. Then $\epsilon \frac{y}{2\|y\|} \in f(B_1(0)_X)$ and $y \in f(X)$.

Since $f$ is continuous the kernel is closed, since $X$ is Banach then $\tilde X :=X/\ker (f)$ is again a Banach space. The map $\tilde f : \tilde X \to Y$ is then a continuous, open linear bijection. The inverse map $\tilde{f}^{\,-1}$ is then also a continuous linear map. For each $y \in Y$ you have

$$\frac{\|\tilde f^{\,-1}(y)\|_X}{\|\tilde f^{\,-1}\|_\infty}≤\|y\|_Y=\left\|(\tilde f \circ \tilde f^{\,-1})(y)\right\|_Y≤\|\tilde f\|_\infty \|\tilde f^{\,-1}(y)\|$$

So the notions of convergence in $Y$ and $\tilde X$ are the same. This makes $Y$ complete since $\tilde X$ is complete.