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Let $X$ and $Y$ be topological space and let $\pi:X\to Y$ be a quotient map (it is surjective map and $Y$ has the quotient topology induced by $\pi$). A subset $U\subset X$ is said to be saturated if $U=\pi^{-1}(\pi(U))$. One easily can checks that the restriction of $\pi$ to any saturated open is a quotient map. Let $U$ be a saturated open set. We can consider two topologies on $\pi(U)$: the subspace topology and the quotient topology $(\pi_{|_U}: U\to \pi(U))$.

What is the relationship between these two topologies?

I managed to prove that the topology quotient is finer than the subspace topology, but just it. I cannot proof that the topologies are the same and I cannot find a counterexample.

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Since $U$ is a saturated open set, the quotient topology on $\pi(U)$ induced by $\pi\lvert_U$ is the same as the subspace topology induced by $Y$.

To see that, we must see that the inclusion $j \colon \pi(U) \hookrightarrow Y$ is open. So let $W\subset \pi(U)$ be an open set. By definition, that means $V := (\pi\lvert_U)^{-1}(W) \subset U$ is open. But $U$ is open, so $V$ is open in $X$. And $U$ is saturated, hence $V$ is saturated, and so $\pi^{-1}(j(W)) = V$ is an open set, hence $j(W)$ is open in $Y$.

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Let $V_s\subset \pi(U)$ a open subset of $\pi(U)$ with the subspace topology. But $\pi(U)$ is open in $Y$, so $V_s$ is open in $Y$. By definition, that means $\pi^{-1}(V_s)$ is open in $X$. Finally, $$(\pi_{|_U})^{-1}(V_s)=\{x\in U;\pi(x)\in V_s\}=U\cap\pi^{-1}(V_s)$$ hence $V_s$ is open in $\pi(U)$ with the quotient topology.

Now, let $V_q\subset \pi(U)$ a open subset of $\pi(U)$ with the quotient topology. So $$V_q\subset \pi(U) \Rightarrow\pi^{-1}(V_q)\subset\pi^{-1}(\pi(U))=U$$ and hence $$(\pi_{|_U})^{-1}(V_q)=U\cap\pi^{-1}(V_q)=\pi^{-1}(V_q)$$ By the definition, that means that $\pi^{-1}(V_q)(=(\pi_{|_U})^{-1}(V_q))$ is open in $U$ and thus $\pi^{-1}(V_q)$ is open in $X$. Again by definition $V_q$ is open in $Y$ and a fortiori $V_q$ is open in $\pi(U)$ with the subspace topology.

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