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The book I'm reading isn't very clear and doesn't provide any concrete examples or definitions as to what a quotient topology is besides this paragraph.

Let $(X,T)$ be any topological space and ~ any equivalence relation on $X$. Let $Y$ be the set of all equivalence classes of ~. We can denote $Y$ by $X/$~. The natural topology to put on the set $Y = X/$~ is the quotient toplogy under the map which identifies the equivalence classes; that is, maps each equivalence class to a point.

I don't understand this, what exactly is a quotient topology? What does it mean by a map that identifies equivalence classes? If a function maps an equivalence class to a point, then isn't it a function from $Y$ to $X$?

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  • $\begingroup$ Have you seen quotients in abstract algebra? $\endgroup$ – user4894 Jul 6 '16 at 21:09
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    $\begingroup$ The relevant section in Munkres' Topology is pretty good. $\endgroup$ – Hoot Jul 6 '16 at 21:11
  • $\begingroup$ This is the first time I've seen quotient spaces. $\endgroup$ – Oliver G Jul 6 '16 at 21:11
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If $X$ is a topological space, and we define an equivalence relation $\sim$ on $X$, then we can construct the quotient topology for $X$ as such. Let $X/\sim$ be the quotient space (i.e. space of equivalence classes), and let $\pi: X \to X/\sim$ be the map $\pi(p) = [p]$ be the map that sends each element of $X$ to its equivalence class in $X/\sim$. We define the quotient topology on $X/\sim$ to be the collection of subsets $U \subseteq X/\sim$ such that $\pi^{-1}(U)$ is open in $X$. Observe that this makes $\pi$ a continuous map.

To illustrate this, imagine Euclidean space $\mathbb{R}^3$. We can define an equivalence relation on it by saying that $(x,y,z) \sim (x',y,',z')$ if and only if $z=z'$. Observe that this automatically satisfies the reflexive, symmetric, and transitive properties. We then see that equivalence classes are of the form $[z_0] = \{(x,y,z) \in \mathbb{R} \; | \; z=z_0\}$, and thus each plane parallel to the $xy$-plane, in a sense, collapses to a single point along the $z$-axis. Therefore we can say that $(\mathbb{R}^3/\sim) \approx \mathbb{R}$.

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  • $\begingroup$ In the last sentence, what do you mean by the approximation symbol? Did you mean homeomorphic? $\endgroup$ – Oliver G Jul 7 '16 at 17:55
  • $\begingroup$ @OliverG Yes. The idea here is that the quotient space generated from the relation given is homeomorphic to the $z$-axis which is just another copy of $\mathbb{R}$. Another popular example is the relation on $\mathbb{R}$ that $x\sim y$ if and only if $x-y \in \mathbb{Z}$. It can be shown that $\mathbb{R}/\mathbb{Z} \approx \mathbb{S}^1$. $\endgroup$ – Mnifldz Jul 7 '16 at 17:58
  • $\begingroup$ So then in order to transform $\Bbb R^3$ into this, there exists a function $f : (\Bbb R^3, T_{Euclidean}) \rightarrow_{onto} (\Bbb R^3 /$ ~$, T_{quotient})$ such that $T_{quotient} = \{ U \subseteq R^3 /$ ~ $ : f^{-1}(U) \in T_{Euclidean}\}$ ? $\endgroup$ – Oliver G Jul 7 '16 at 18:05
  • $\begingroup$ @OliverG Yes, though I would be careful and make sure to put braces around your expression for $T_{quotient}$. Try the map $\pi:\mathbb{R}^3 \to (\mathbb{R}^3/\sim)$ given by $\pi(x,y,z) = [z]$. $\endgroup$ – Mnifldz Jul 7 '16 at 18:07
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    $\begingroup$ That's exactly what I was trying to understand, thank you. The canonical representation of each equivalence class it what I was confused about. $\endgroup$ – Oliver G Jul 7 '16 at 22:48
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For topological spaces $X$ and $Y$, a map (of sets) $f\colon X\to Y$ is an identification if $f$ is surjective and $Y$ carries the coinduced topology, i.e. $U \subset Y$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. Note that this topology on $Y$ makes the map $f$ continuous.
Even more is true: If $f$ as above is an identification, then for each topological space $T$ a map $g\colon Y\to T$ is continuous if and only if the composition $gf\colon X\to T$ is continuous.

In particular, for an equivalence relation $\sim$ on $X$ you may form the set $Y=X/\sim$ and endow it with the coinduced topology with respect to the projection $\pi\colon X\to X/\sim$. Since this map is surjective, it is an identification. In this particular case, the topology on $X/\sim$ is also called quotient topology.

I guess the name "identification" comes from these sort of application, where you make points equal in the sense that points with $x\sim y$ are mapped to the same point $[y]=[x] \in X/\sim$.

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  • $\begingroup$ Since the OP hasn't seen quotient spaces before, and since his book uses a different terminology, it's worthwhile to point out that what you call the coinduced topology is what the book calls the quotient topology. $\endgroup$ – Andreas Blass Jul 6 '16 at 22:04
  • $\begingroup$ @AndreasBlass Right, I should have said that. $\endgroup$ – Plankton Jul 6 '16 at 22:12

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