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In this question:
universal property in quotient topology
I saw the following theorem:

Let $X$ be a topological space and $\sim$ an equivalence relation on $X$. Let $\pi: X\to X/{\sim}$ be the canonical projection. If $g : X → Z$ is a continuous map such that $a \sim b$ implies $g(a) = g(b)$ for all $a$ and $b$ in $X$, then there exists a unique continuous map $f : X/{\sim} → Z$ such that $g = f ∘ \pi$.

I was wondering how one would prove this.

Community
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new guy
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  • Can you define such a map $f$ first? Since $\pi$ is a quotient map, you can use that $f\circ \pi$ is continuous iff $f$ is. – sqtrat Mar 14 '16 at 12:50
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    Notice this conspicuous difference: $$f : X/\sim \to Z$$ $$f : X/{\sim} \to Z$$ Since "$\sim$" is a binary relation symbol, a certain amount of space appears to its left and right in things like $a\sim b$. That spacing is inappropriate in $f : X/{\sim} \to Z$, and is avoided by coding it as f : X/{\sim} \to Z. That way there's nothing to its left or right, so that spacing isn't there. I edited the question accordingly. $\qquad$ – Michael Hardy Mar 14 '16 at 12:51
  • @MichaelHardy Gee, I hadnt noticed that. Thanks! – new guy Mar 14 '16 at 12:54
  • It seems clear what the function would be; the problem then is to prove it's continuous. $\qquad$ – Michael Hardy Mar 14 '16 at 12:54
  • @sqtrat I think that $f(\pi(a))=g(a)$ – new guy Mar 14 '16 at 12:56
  • @newguy Yes, see answer by Brian M. Scott below. – sqtrat Mar 14 '16 at 12:58
  • As always, first try a direct approach based on the def'ns.It often works. In this case, the def'n of the quotient topology. See Brian M. Scott's A. – DanielWainfleet Mar 14 '16 at 21:43

1 Answers1

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For $x\in X$ let $[x]$ denote that $\sim$-equivalence class of $x$; $X/{\sim}=\{[x]:x\in X\}$. To show that such an $f$ exists, we simply define it: for $[x]\in X/{\sim}$ let $f([x])=g(x)$. Now use the fact that $g$ is constant on $[x]$ to show that $f$ is well-defined.

To show that $f$ is unique, suppose that $h:X/{\sim}\to Z$ is continuous and satisfies $g=h\circ\pi$. Let $[x]\in X/{\sim}$ be arbitrary. Then

$$f([x])=(f\circ\pi)(x)=g(x)=(h\circ\pi)(x)=h([x])\;,$$

and hence $f=h$.

To show that $f$ is continuous, let $U$ be an open set in $Z$. Show that $$f^{-1}[U]=\{[x]\in X/{\sim}:x\in g^{-1}[U]\}\;,$$ and then use the fact that $X/{\sim}$ bears the quotient topology to conclude that $f^{-1}[U]$ is open in $X/{\sim}$.

Michael Hardy
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Brian M. Scott
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  • Hi, i think you might still need to show that $f$ must be continuous – KnobbyWan Mar 14 '16 at 12:57
  • @Khoria Not necessary, since $\pi$ is a quotient map. – sqtrat Mar 14 '16 at 12:58
  • And how would I show continuity? – new guy Mar 14 '16 at 12:58
  • @newguy: Sorry: I got called away before I was finished. I’ve added a sketch of that argument to my answer. – Brian M. Scott Mar 14 '16 at 13:02
  • @BrianM.Scott Could you please clarify this statement: Now use the fact that is constant on [] to show that is well-defined.? I have not really understood it. – rainman Aug 15 '20 at 11:02
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    @rainman: I defined $f([x])$ to be $g(x)$. However, if $y\in[x]$, then $[y]=[x]$, and according to my definition $f([y])=g(y)$. Since $[x]=[y]$, this means that $f([x])$ has to be $g(y)$, too. And that means that my definition makes no sense if $g(x)\ne g(y)$. That is, $f$ is not well-defined if different representatives of a $\sim$-equivalence class can be treated differently by $g$. – Brian M. Scott Aug 15 '20 at 16:31
  • @Brian M. Scott: Thank you so much. I appreciate it. – rainman Aug 15 '20 at 19:07
  • @rainman: You’re very welcome. – Brian M. Scott Aug 15 '20 at 19:08