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Let $X$ be a topological vector space over the field $K$, where $K=\mathbb R$ or $K= \mathbb C$, and let $f:X\rightarrow K^n$ ($n \in \mathbb N$) be a linear and surjective functional. How to prove that $f$ is an open mapping?


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up vote 3 down vote accepted

At least for $X$ locally convex, with $K$ being $\mathbb R$ or $\mathbb C$, for $x_1,\ldots,x_n$ in $X$ mapping to the standard basis $e_1,\ldots,e_n$ of $K^n$, the set $\{x_1,\ldots,x_n\}$ is bounded in $X$, so, given a convex balanced open nbd $U$ of $0$ in $X$, there is $t_o$ such that $tU\supset\{x_1,\ldots,x_n\}$ for all $t\ge t_o$. Thus, $f(U)=t^{-1}\cdot f(tU)$ which contains $t^{-1}$ times the convex hull of all points $z_1e_1,\ldots, z_ne_n$ for $z\in K$ with $|z|\le 1$. This seems not to use continuity-or-not of $f$.


So, yes, without continuity, but using local convexity.

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Since $f$ is continuous, we have that $$ \ker(f) := f^{-1}\{0 \} $$ is closed. Therefore $$ X/\ker(f) $$ is a Hausdorff topological vector space of dimension $n$, which is isomorph to $K^n$. Let $\phi$ be the quotient mapping $$ \phi\colon X \to X/\ker(f). $$ Define $f'$ via $$ f = f' \circ \phi. $$ As a linear surjective map of finite dimensional vector spaces of equal dimensions, the mapping $f'$ is open, as is the quotient mapping. Therefore $f$ is open, too. (The general open mapping theorem does not use convexity in either domain or codomain, so we should not expect to have to use convexity in our assumptions).

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