Given a topological space $(X,\tau)$ and an equivalence relation ~ in $X$ we can define a topology $\tau'$ in the quotient space $X/$~ as follows: $$A \in \tau' \iff \pi^{-1}(A) \in \tau $$ This is how we defined the quotient topology in our course, but there is this excercise which asks me to prove that given a function $f:X/$~ $\to Y$, $f$ is continuous iff $f○\pi$ is continuous.

$\pi$ is continuous, so if $f$ is also continuous, so it is their composition. To prove the other direction lets take an open set $A \subset Y$ and, assuming the composition is continuous, we have $(f○\pi)^{-1}(A)=\pi^{-1}(f^{-1}(A))$ is open in $X$. But I can't see how this implies $f^{-1}(A)$ is open, the converse holds because of the continuity of $\pi$, but it doesn't seem to me that $\pi$ is an open function. If that were the case then for an open set $A$ in $X$ which is a union of equivalence classes, any set $B$ consisting of one point of each of those equivalences classes would also be open because $A$ and $B$ would have the same image under $\pi$.

Is this reasoning right? If so, is $\pi$ an open function and in that case how can I prove it?


  • $\begingroup$ In general $\pi$ doesn't have to be open. Here you need additional assumptions on your equivalence classes. The most likely partition is $\{f^{-1}(y):y\in Y\}$. $\endgroup$ – Qiyu Wen Jul 5 '16 at 19:55

Refer to your equivalence \begin{equation} A \in \tau' \Longleftrightarrow \pi^{-1}(A) \in \tau \end{equation} and apply it to $f^{-1}(A)$. In particular, you see that $\pi^{-1}(f^{-1}(A))$ is open in $X$ so $f^{-1}(A)$ is open in $X / \sim$.

Moreover, not all quotient maps are open. Consider the relation on $\mathbb R$ where the equivalence classes are $(-\infty, 0)$ and $[0,\infty)$. Then the image of the set $(1,2)$ is not an open set in the quotient.

  • $\begingroup$ Ah ok , now I see it $\endgroup$ – la flaca Jul 5 '16 at 20:02
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    $\begingroup$ @Eliana: Maybe it will be easier to see if you give $f^{-1}[A]$ a name; call it $B$. Now you know that $\pi^{-1}[B]\in\tau$, so by definition $B\in\tau'$. $\endgroup$ – Brian M. Scott Jul 5 '16 at 20:05
  • $\begingroup$ Of course, I feel so silly. Thank you both $\endgroup$ – la flaca Jul 5 '16 at 20:06

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