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Let $T:X\to Y$ be a linear continuous surjection between Banach spaces $X$ and $Y$. By the open mapping theorem, we have $T$ is open. Now let $C$ be a closed convex subset of $X$ satisfying that $T(C)$ is also closed. Does $T:C\to T(C)$ is open?

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  • $\begingroup$ You are right about my first answer, it was wrong. $\endgroup$ – Sinusx Apr 18 '14 at 19:44
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Note:

This does not answer the question, because I overlooked the convexity assumption.

Now I think that the answer is negative in general case, without further conditions on $C$. Here is an counterexample. Take $X=\mathbb{R}^2$, $X=\mathbb{R}$ and let $T$ be the projection on one of the axis, say $T(x,y)=x$. This map is open. Let $C_1$ and $C_2$ be closed triangles, with vertices in $(2,2),(5,2),(5,5)$ and $(11,2),(8,2),(8,5)$ respectively, and $C_3$ be half of the strip

$$C_3 :=\{(x,y):5\leq x \leq 8, y \geq 2 \},$$ and define $C =C_1 \cup C_2\cup C_3$. Consider an open in the relative topology of $T$ set $U$, which lies ''high enough'', say

$$ U = \{(x,y):5\leq x \leq 8, 10 < y < 12 \} $$

Its projection is not open in the relative topology of $T(C)= [2;11]$, since $T(U)=[5;8]$.

Here is a visual drawing.

Here is an illustrating image

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  • $\begingroup$ Thank for your good idea and enthusiasm. However, in your counterexample, $C$ is not convex. Although I hope the answer of this question is positive, it may be negative and we need some further conditions on $C$. $\endgroup$ – luozhenghua Apr 19 '14 at 1:02

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