# A question regarding the property of a quotient map.

Let $X$ and $Y$ be two topological spaces. Suppose $f:X \longrightarrow Y$ be a quotient map. Consider the equivalence class of $X$ formed by the disjoint non-empty fibres of $f$. Let $X/{{\sim}_f}$ be the set of all these non-empty disjoint classes of $X$. Consider the map $\tilde {f} : X/{{\sim}_f} \longrightarrow Y$ defined by $\tilde {f} ([x]) = f(x),$ $x \in X$. Then this map is well-defined and one-to-one by the construction and also since $f$ is onto, $\tilde {f}$ is onto. Let $p : X \longrightarrow X/{{\sim}_f}$ be the quotient map where $X/{{\sim}_f}$ is endowed with quotient topology. Then it is not hard to see that $\tilde {f} \circ p = f$. SInce $f$ is continuous so by Universal Mapping Property we can say that $\tilde {f}$ is continuous. Now my question is can we say that $\tilde {f}$ is a homeomorphism? My instructor said that it is true and we were given it to prove.

What I have tried is as follows $:$

In order to show that $\tilde {f}$ is a homeomorphism we need only to prove that it is open. So I take an open subset $\tilde {V} \subset X/{{\sim}_f}$. We need to prove that $\tilde {f} (\tilde {V})$ is open in $Y$. Now what is $\tilde {f} (\tilde {V})$? I have found that $\tilde {f} (\tilde {V}) = f(S)$. Where $$S = \cup_{A \in \tilde {V}} A.$$ Now since $\tilde {V}$ is open in $X/{{\sim}_f}$ so $p^{-1} ( \tilde {V} )$ is open in $X$. Now what is $p^{-1} (\tilde{V})$? I have found that $p^{-1} (\tilde{V}) = S$. So we have to prove that $f(S)$ is an open subset of $Y$, where $S$ is an open subset of $X$.

I tried to prove it by contradiction. If not let $f(S)$ is not an open subset of $Y$. But that implies $f^{-1}(f(S))$ is not an open subset of $X$. If $f$ is one-to-one we are clearly done since then $f^{-1}(f(S))=S$ which leads us to a contradiction because $S$ was an open subset of $X$. But if $f$ is not one-to-one I don't know how to proceed. Please help me in this regard.

Thank you very much.

You prove that $\tilde{f}$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X \to Y$, $p : X \to X / \sim f$ and a bijection $\tilde{f} : X / \sim f \to Y$ such that $\tilde{f} \circ p = f$. You correctly conclude that $\tilde{f}$ is continuous by the universal property of the quotient map $p$. The same argument shows that $\tilde{f}^{-1}$ is continuous because $\tilde{f}^{-1} \circ f = p$.
• So it follows from what I proved that if $f$ is open,continuous and onto then $\tilde {f}$ is a homeomorphism. Isn't it so @Paul Frost? Commented Aug 5, 2018 at 16:32
• Yes, by construction $\tilde{f}$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $\tilde{V}$ be open in $X / \sim f$. Then $f^{-1}(\tilde{f}(\tilde{V})) = (\tilde{f} \circ p)^{-1}(\tilde{f}(\tilde{V})) = p^{-1}(\tilde{f}^{-1}(\tilde{f}(\tilde{V}))) = p^{-1}(\tilde{V})$ which is open in $X$. Therefore $\tilde{f}(\tilde{V})$ is open in $Y$. Commented Aug 5, 2018 at 16:56