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I am asked to prove that:

Given $f:X/{\sim}\to Z$ and $q:X \to X/{\sim}$, then the quotient topology (on $X/{\sim}$) is the minimal topology such that $f:X/{\sim}\to Z$ is continuous $\iff f\circ q:X\to Z$ is continuous.

I have tried to put $M=\{f^{-1}(V):V\in\mathscr{T}_Z\}$ , where $\mathscr{T}_Z$ is the topology of Z, and try to show the quotient topology $\mathscr{T}_q$ of $X/{\sim}$ is equal to $M$. But I can only prove that $M\subseteq\mathscr{T}_q$.

Could anyone please give some hint to me? (Or even if that statement is true)

egreg
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Longitude
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    The quotient topology is the minimal topology such that, for any space $Z$ and function $f:X/{\sim}\to Z$, the continuity of $fq$ implies that of $f$. See http://math.stackexchange.com/questions/227476/request-for-gentle-explanation-of-defining-a-topology-with-its-universal-propert – Stefan Hamcke Mar 20 '16 at 11:43
  • thanks a lots! Seems like I misunderstood the problem – Longitude Mar 20 '16 at 12:34

1 Answers1

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The statement, as written, is false. Consider $Z=\{z\}$, a singleton. Then every map $X\to Z$, as well as every map $X/{\sim}\to Z$, is continuous, no matter what topology we use on $X/{\sim}$.

Thus the minimal topology on $X/{\sim}$ satisfying the property that a map $f\colon X/{\sim}\to Z$ is continuous if and only if $f\circ q\colon X\to Z$ is continuous is the indiscrete topology.

As soon as the quotient topology is not indiscrete, you have a counterexample.

egreg
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