Let $X,Y$ be topological spaces, $\sim$ an equivalence relation on $X$ and $X/{\sim}$ the quotient space equipped with the quotient topology. Moreover, let $f:X\to Y$ be a map such that $f(x) = f(y)$ if $x \sim y$. Show that the map $g:X /{\sim} \to Y, [x] \to f(x)$ is continuous if and only if $f$ is continuous.

I started with defining a map $h:X \to X /{\sim}$, $x \to [x]$ $g(h(x)) = f(x)$. Then

$$ f^{-1}(U) = \{x \in X :f(x) \in U \} = \{x \in X : g(h(x)) \in U \} = (g \circ h)^{-1}(U) $$

Now $h$ continuous, so we have the following equivalence.

$f$ also continuous $\Leftrightarrow$ $g \circ h$ continuous $\Leftrightarrow$ g continuous

  • $\begingroup$ Are you sure about last equivalence? For example, in $\mathbb{R}$ if $h$ is a constant function and $g$ is any function, then $g\circ h$ is continuous (as it is constant) but $g$ need not be continuous. You need to argue a bit differently $\endgroup$ – user160738 Jun 15 '17 at 20:35
  • $\begingroup$ How would you argue? @user160738 $\endgroup$ – Olba12 Jun 15 '17 at 20:55
  • $\begingroup$ In my lecture notes I have "Let $(X_i)_{i \in I}$ be top. spaces, $Y$ a set, $f_i: X_i \to Y$ maps. For any top.space $Z$ and any map $g:Y \to Z$, $g$ is continuous $\Leftrightarrow g \circ f_i$ is continuous $\forall i$." $\endgroup$ – Olba12 Jun 15 '17 at 21:00
  • $\begingroup$ And the topology on $Y$ should be the maximal topology which is induced by those $f_i$. So your argument is absolutly right. $\endgroup$ – C.Ding Jun 16 '17 at 0:00

It seems that you're implying that if $h:X\to Y$ and $g:Y\to Z$ such that $h$ is continuous, then $g\circ h$ is continuous if and only if $g$ is continuous. This is simply not true (if I am misinterpreting what you say, then let me know).

Note that for any open set $U\subset Y$, $f^{-1}(U)=h^{-1}(g^{-1}(U))$. Now try using the fact that by the definition of the quotient topology, $g^{-1}(U)$ is open in $X/{\sim}$ if and only if $h^{-1}(g^{-1}(U))$ is open in $X$.

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  • $\begingroup$ Is $h^{-1}(g^{-1}(U)) = (g \circ h) ^{-1} (U)$, or what, becuase we dont have same notation? $\endgroup$ – Olba12 Jun 15 '17 at 21:59
  • $\begingroup$ So the OP is right by your proof! $\endgroup$ – C.Ding Jun 16 '17 at 0:06
  • $\begingroup$ @Olba12 Ya that's correct. The way I wrote it I thought would make it easier to see, since by definition of the quotient topology, a subset $V\subset X/{\sim}$ is open if and only if $h^{-1}(V)$ is open. $\endgroup$ – Alex Mathers Jun 16 '17 at 0:21
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    $\begingroup$ @C.Ding The OP didn't reach the conclusion in the same way. They're trying to state that $g\circ h$ is continuous if and only if $g$ is continuous for any $g$ since $h$ is continuous, which is untrue. $\endgroup$ – Alex Mathers Jun 16 '17 at 0:22

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