This is problem 6.1.8 of S. Morris's "Topology without Tears":

Let $f$ be a continuous mapping of a metrizable space $(X,\tau)$ onto a topological space $(Y,\tau_1)$. Is $(Y, \tau_1)$ necessarily metrizable?

My Solution (by Counterexample):

Let $X$ be the Real line with the standard Euclidean topology. Clearly, $f$ is metrizable.

Let $Y = \{0,1\}$ with the indiscrete topology, i.e., $\tau_1 = \{\emptyset, \{0,1\}\}$. An indiscrete space with at least two points is not metrizable

Define $f$ as follows:

$ f(x)= \begin{cases} 0&\text{if}\, x\in \mathbb{Q}\\ 1&\text{if}\, x\in \mathbb{I} \end{cases} $

So, every open interval in $\mathbb{R}$ maps to $\{0,1\}$, and the inverse image of $\{0,1\}$ is $\mathbb{R}$. So, $f$ maps open sets to open sets.

Therefore, $f$ is a continuous function from a metrizable space to a non-metrizable space.

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    $\begingroup$ You mean "$\{0,1\}$ with antidiscrete topology". Yes, you're counterexample is correct. $\endgroup$ – freakish Dec 17 '18 at 18:02
  • $\begingroup$ In the definition of $f$, you mean $\mathbb I\setminus \mathbb Q$, I take it. It is also called the trivial topology, btw. Oh, $\mathbb I$ means irrational. I got it. $\endgroup$ – Chris Custer Dec 17 '18 at 18:45
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    $\begingroup$ Note that any surjective map $\Bbb R\to \{0,1\}$ will do, since a map to an indiscrete space is always continuous. $\endgroup$ – Christoph Dec 17 '18 at 18:57
  • $\begingroup$ Oops! I typed "discrete" for indiscrete in the first instance. Thank you for the correction! $\endgroup$ – Cassius12 Dec 17 '18 at 19:25
  • $\begingroup$ You can also get examples using the discrete topology. Let Y be any non-metrisable space, X the same set with the discrete topology and f the identity mapping from X to Y. $\endgroup$ – David Hartley Dec 17 '18 at 21:10

Your use of the word discrete in "Let $Y=\{0,1\}$ with the discrete topology" is the opposite of what you mean.

Also, "So, $f$ maps open sets to open sets." is not what defines the continuity of $f$. You have to check that $f^{-1}$ maps open sets to open sets (which you did).

All in all, the counterexample is correct.

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    $\begingroup$ The OP also checked that $f$ maps open sets to open sets (in other words that the map is open). A much simpler map could also have been used that was not open. Still, it's a fine counterexample, and even a counterexample to the weaker statement that every open continuous map preserves metrizability. $\endgroup$ – Toby Bartels Dec 17 '18 at 18:29
  • $\begingroup$ Thanks! I fixed the typo on "discrete" to "indiscrete". Also, I appreciate your pointing out that the key thing to check is the inverse map. $\endgroup$ – Cassius12 Dec 17 '18 at 19:29

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