Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping.

Question

Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping.

I think if I can show that $T(B_X)$ contains an open ball then I am done where $B_X$ is the unit ball in $X$. But I am unable to show that. Need some help...

Answer 1

1. If $T$ is continuous, then $\ker(T)$ is closed and the quotient map $$ \pi : X \to X/\ker(T) $$ is an open map. Furthermore, $T$ induces an injective map $$ S : X/\ker(T) \to Y $$ Since $Y$ is finite dimensional, so is $X/\ker(T)$, and so $S$ (whose range is $Y$) is now a homeomorphism. In particular, $S$ is an open map, so $$ T = S\circ \pi $$ is also open.

2. If $\dim(Y) = 1$, then it follows from an earlier question that if $U$ is a non-empty open set, then $T(U) = \mathbb{C}$, so it is, in particular, an open map.

3. Not sure about the general case (if $T$ is discontinuous and $\dim(Y) > 1$), but perhaps someone else can complete that case (I don't think induction works, but perhaps it could)