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If I am understanding correctly, a quotient map can be defined in this way (actually I quoted the following from Munkres):

Let $X$ and $Y$ be topological spaces; let $p:X \rightarrow Y$ be a surjective map. The map $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

Or in this way:

Let $C$ be an equivalence relation in $X \times X$. The map $p:X \rightarrow X/C$ s.t. $p(x)=[x]$ ($x \in X, [x]$ is an equivalence class) is defined to be a quotient map.

I understood the concept from the second one, since equivalence classes partition a set, which is very easy to visualize. (Can anyone answer this?) My lecturer introduced equivalence relation and quotient space first also. Only when I need the strong continuity I will consider the first one, which is convenient. However when I read Munkres, I have no idea why those two describe the same thing (at least from my opinion). Can someone explain why to me?

Community
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MonkeyKing
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    I think there is an error in the link. The quotient topology is not the coarsest but the finest topology that makes $p$ continuous because the trivial topology ${\varnothing, X}$ is the coarsest one that makes $p$ continuous. – Hermis14 Mar 07 '22 at 09:56
  • I am slow in getting the second definition. I think a quotient map must be continuous? – High GPA Nov 10 '23 at 23:01

1 Answers1

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Note that you didn't provide a topology in the second case. Actually it is defined the same way as in the first case, that $U\in X/C$ is open iff $p^{-1}(U)$ is open.

And in the first case, a surjective map naturally induces an equivalence relation on $X$ with $x_1\sim x_2 \Leftrightarrow f(x_1)=f(x_2)$ with exactly $Y$ the set of equivalence classes.

Colliot
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  • So I consider union of all open sets $U_i$ (since each is open in $X/C$, it is open in $X$) which are equivalent to each other, and this will be open since arbitrary union of open set is open? – MonkeyKing Apr 23 '15 at 02:13
  • @MonkeyKing, what does it mean for open sets to be equivalent to each other? – Colliot Apr 23 '15 at 02:22
  • Ah, bad expression. I attempt to mean I want to find $U_1$ and $U_2$, like $x_1 \sim y_1, x_2 \sim y_2 \dots$ happen. Any way, if this is wrong expression, how do I prove $U \in X/C$ is open iff $p^{-1}(U)$ is open? – MonkeyKing Apr 23 '15 at 02:29
  • @MonkeyKing, you don't need a proof. That is the definition. Otherwise where does the topology on $X/C$ come from in the second case? – Colliot Apr 23 '15 at 02:35
  • Oh right, quotient topology. Thank you very much for the patience. – MonkeyKing Apr 23 '15 at 02:39