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

Thanks.

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    $\begingroup$ 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 $X$. $\endgroup$
    – Andrea
    Sep 11, 2011 at 11:47
  • $\begingroup$ @Andrea: it is another definition. In "my" construction points of $Y$ was identified. (e.g. $D^{n}/\partial D^{n}\thickapprox S^{n}$) $\endgroup$
    – Aspirin
    Sep 11, 2011 at 12:45
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    $\begingroup$ 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 $X$. $\endgroup$ Sep 11, 2011 at 13:44

2 Answers 2

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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 $X$."

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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 $\tilde{g} \circ q = g$.

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