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Let $X$ and $Y$ be two topological spaces and $p: X\to Y$ be a quotient map.

If $A$ is a subspace of $X$, then the map $q:A\to p(A)$ obtained by restricting $p$ need not be a quotient map. Could you give me an example when $q$ is not a quotient map?

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Standard counter-example:

Let $X=[0, 2\pi]$ and $f: X\to \mathbb{S}^1$ where $f(x) = (\cos(x), \sin(x))$

Then $f$ is surjective and closed (why?) and thus a quotient map. But the restriction of $f$ to $[0, 2\pi)$ cannot be a quotient map as $[0,2\pi)$ is not homeomorphic to the circle.

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  • $\begingroup$ Hi M.B. To my understanding you are wording this as $[0,2\pi]$ is homemorphic to the circle which its not? You write "the restriction of $f$ to $[0,2\pi)$ cannot be a quotient map as $[0,2\pi)$ is not homeomorphic to the circle." Do you mean that $[0,2\pi ]$ can be made homemorphic to the circle if you identify the endpoints and this cannot be made for $[0,2\pi)$? Is that your argument? Thank you if you have time to answer. I am new to topology. $\endgroup$ – JKnecht Mar 10 '16 at 21:55
  • $\begingroup$ If $[a,b]$ were homeomorphic to the circle, then there exists a homeomorphism between them, say $\phi : [a,b]\rightarrow S^1$. Then, when $g=\phi$ wherever it is defined, the map $g:[a,b)\rightarrow S^1\setminus \{\phi (b)\}$ is also a homemorphism (why?). Since connectedness is a topological property, we expect that $[a,b)$ connected implies $S^1\setminus \{\phi (b)\}$ connected. Yet, if we remove a point from the circle, we see that it is not connected. @JKnecht $\endgroup$ – user322548 Nov 15 '17 at 2:40
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M.B. has given an example that shows that the restriction of a quotient map to an open subspace need not be a quotient map. Here is an example showing that the restriction to a saturated set also need not. Restrictions to an open saturated set however always works.

Let $X=[0,2],\ \ B=(0,1],\ \ A=\{0\}\cup(1,2]$ with the euclidean topology. Then the identification $q:X\to X/A$ is a quotient map, but $q:B\to q(B)$ is not.

Here is why: As Ronald Brown wrote, $q:B\to q(B)$ is a quotient map iff each subset of $B$ which is saturated and open in $B$ is the intersection of $B$ with a saturated open set in $X$. Now $U=\left(\frac12,1\right]$ is open and saturated in $B=(0,1]$, but if it were the intersection of $B$ and an open saturated $V$, then this $V$ would intersect $A$ and, since it is saturated, it would contain $0$, and then again by openness it had to contain $[0,\epsilon)$ for some $\epsilon>0$. So the intersection of $V$ and $B$ can never be $U$.

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  • $\begingroup$ What do you mean by the "identification" $q:X\to X\setminus A$? Could you give a bit more detail? $\endgroup$ – Xena Jun 9 '13 at 17:57
  • $\begingroup$ Oh I looked it up; it means "quotient map"... But still, how is $q$ defined? $\endgroup$ – Xena Jun 9 '13 at 18:01
  • $\begingroup$ If you have an equivalence relation $\sim$ on the set $X$, then $X/\sim$ is the set of equivalence classes. This space is the quotient space under the relation $\sim$ and is given the final topology of the map $q$ which sends each point to its equivalence class. A particular simple relation is given by $x\sim y$ if $x,y\in A$ or $x=y$. Then we write $X/A$ instead of $X/\sim$. This quotient space has one class for each point outside of $A$ and one class for $A$ itself. In other words, we identify $A$ to a single point. $\endgroup$ – Stefan Hamcke Jun 9 '13 at 18:12
  • $\begingroup$ Thank you! So $q:[0,2]\to (0,1]\cup \{0\}\cup(1,2]$ is a quotient map but I cannot see why $q':(0,1]\to (0,1]$ ($q(B)=(0,1]$, right?) is not a quotient map... $\endgroup$ – Xena Jun 9 '13 at 18:27
  • $\begingroup$ @Bedi: I have edited my answer. If you are not familiar with the terminology: A set $U$ is saturated if $q^{-1}(q(A))$. A map $q:X\to Y$ is a quotient map iff the image of each saturated open $U$ is open in $Y$. $\endgroup$ – Stefan Hamcke Jun 9 '13 at 21:46
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It may be helpful to have the general result. The following is 4.3.1 of Topology and Groupoids in the above notation:

Suppose $p: X \to Y$ is an identification map, and $A$ is a subspace of $X$.

The following conditions are equivalent.

(a) the restriction $q: A \to p(A)$ of $p$ is an identification map;

(b) each $q$-saturated set which is open in $A$ is the intersection of $A$ with a $p$-saturated set open in $X$;

(c) as for (b) but with 'open' replaced by 'closed'.

This result has the usual corollaries, for example as stated above.

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  • $\begingroup$ Dear Ronnie, your URL seems to be invalid? $\endgroup$ – M.B. Jun 9 '13 at 18:53
  • $\begingroup$ @M.B.: Thanks! Corrected. $\endgroup$ – Ronnie Brown Jun 9 '13 at 21:15

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