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Munkres' "Topology" (Second edition) says the following:

Let $p:X\to Y$ be a quotient map; let $A$ be a subspace of $X$ that is saturated with respect to $p$; let $q:A\to p(A)$ be the map obtained by restricting $p$. If $A$ is either open or closed in $X$, then $q$ is a quotient map.

  1. Isn't $A=p^{-1}(Y)$, considering $A$ is saturated with respect to $p$, and $p$ is surjective because it is a quotient map?

  2. If (1) is true, isn't $A=X$, and hence automatically closed (and open)?

Thanks in advance!

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    $\begingroup$ Saturated means that $A=p^{-1}(p(A))$. $\endgroup$ – Stefan Hamcke Feb 26 '14 at 17:00
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According to Munkre's saturated means that either $p^{-1}(\{y\}) \subset A$ or $p^{-1}(\{y\}) \subset A^c$ for every $y \in Y$. You only get that $A = p^{-1}(p(A))$.

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  1. is false, you only have $A=p^{-1}(Z)$ with $Z$ a certain subset of $Y$.
  2. also if false.

You have to show that $q$ is continuous (and it is beacuse it is obtained form a continuous map by restricting domain and codomain) and surjective (and it is because $p(A)=q(A)$ by definition of $q$) and that if $q^{-1}(T)$ is open in $A$, then $T$ is open in $q(A)$. This is true since: (suppose $A$ is open in $X$) $q^{-1}(T)$ is open in $X$. Since $q^{-1}(T)=p^{-1}(T) \cap A=p^{-1}(Z \cap T)$ we have $Z \cap T$ is open in $Y$. But $Z \cap T=T$ so $T$ is open in $q(A)$ .

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