# A surjective linear map into a finite dimensional space is open

I'm in search of different proofs of the following proposition:

$\bf{Proposition}$: Suppose $X$ and $Y$ be topological vector spaces, $\text{dim }Y<\infty$, and $\Lambda:X\to Y$ is a surjective linear map. Then $\Lambda$ is open.

Any and all proofs are welcomed.

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I may be missing something, but why does it follow that $\Lambda$ is open if the $\Lambda_i$ are? –  copper.hat Jan 9 '13 at 1:58
@copper.hat The topology on $K^n$ is the product topology. –  Benji Jan 9 '13 at 2:02
I realize that, but for example, the map $x \mapsto (x,x)$ satisfies the condition that the components are open, but the ensemble is not. (I realize that this map is not surjective, but this is the source of my confusion.) I think you cannot just focus on each map separately, unless you have somehow solved the 'independence' problem first? –  copper.hat Jan 9 '13 at 2:05
@copper.hat You are correct. I was mistaken. For some reason I imagined the image of a set to be the product of the images, which is clearly wrong. I will edit my question accordingly. –  Benji Jan 9 '13 at 2:10

How about this approach: Let $e_i$ be basis vectors of $\mathbb{K}^n$ (ie, $Y$), and choose $x_i \in X$ such that $\Lambda x_i = e_i$. Now consider the map $\phi: \mathbb{K}^n \to X$ given by $\phi(\alpha) = \sum \alpha_i x_i$. $\phi$ is continuous since $X$ is a tvs. Also, we note that $\Lambda \circ \phi$ is the identity mapping. Let $U \subset X$ be open, then by continuity, $\phi^{-1} U$ is open, and hence $\Lambda U = \Lambda \circ \phi (\phi^{-1} U)$ is open.
We can only say $\Lambda U \supset \Lambda \circ \phi (\phi^{-1} U)$, but then for every point of $\Lambda U$ can have a neighborhood belonged to $\Lambda U$. So $\Lambda U$ is open. –  Danielsen Jul 24 '13 at 12:33