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I quote a previously asked question :

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

The original answer given there assumes $X$ is locally convex. Is this a necessary condition?

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According to the second answer, it's seems it's not the case. – Davide Giraudo Nov 28 '12 at 20:20
up vote 2 down vote accepted

Let $G$ be an open subset of $X$ and assume that $f(G)$ is not open. Denoting by $B_m$ the ball with radius $1/m$ around $0$ in $K^n$, this means that there is a $g \in G$ with $ [f(g) + B_m] \cap [K^n\setminus f(G)] \neq \emptyset$ for all $m$, i.e. there is a $c_m = (c_m^{(1)}, ..., c_m^{(n)}) \in B_m$ with $f(g) + c_m \notin f(G)$.

By the surjectivity of $f$, there is some $x_i$ with $f(x_i)=(...,0,1,0,...)$, being $1$ at the $i$-th coordinate. Let now $h_m = g + \sum_{i=1}^n c_m^{(i)} x_i$. Then $f(h_m) = f(g) + c_m \notin f(G)$ for all $m$.

On the other hand, as $(c_m)_m$ converges to $0$ in $K^n$, $(h_m)_m$ converges to $g$ in $X$. As $G$ is an open neighbourhood of $g$, $h_m$ is eventually in $G$ and hence $f(h_m)$ eventually in $f(G)$, contradiction.

To summarize, neither $X$ locally convex nor $f$ continuous is necessary.

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