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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.

Benji
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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 09 '13 at 01:58
  • @copper.hat The topology on $K^n$ is the product topology. – Benji Jan 09 '13 at 02:02
  • 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 09 '13 at 02:05
  • @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 09 '13 at 02:10

3 Answers3

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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.

copper.hat
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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 '15 at 02:49
  • @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 '15 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 '15 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 '15 at 01:55
  • Since $\Lambda \circ \phi$ is the identify map, then equality follows. I'm not sure why Danielsen suggested that equality is not valid. – copper.hat Sep 18 '15 at 07:15
  • @copper.hat $U$ may not be included in the range of $\phi$, so as Pedro said we only have $\phi(\phi^{-1}U)\subset U$. – Danielsen Sep 19 '15 at 02:13
  • @Danielsen: Thanks for catching that, I missed that completely. I can't delete my answer. – copper.hat Sep 19 '15 at 02:23
  • @copper.hat: You are welcome. You need not delete your answer, it's almost complete. – Danielsen Sep 19 '15 at 02:26
  • @Danielsen: I prefer to delete it while fixing it and it will be a few hours before I can fix it. – copper.hat Sep 19 '15 at 02:45
  • @copper.hat: Maybe edit your answer is alright, not need to delete it. Just a suggestion, any choice of you is alright. – Danielsen Sep 19 '15 at 02:49
  • @Pedro: I have corrected my answer. Thanks to Danielsen for being quietly persistent. – copper.hat Sep 19 '15 at 05:15
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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 (and the topology thus obtained 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.

caffeinemachine
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My proof only works if we assume in addition that $Y$ is Hausdorff, so that $Y$ is actually $\mathbb K^n$. As pointed out in copper.hat's proof (which doesn't have this limitation), it suffices to show that for every open neighborhood $U$ of $0_X$, $\Lambda U$ contains an open neighborhood of $0_Y$. Let $\{e_1,\dots,e_n\}$ be a basis of $Y$ and $e_i=\Lambda x_i$. Since addition of $n$ elements in $X$ is continuous, and since the sum of $n$ zeros is zero, there exists a neighborhood $V$ of $0$ such that $$\underbrace{V+V+\cdots+V}_n\subset U.$$ Next, since $0x_i=0$, by continuity of scalar multiplication there exists $\epsilon>0$ such that $\alpha_ix_i\in V$ whenever $|\alpha_i|<\epsilon$ and $1\le i\le n$. Therefore, whenever $|\alpha_i|<\epsilon$ for each $1\le i\le n$ we have $$\alpha_1x_1+\alpha_2x_2+\cdots+\alpha_nx_n\in U,$$ thus $$\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n\in\Lambda U.$$ But since $Y$ is finite-dimensional and Hausdorff, it is isomorphic to the topological vector space $\mathbb K^n$ and the vectors $\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n$ with $|\alpha_i|<\epsilon$ form an open neighborhood of $0$, since they correspond to an open cube in $\mathbb K^n$. So the proof is complete.

Noiril
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