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Is the following theorem true? It seems straightforward but I haven't seen it published anywhere, not even as a corollary, so I'm concerned I've missed something. Discussions that introduce quotient spaces all seem to dance around this very simple and useful fact. Why don't they just come right out and say it?

Let $X$ and $Y$ be a topological spaces. Let $\sim$ be an equivalence relation on $X$. Then $Y$ is homeomorphic to the quotient space $X/{\sim}$ iff there exists a quotient map $f:X \to Y$ that induces the same partition as $\sim$.

TJCrow
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    Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong. – Bach Jan 06 '19 at 16:12
  • Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post. – TJCrow Jan 06 '19 at 17:08
  • That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think. – Bach Jan 06 '19 at 17:18
  • @Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation. – TJCrow Jan 06 '19 at 17:34
  • Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted. – Bach Jan 07 '19 at 00:37
  • Right, I understand. I know I'm being picky, but I have my reasons. See my comment to @Paul's answer. Cheers! – TJCrow Jan 08 '19 at 03:11

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It is well known (and stated in most textbooks) that a continuous surjection $f : X \to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y \to Z$:

$g$ is continuous if and only if $g \circ f : X \to Z$ is continuous.

An obvious corollary is this.

Given two quotient maps $f : X \to Y, f' : X \to Y'$ and a bijection $\phi : Y \to Y'$ such that $\phi \circ f = f'$. Then $\phi$ is a homeomorphism.

Each quotient map $f : X \to Y$ induces an equivalence relation on $X$ by defining $x \sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X \to X/\sim$.

In fact, this does not frequently occur as an explicit statement in the literature.

Paul Frost
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  • Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{\sim}$ where $\sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] \to S^1$, $f(t)=(\cos t, \sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $\sim$ does and (2) that $f$ is a quotient map. – TJCrow Jan 06 '19 at 18:19
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I just learned that it occurs in Munkres. See Xiang Yu's answer to How does the quotient $\mathbb{R}/\mathbb{Z}$ become the circle $S^1$?.

Paul Frost
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  • Thanks again, @Paul Frost. But, I'll make the same quibble that I made with user549397. The theorem in Xiang Yu's answer states that the codomain of a quotient map is homeomorphic to the quotient space induced by the map. The theorem in the op is a bit stronger. It gives conditions under which any quotient space is homeomorphic to the codomain of a quotient map. It makes explicit the bijection between the partition created by the "gluing instructions" and the partition created by the quotient map... a distinction I consider important to explicate though most seem content to gloss over. – TJCrow Feb 10 '19 at 15:50