# A question on open mappings

Let $X$ be a metric space. $f: X \to Y=f(X)$ is open and continuous mapping. Must $Y$ be a metric space? Thanks for your help.

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What topology do you have on $Y$? –  copper.hat Apr 26 '13 at 6:54
If I know the topology on $Y$, then I could judge whether it is a metric space. –  Paul Apr 26 '13 at 6:57
Well, how can it be open and continuous without some metric? If you take the discrete topology on $Y$, then any function is open and continuous, and you can use the metric $d(x,y) = \begin{cases} 0 & x=y \\ 1 & x \neq y \end{cases}$. –  copper.hat Apr 26 '13 at 7:07
Also, the topology doesn't define a metric. –  copper.hat Apr 26 '13 at 7:10
I don't think so. –  Paul Apr 26 '13 at 7:24

No. Let $X = \mathbb{R}$ and let $Y = \{a, b\}$ have the topology $\{\emptyset, \{a\}, \{a, b\}\}$. Define the map $f: \mathbb{R} \to Y$ by $f(0) = b$ and $f(x) = a$ for all $x \in \mathbb{R} \setminus \{0\}$. Then, $f$ is open, continuous, and onto, but $Y$ is not metrizable.