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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

2 Answers 2

up vote 4 down vote accepted

Thanks to Danielsen for catching an error in my previous proof.

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.

Suppose $U \subset X$, then since $\phi (\phi^{-1} U) \subset U$, we see that $\phi^{-1} U = \Lambda \circ \phi (\phi^{-1} U) \subset \Lambda U$.

Let $U \subset X$ be an open neighbourhood of $0 \in X$. Then, by continuity, $\phi^{-1} U$ is open, $0 \in \phi^{-1} U$, and hence $\Lambda U$ contains an open neighbourhood of $0 \in Y$.

Now suppose $U \subset X$ is open, and $y_0 \in \Lambda U$. Then $y_0 = \Lambda x_0$ for some $x_0 \in U$. Let $U' = U -\{x_0\}$, which is an open neighbourhood of $0$. Then $\Lambda U'$ contains an open neighbourhood $V' \subset Y$ of $0$. Then $V = V'+\{\Lambda x_0\} = V' + \{y_0\} \subset \Lambda U$ is an open neighbourhood of $y_0$. Hence $\Lambda$ is an open map.

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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
copper.hat and @Danielsen's, could you explain the above comment? I didn't understand the end of the proof if the equality is not valid. So, some details will be appreciated. – Pedro Sep 17 at 2:49
@Pedro The comment is two years ago, I don't even remember have written it. I think what I meant is that to fix $x\in U$ we can choose the map $\phi$ (or in other word choose the $\mathbb{K}^n$) such that $x\in \phi (\mathbb{K}^n)$, then $\Lambda \circ \phi (\phi^{-1} U)$ is the neighborhood of $x$ we want. – Danielsen Sep 17 at 10:43
@Pedro: I'm not sure what Danielsen's point was. The proof shows that $\phi^{-1}U$ is open, and since $\Lambda U = \phi^{-1}U$, we see that $\Lambda U$ is open, hence $\Lambda$ is open. – copper.hat Sep 17 at 21:16
copper.hat, @Danielsen pointed out that the equality $\Lambda U =\phi^{-1}U$ is not valid. We have $\phi^{-1}U=\Lambda \circ \phi (\phi^{-1} U) \subset \Lambda U$ because $\phi(\phi^{-1}U)\subset U$. How to get the other inclusion? – Pedro Sep 18 at 1:55

I don't know what topological vector spaces are. But what I know is that any finite dimensional real vector space can be made into a topological space by choosing any norm on it which does not depend on the choice of the norm.

Now if $V$ is any finite dimensional real vector space and $W$ is a linear subspace of $V$, the projection map $\pi:V\to V/W$ can be easily verified to be an open map.

Any surjective linear map $T:V\to Y$ factors uniquely through a linear isomorphism $\bar T:V/\ker T\to Y$.

Thus $T$ is a composition of two open maps and thus itself is open.

Hope this at least throws some light on the problem.

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