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Is it true that quotient space of a Hausdorff space is necessarily Hausdorff?


In the book "Algebraic Curves and Riemann Surfaces", by Miranda, the author writes:

"$\mathbb{}P^2$ can be viewed as the quotient space of $\mathbb{C}^3-\{0\}$ by the multiplicative action of $\mathbb{C}^*$. In this way, $\mathbb{}P^2$ inherits a Hausdorff topology, which is the quotient topology from the natural map from $\mathbb{C}^3-\{0\}$ to $\mathbb{}P^2$"

It is true that the complex projective plane $\mathbb{}P^2$ is Hausdorff, but the above reasoning by Miranda will be true if the statement in the question is true.

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    [Wikipedia](http://en.wikipedia.org/wiki/Hausdorff_space#Properties) mentions that the answer is no...2011-07-07
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    You can get the Mobius strip as a (topological) quotient, by identifying the edges of a square in the right way (and, of course, each point is in its own class), but I don't know if this topological quotient can also be made into a group quotient. So at least in the case when the quotient is a topological one, the answer is no.2011-07-07
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    As Zev says, the general statement is false. Also, you can't read this statement out of what Miranda is saying. He only says that the usual topology on $P^2$ *is* the quotient topology you describe.2011-07-07
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    @Theo, Zev: but I think the article Zev links to refers to topological quotient, and not group quotients, i.e., spaces that result from group actions.2011-07-07
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    @gary: The Möbius strip is pretty Hausdorff... A better example would be the line with two origins (let $\mathbb{Z}/2$ act on the two lines by switching $(0,t)$ with $(1,t)$ except if $t = 0$) or, more drastically: consider $\mathbb{R}/\mathbb{Q}$.2011-07-07
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    @gary: [The quotient of a topological space by a group action is a special case of the topological quotient](http://en.wikipedia.org/wiki/Quotient_space#Examples) (the equivalence classes of the topological quotient are the orbits of the action).2011-07-07
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    In fact, it is easy to see that in all non-trivial cases, we can find an equivalence relation on a topological space $X$ such that the quotient topology is not Hausdorff. For example, let $X$ be a non-discrete topological space and choose a subset $A\subseteq X$ such that $A$ is not open in $X$. Define an equivalence relation on $X$ such that the equivalence classes are precisely $A$ and $X\setminus A$. The resulting quotient space is clearly not Hausdorff since the point corresponding to the equivalence class $X\setminus A$ is not a closed point.2011-07-07
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    @Zev: Yes, but the objection is (I think): given that you know that quotient's aren't Hausdorff in general, you cannot infer that in the special situation of group actions the statement doesn't hold.2011-07-07
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    @Theo: Ah, a fair point. Still, it is easy to make Amitesh's example one where the equivalence classes are the orbits of some group action, which would provide a counterexample.2011-07-07
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    @Zev: sure, I gave two examples in my second comment. I'm out of this discussion now.2011-07-07
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    @Theo: Sorry, I hadn't read your entire comment; your examples are very illustrative.2011-07-07
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    @Theo: you are absolutely right; the Mobius strip is even a manifold. I was thinking of orientability, not Hausdorff.2011-07-07
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    @user8186 I came here after reading the exact same line in the exact same book as you.2017-06-13

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