Is a continuous image of a normal space normal?

Question

Problem Let $f:X\rightarrow Y$ be closed continuous surjective map. Show that if $X$ is normal then So is $Y$.

What if we drop the 'closed' condition? I want a counter example. I know the proof of this theorem.

score 5 · Accepted Answer · answered Jun 23 '20 at 03:19

5

Let $X$ be $\Bbb N$ with the discrete topology, let $Y$ be $\Bbb N$ with the cofinite topology, and let $f:X\to Y$ be the identity map. Clearly $X$ is normal and $f$ is continuous and surjective. However, $Y$ is $T_1$ but not Hausdorff, so it clearly is not normal.

answered Jun 23 '20 at 03:19

Brian M. Scott

616,228

Can you explain $\Bbb{N}$ with the cofinite topology? Please. – Jun 23 '20 at 03:50
@Joseph: If $X$ is any set, the cofinite topology on $X$ is $${\varnothing}\cup{X\setminus F:F\text{ is finite}};.$$ That is, the open sets are the empty set and the sets whose complements are finite. Equivalently, the closed sets are $X$ and the finite sets. – Brian M. Scott Jun 23 '20 at 03:52
$\Bbb{N}$ with cofinite topology ,$\Bbb{N}$ is only closed set. – Jun 23 '20 at 03:59
@Joseph: No, all of the finite subsets of $\Bbb N$ are also closed. – Brian M. Scott Jun 23 '20 at 04:00
Ahh yes. How can I comment(previous comment) like that? Thanks a lot Professor. – Jun 23 '20 at 04:03
@Joseph: You’re very welcome. – Brian M. Scott Jun 23 '20 at 04:04

score 3 · Answer 2 · answered Jun 23 '20 at 03:24

3

Consider $f\colon \Bbb R\amalg\Bbb R\to \Bbb R\amalg\Bbb R/\sim$, where $(x,0)\sim(x,1)$ for all $x\neq 0$. That is, $f$ is the quotient map to the line with a doubled-origin. Clearly $f$ is continuous and surjective (as are all quotient maps), but $f$ is not closed: take an interval $[0,1]$ in the first copy of $\Bbb R$. Then $f([0,1])$ is the interval $(0,1]$ plus one of the two origin points. This set is not closed since the preimage consists of $[0,1]$ in the first copy of $\Bbb R$ and $(0,1]$ in the second copy.

answered Jun 23 '20 at 03:24

pancini

19,216

But why is the quotient space here non normal? – murray Apr 17 '21 at 19:07
The two origins can't be covered with disjoint open sets – pancini Apr 17 '21 at 20:14

Eric Wofsey · Answer 3 · 2020-06-23T03:59:35.497

3

Let $Y$ be any topological space and let $X$ be $Y$ with the discrete topology. Then $X$ is normal, and the identity map $X\to Y$ is a continuous surjection. So, any non-normal space $Y$ at all will give a counterexample in this way.

(In fact, more strongly, every topological space is a quotient of a normal space. See for instance the construction in this answer of mine.)

edited Jun 23 '20 at 03:59

answered Jun 23 '20 at 03:53

Eric Wofsey

330,363

Is a continuous image of a normal space normal?

3 Answers3