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Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)<\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.

The second assertion is easier since the quotient space $X/N$ has the universal mapping property.

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By some of the isomorphism theorems $X/N$ is isomorphic to $f(X)=Y$. Now the canonical projection $\pi:X\to X/N;x\mapsto x+N$ is open. Indeed, if $U$ is an open set of $X$, then $\pi^{-1}(\pi(U))=\bigcup_{x\in U} x+U$ where each of the $x+U$ is open because is the image of an open set by an homeomorphism. Then it is enough to show that the isomorphism between $X/N$ and $Y$ is an homeomorphism. – leo Jun 21 '13 at 2:48
To expend the leo's comment, a linear isomorphism between topological vector spaces is actually a homeomorphism. To prove this, you can reduce to the case that one of these two spaces is $R^n$. – lee Jun 21 '13 at 4:09
@lee.I think your comment isn't general true.That an isomorphism between topological vector spaces means it is also linear,one-to-one,and surjective but may be not continuous. – mathon Jun 21 '13 at 4:59
That's why I didn't go further. Because I wasn't sure what is the default topology considered in a finite dimensional vector espace $W$. But, if it is the topology induced by $\Bbb R^{\dim W}$ then everything follows as lee indicated. – leo Jun 21 '13 at 17:06
@leo:we may get a compatible topology induced by $\mathbb{R}^{dimW}$. – mathon Jun 22 '13 at 1:48

Here is a tedious proof of the first assertion:

To show that $f$ is open, it is sufficient to show that the image of a neighborhood of $0 \in X$ contains a neighbourhood of $0 \in Y$.

Let $y_1,...,y_n$ be a basis for $Y$ and let $x_i \in f^{-1} \{ y_i \}$. Note that the $x_i$ are linearly independent. Define $\Lambda : \mathbb{F}^n \to X$ by $\Lambda \alpha = \sum_k \alpha_k x_k$. $\Lambda$ is linear, continuous and $\ker \Lambda = \{ 0\}$.

Let $U$ be a neighborhood of $0 \in X$. By continuity of $\Lambda$, $\Lambda^{-1} U$ is open, hence contains $B(0, \delta) \subset \mathbb{F}^n$ for some $\delta>0$.

Define $\Phi:\mathbb{F}^n \to Y$ by $\Phi = f \circ \Lambda$, and note that $\Phi$ is bijective, hence $\Phi(B(0, \delta))$ is open (and contains $0$). Since $0 \in \Phi(B(0, \delta)) = f (\Lambda (B(0, \delta))) \subset f(U)$, we see that $f(U)$ contains a neighborhood of $0 \in Y$. Hence $f$ is open.

For the second part, let $\pi: X \to X/\ker f$ be the quotient map. Note that $\ker f$ is closed, hence $X/\ker f$ is a topological vector space (in fact this is iff, closedness is needed to ensure that $X/\ker f$ is Hausdorff). Furthermore, $\pi$ is (by definition) continuous. The map $\tilde{f}: X/\ker \to Y$ is a bijection, hence $X/\ker$ is finite dimensional, and so$\tilde{f}$ is continuous. Since $f = \tilde{f} \circ \pi$, it follows that $f$ is continuous.

Alternatively, one could show that a functional is continuous iff its kernel is closed (Rudin's route involves showing that $\ker f$ closed implies $\ker f$ is not dense, and that the latter implies that $f$ is bounded, and continuity follows from this). Continuity of linear maps between finite dimensional topological vector spaces finishes the proof.

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Thank you.Your 'tedious' proof is clearly easy to understand,but I'm still puzzled by $\parallel\cdot\parallel_{\infty}$. – mathon Jun 21 '13 at 5:10
@mathon: Sorry, I had a more complicated proof, then simplified the complicated part and forgot to prune completely. I will remove that part. All norms are equivalent on $\mathbb{F}^n$ (which I am assuming to be either real or complex), so it doesn't matter. – copper.hat Jun 21 '13 at 6:14
@mathon: I added two 'proofs' for the second assertion. I quoted proofs because in both cases, I implicitly rely on the fact that a linear map between two finite dimensional topological vector spaces is continuous. – copper.hat Jun 21 '13 at 7:37
@Higgins:Let me make a detail about your alternative proof about the second assertion.For every topological vector space $X$ there induced a functional $\Phi:X\rightarrow C$,since $N$ is closed,then $\Phi$ is continuous.Since there is a continuous map $g:C\rightarrow Y$,then $f=g\circ\Phi$ is continuous. – mathon Jun 22 '13 at 1:36
@DavidChan: Note that $\Phi(\alpha) = \sum_k \alpha_k f(x_k) = \sum_k \alpha_k y_k$, and $y_k$ forms basis for $Y$. If $\Phi(\alpha) = 0$, then since $y_k$ are a basis, we must have $\alpha = 0$, hence $\Phi$ is injective. Since the $y_k$ form a basis, then any $y$ can be written as $y = \sum_k \beta_k y_k$, and so $\Phi(\beta) = y$. Hence $\Phi$ is surjective. – copper.hat Feb 4 '14 at 21:48

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