A closed set $A\subset Y$ is closed on $Y$ iff $\pi^{-1}(A)$ is closed on $X$.

Question

Let $X$ be a topological space, $Y$ be a quotient space of X and $\pi:X\rightarrow Y$ be a quotient function.

How to prove the following: A closed set $A\subset Y$ is closed on $Y$ iff $\pi^{-1}(A)$ is closed on $X$.

My attempt: $(\Rightarrow)$ Suppose that $A\subset Y$ is closed on $Y$. Then $Y-A$ is open on $Y$. How $\pi$ is the quotient function, $\pi^{-1}(Y-A)=\pi^{-1}(Y)-\pi{-1}(A)=X-\pi^{-1}(A)$ is open on $X \Leftrightarrow \pi^{-1}(A)$ is closed on $X$.

It's fine? How to prove the other inclusion?

Answer 1

Characteristics of a quotient map $\pi:X\to Y$ are:

• $\pi$ is continuous.
• $\pi$ is surjective.
• If the preimage of $A\subseteq Y$ wrt to $\pi$ is open then $A$ is open.

From the fact that $\pi$ is continuous it follows that $\pi^{-1}(A)$ is closed whenever $A\subseteq Y$ is closed.

(That is the part you proved yourself. Only continuity is used there).

For the other side: if $\pi^{-1}(A)$ is closed then its complement is open, and this with: $$\pi^{-1}(A)^{\complement}=\pi^{-1}(A^{\complement})$$From the fact that $\pi^{-1}(A^{\complement})$ is open we conclude that $A^{\complement}$ is an open set, hence $A$ is a closed set.

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remark:

The third characteristic can also be interchanged by:

• If the preimage of $A\subseteq Y$ wrt to $\pi$ is closed then $A$ is closed.