Does a continuous function preserve metrizability?

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.

• You mean "$\{0,1\}$ with antidiscrete topology". Yes, you're counterexample is correct. – freakish Dec 17 '18 at 18:02
• 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. – Chris Custer Dec 17 '18 at 18:45
• Note that any surjective map $\Bbb R\to \{0,1\}$ will do, since a map to an indiscrete space is always continuous. – Christoph Dec 17 '18 at 18:57
• Oops! I typed "discrete" for indiscrete in the first instance. Thank you for the correction! – Cassius12 Dec 17 '18 at 19:25
• 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. – 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).
• 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. – Toby Bartels Dec 17 '18 at 18:29