# A question about the restriction of quotient maps to subsets of domain.

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)?

• Saturated means that $A=p^{-1}(p(A))$. – Stefan Hamcke Feb 26 '14 at 17:00
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 $A = p^{-1}(p(A))$.
1. is false, you only have $$A=p^{-1}(Z)$$ with $$Z$$ a certain subset of $$Y$$.
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 $$q(A)$$ .