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Question Let $X$ be a set and let $(Y, \tau)$ be a topological space. Let $g:X \to Y$ be a given map. Define $$\tau'=\{U \subset X~|~ U=g^{-1}(V) ~\text{for some} ~V \in \tau\}.$$ Which of the following statements are true?

  1. $\tau'$ defines a topology on $X$.

  2. $\tau'$ defines a topology on $X$ only if $g$ is onto.

  3. Let $g$ be onto. Then the quotient space of $X$ with respect to this relation, with the topology inherited from $\tau'$, is homeomorphic to $(Y, \tau)$.

My Attempt.

  1. True. (I just check the axioms of topology and it satisfied by $\tau'$)

  2. False. (It doesn't require "onto" ness anywhere in the proof of 1. Although I don't have any counter example).

  3. Let $g$ be Onto. $x \sim y\text{ iff }g(x)=g(y) $ Then $\sim$ defines an equivalence relation on $X$. Let the set of all $\sim$equivalence classes is $X^*=\{[x]~|~x \in X\}$. Give the quotient topology on $X^*$. And $p:X \to X^*$ defined by $x\mapsto [x]$ be the quotient map.

I try to find out an Homeomorphism between $X^*$ to $(Y,\tau)$.

Define $~f: X^* \to (Y,\tau)$ by following rule: for any $[x]\in X^*, p^{-1}([x])$ is basically all elements in $X$ that are equivalent to $x$ w.r.to $\sim$. Hence $g(p^{-1}([x]))$ is a singleton set in $Y$. Call that element to be $f([x])$. Also $f$ is well defined here can be verified using the fact that $x \sim y ~iff~ g(x)=g(y) $. Here also $fp=g$. And since $g$ is continuous surjection so $f$ becomes continuous and bijection: In fact, Let $U $ be open in $Y$, then $p^{-1}(f^{-1}(U))=g^{-1}(U)$ is open in X. Since $p$ is quotient map then $f^{-1}(U)$ is open in $X^*$. Hence $f$ is continuous. Also it can be proved that $f$ is bijection. Now I try to prove $f$ to be open map but I can't. I think it require $g$ to be a quotient map...

How can I proceed further . ..? Thanks .

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2 Answers 2

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  1. Correct.
  2. Correct, and no counterexample is needed.
  3. In order to prove that $f$ is open, let $A$ be an open subset of $X/\sim$. This means that the set$$A^\star=\{x\in X\,|\,[x]\in A\}$$is an open subset of $X$. But $f(A)=g(A^\star)$ and this set is open, because\begin{align}g(A^\star)\in\tau&\iff g^{-1}\bigl(g(A^\star)\bigr)\in\tau'\\&\iff A^\star\in\tau',\end{align}since $A^\star=g^{-1}\bigl(g(A^\star)\bigr)$.
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  • $\begingroup$ Okay I got it....Since $A^*\in \tau'$ there exists an $V \in \tau $ such that $A^*=g^{-1}(V)$ now $g $ is onto so $g(A^*)=V \in \tau$...right...?? $\endgroup$
    – sigma
    Aug 23, 2018 at 16:21
  • $\begingroup$ @IndrajitGhosh There was an error in my answer and I've edited it. I suppose that it is fine now. $\endgroup$ Aug 23, 2018 at 16:23
  • $\begingroup$ @ Santos....Is my argument is right as I mention in the last comment... $\endgroup$
    – sigma
    Aug 23, 2018 at 16:25
  • $\begingroup$ @IndrajitGhosh It looks fine. $\endgroup$ Aug 23, 2018 at 16:26
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  1. You are right. The set $\tau'$ only depends on the subspace $g(X) \subset Y$. In fact, the open sets of $g(X)$ are precisely the sets of the form $g(X) \cap V$ with $V \in \tau$ and obviuosly $g^{-1}(g(X) \cap V) = g^{-1}(V)$.

  2. Your approach is okay, but you can do it easier. By definition $g : X \to Y$ is a quotient map (or identification map) because it is onto and $V \in \tau$ if and only if $g^{-1}(V) \in \tau'$ (this is the definition of $\tau')$. Now you have two quotient maps $g: X \to Y$, $p : X \to X^\ast$ and a bijection $f : X^\ast \to Y$ such that $f \circ p = g$. This automatically implies that $f$ is a homeomorphism - simply use the universal property of the quotient topology. See e.g. my answer to A question regarding the property of a quotient map..

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