Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Lets $X$ is a topological space and $Y$ is some subset of $X$. How we define topology of quotient space $$X/Y=\{x\in X~|~x\sim y\Leftrightarrow x, y\in Y\}.$$


share|cite|improve this question
There's something wrong in the formula. I think you want to collapse $Y$, so the equivalence relation you must consider is defined by: $x \sim x'$ iff $x = x'$ or $x, x' \in Y$. Now consider the quotient projection $p : X \rightarrow X / \sim$ and define the topology of $X/ \sim$ saying that a subset $A \subset X/ \sim$ is open iff $p^{-1}(A)$ is open in $X$. – Andrea Sep 11 '11 at 11:47
@Andrea: it is another definition. In "my" construction points of $Y$ was identified. (e.g. $D^{n}/\partial D^{n}\thickapprox S^{n}$) – Aspirin Sep 11 '11 at 12:45
I think Andrea's answer still applies. Given a set $A$ in $X/Y$, either it contains a point of $Y$, or it doesn't. If it doesn't, it's open in $X/Y$ iff it's open in $X$. If it does, it's open in $X/Y$ if and only if $S\cup Y$ is open in $X$. – Gerry Myerson Sep 11 '11 at 13:44

For a set $X$ and an equivalence relation $R$, the quotient set $X/R$ is the set of equivalence classes $[x]$ of $R$. The quotient set comes with a quotient map $\pi:X\rightarrow X/R$, naturally defined by sending $x$ to its equivalence class $[x]$. The "quotient topology" on $X/R$ is defined by saying "$U\subset X/R$ is open iff $\pi^{-1}(U)$ is open in $X$." (the quotient topology itself is a special case of two things: 1) the weak topology induced by a famaily of maps too/from your set, and 2) a pushout diagram. You should check these things out, they're cool.)

Your example, quotienting by a subset, is a special case of a quotient set. To make your relation an equivalence relation, just add the diagonal $\Delta_X= \{ (x,x)\in X\times X\ |\ x\in X\}$. An element $[x]\in$"$X/Y$" is either a singleton $[x]=\{x\}$ (where $x\not\in Y$), or $[x]=Y$. Hence the quotient map is a bijection on $X-Y$.

If $U\subset X/Y$, then $U$ is open iff $\pi^{-1}(U)=\{x\in X\ |\ [x]\in U\}$ is open in $X$. We can write any $U$ as $(U-\{Y\})\ \dot{\cup}\ (\{Y\}\cap U)$, so in general $U$ is open iff $\pi^{-1}(U)=\pi^{-1}(U-\{Y\})\ \dot{\cup}\ \pi^{-1}(\{Y\}\cap U)$ is open in X. Thus for $U\subset X/Y$: if $Y\not\in U$ then $\pi^{-1}(U)$ is disjoint from $Y\subset X$ so the quotient map is a bijection; if $Y\in U$, then we must check that $\{x\in X-Y\ |\ [x]\in U\}\cup Y$ is open in X.

So if you want to think of $X/Y$ as "$X$, except $Y$ has collapsed" you can think "$U$ is open iff $U$ is open in $X$ and doesn't intersect $Y$, or $U\cup Y$ is open in $X$."

share|cite|improve this answer

Just as in the case of groups, as a set the quotient $X/Y$ is the set where $Y$ has been collapsed to a point. So define an equivalence relation on $X$ by $x_1 \sim x_2$ iff $ x_1$ and $x_2$ belong to $Y$. So now all the points of $Y$ are identified with each other, and all the points disjoint from $Y$ contain only themselves in their corresponding classes. Then $X/Y:= X/\sim$.

Now, we have an obvious surjective map $q: X \to X/\sim$ which sends each point of $X$ to its corresponding class. We want to give $X/\sim$ a topology so that $q$ is continuous. The obvious choice is the final topology: that is, the finest topology such that $q$ is still continuous.

The reason we make this choice is because then the quotient satisfies the universal property that any continuous function $g: X \to Z$ which makes the same identifications as $q$ passes to a unique continuous map $\tilde{g} : X/\sim \to Z$ such that $\tilde{g} \circ q = g$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.