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As far as I know, a quotient space of a topological space must be defined wrt an equivalence relation on the topological space.

  1. But then I wonder what the equivalence relation is in the following example from Wikipedia:

    In topology, especially algebraic topology, the cone $CX$ of a topological space $X$ is the quotient space: $$ CX = (X \times I)/(X \times \{0\})\, $$ of the product of $X$ with the unit interval $I = [0, 1]$. Intuitively we make $X$ into a cylinder and collapse one end of the cylinder to a point.

  2. Is a subset of a topological space able to induce an equivalence relation on the topological space, and to induce a quotient space? I know it is true for a subspace of a vector space.

Thanks and regards!

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up vote 7 down vote accepted

Let $X$ be a topological space, and let $A\subseteq X$. $A$ defines the following equivalence relation $\overset{A}\sim$ on $X$: for $x,y\in X$, $x\overset{A}\sim y$ iff either $x=y$, or $\{x,y\}\subseteq A$. The quotient space $X/A$ is defined to be the same as the quotient space $X/\overset{A}\sim$.

Added: To put it a little differently, the partition of $X$ induced by $\overset{A}\sim$ is $$\{A\}\cup\big\{\{x\}:x\in X\setminus A\big\}\;.$$ The topology of of the quotient space is defined as follows: $V\in X/A$ is open iff $\pi^{-1}[V]$ is open in $X$, where $\pi:X\to X/A$ is the obvious quotient map.

As Alex Becker mentions in the comments, this notion of quotient is somewhat different from the notion used when some algebraic structure is present, as in the case of quotients of groups, rings, vector spaces, etc.

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THanks! (1) It seems that the equivalence classes are $A$, and the singleton subsets of $A^c$? (2) This definition doesn't depend on topology at all, actually not assuming any structure on a set $X$? (3) When $X$ is a vector space, and $A$ is a subspace of $X$, the quotient space in the vector space sense doesn't coincide with the definition in your reply. Does it? – Tim Apr 3 '12 at 22:58
It is worth noting that when you have additional structure on your topological space, e.g. a vector space, this is not the same equivalence relation as the one you would typically talk about being induced by certain subsets, e.g. by a subspace. – Alex Becker Apr 3 '12 at 22:59
@Tim: I’ve added a bit in response to your comment; I think that I’ve taken care of everything that you asked about. – Brian M. Scott Apr 3 '12 at 23:05
@BrianM.Scott Q.3. in this note says that the quotient space $X/A =(X\setminus A)\coprod \{*\}$. I cant find the basic lecture and i guess $\coprod$ is disjoint union but its so complicated for me to show that $X/A$ and $(X\setminus A)\coprod \{*\}$ are equal as topological spaces and then how we can clarify the open sets. could you help me, please? – M.A. Apr 29 at 15:44
@M.A.: Yes, that’s the disjoint union. That description corresponds to the partition in the Added part of my answer: $\{*\}$ represents the piece $A$ of the partition, and each point $x\in X\setminus A$ corresponds to the piece $\{x\}$ of the partition. For the topology, let $U$ be an open set in $X$. If $U\cap A=\varnothing$, then $U$ is open in $X/A$. If $U\supseteq A$, then $U/A$ is open in $X/A$, where $U/A=(U\setminus A)\cup\{*\}$, the set obtained by collapsing $A$ to a single point. Every open set in $X/A$ is of one of these forms. – Brian M. Scott Apr 29 at 22:13

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