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I am a bit confused about this concept because I have read that the quotient space is second countable if the quotient map is open. However, I thought the definition of a quotient map was a surjective, continuous, open mapping.

Suppose that $X$ is a second countable topological space, and ~ is an equivalence relation. The canonical mapping $q: X \rightarrow X$/~ is a quotient map, so $X$/~ would also be second countable.

Where am I going wrong with this idea?

Martin Sleziak
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user190570
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    Quotient maps don't have to be open, though I don't have an example handy. – Matt Samuel Jan 20 '15 at 04:30
  • @MattSamuel Is the example I gave of the quotient map induced by an equivalence relation open? – user190570 Jan 20 '15 at 04:36
  • not necessarily. I think I have an example, still pondering it though. – Matt Samuel Jan 20 '15 at 04:38
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    @Matt: Identify the integers in $\Bbb R$ to a single point; the quotient map from $\Bbb R$ to the quotient space is not open. For instance, the image of $\left(-\frac12,\frac12\right)$ is not open in the quotient, since it does not contain a nbhd of the point representing the collapsed integers. – Brian M. Scott Jan 20 '15 at 04:44
  • Suppose $X\rightarrow Y$ is a quotient map and $Y$ is metrizable. Then if $X$ is second countable it must have a dense set. By projecting this to $Y$ you obtain a dense set there, and as $Y$ is metrizable it must be second countable. You can generalize this a bit to the case in which $Y$ has a countable cover by metrizable opens (as if you want to prove that the projective space is second countable for example). – nowhere dense Aug 17 '19 at 20:00

2 Answers2

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Let $\sim$ be the following equivalence relation on $\Bbb R$: $x\sim y$ if and only if $x=y$, or $x,y\in\Bbb Z$. In other words, the $\sim$-equivalence classes are $\Bbb Z$ and the singletons $\{x\}$ for $x\in\Bbb R\setminus\Bbb Z$. Let $z$ be the point of the quotient space corresponding to $\Bbb Z$; then the quotient is not first countable at $z$, so it’s not second countable, even though $\Bbb R$ is. Also, the quotient map is not open, since the image of $\left(-\frac12,\frac12\right)$ does not contain a nbhd of $z$.

A map $q:X\to Y$ is a quotient map if it’s surjective, and $U\subseteq Y$ is open in $Y$ if and only if $q^{-1}[U]$ is open in $X$. This condition ensures that $q$ is continuous, but it does not require $q$ to be open (or closed).

Brian M. Scott
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Take $\mathbb R$ and identify $\mathbb Z$ to a point. This will give you a quotient space that is a point with countably infinitely many loops coming out of it. A set in the quotient is open if and only if its pullback via the quotient map is open. In particular, any open neighborhood of the central point must contain a point from each one of the loops, for otherwise the pullback will have isolated points and hence not be open.

The quotient map is not open, as for example $(-1/2,1/2)$ does not map to an open set because the image contains the central point but does not include any points from all but 2 of the loops. The quotient space also does not have a countable basis since it is not even first countable: given any countable collection of open sets containing the central point we may, using a diagonal argument, construct an open set containing the central point that does not contain any of them.

Matt Samuel
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    I arrived at this answer independently. I was typing on my phone so I didn't see any activity while I was typing it. – Matt Samuel Jan 20 '15 at 04:58