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As the topic, Is a continuous mapping which is also open (closed) must be closed (open)?

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In one direction: No. The projection $\pi \colon \mathbb R^2 \to \mathbb R$, $x \mapsto x_1$ is continuous and open, but maps $\{x : x_1x_2 = 1\}$, a closed set onto $\mathbb R \setminus \{0\}$, which is not closed. –  martini Nov 21 '12 at 14:59
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No. The projection $\pi \colon \mathbb R^2 \to \mathbb R$, $x \mapsto x_1$ is continuous and open, but maps $\{x : x_1x_2 = 1\}$, a closed set onto $\mathbb R \setminus \{0\}$, which is not closed.

In the other direction: The very same map with other domain $\pi\colon S^1 \to \mathbb R$, $x \mapsto x_1$ is closed and continuous, but not open as its image is not open in $\mathbb R$.

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A minimal example: Let $X=\{0,1\}$ with the topology $\{\emptyset,\{1\},X\}$ (i.e. the Sierpiński space). Then the constant map $f:X\to X$, $f\equiv 1$ is open but not closed, and $g\equiv 0$ is closed but not open.

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