Characterization of Continuous, Closed and Open maps

closed-mapgeneral-topologymanifoldsopen-mapproof-writing

I've been strugrilling trying to prove the following results from Lee's book on Topological Manifolds.

Proposition 1. Let $f:X\to Y$ be a function between topological spaces.

  1. $f$ is continuous if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for all $A \subseteq X$.

  2. $f$ is closed if and only if $\overline{f(A)} \subseteq f(\overline{A})$
    for all $A \subseteq X$.

  3. $f$ is continuous if and only if $f^{-1}(\mbox{int}B) \subseteq \mbox{int} f^{-1}(B)$ for all $B \subseteq Y$.

  4. $f$ is open if and only if $\mbox{int} f^{-1}(B) \subseteq f^{-1}(\mbox{int}B)$ for all $B \subseteq Y$.

For the first one, we need the following result

Proposition 2. Let $f: X \to Y$ be a function between topological spaces. The following are equivalent

  1. $f$ is continuous.

  2. For each $x \in X$ and every neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subseteq V$.

  3. For every subset $B$ of $Y$, $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$.

To prove the first part of proposition 1. ($\Rightarrow$) Let $A \subseteq X$ and let $x \in \overline{A}$. Let $V$ be a neighborhood of $f(x)$ and let $U$ be a neighborhood of $x$ such that $f(U) \subseteq V$. Since $x \in \overline{A}$, $U \cap A \neq \emptyset$ and hence $\emptyset \neq f(U \cap A) \subseteq f(U)\cap f(A) \subseteq V \cap f(A)$. Thus, every neightborhood of $f(x) \in \overline{f(A)}$ and $f(\overline{A}) \subseteq \overline{f(A)}$.

($\Leftarrow$) Let $A = f^{-1}(B)$. Then, $f(\overline{A}) \subseteq \overline{f(A)} = \overline{f(f^{-1}(B))} = \overline{B}$. Hence, $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$. Which is equivalent to $f$ being continuous. $\Box$

I'm unsure how to proceed with the other statements. Thanks in advance!

Best Answer

The following facts may be helpful: for any sets $A \subseteq X$ and $B \subseteq Y$,

  • $A$ is closed if and only if $A = \overline{A}$.
  • $A$ is open if and only if $A = {\rm int} \, A$.
  • $A \subseteq f^{-1}(f(A))$ and $f(f^{-1}(B)) \subseteq B$.

Proposition 1.2: ($\implies$): Suppose $f$ is closed. Then $f(\overline A)$ is closed for any $A \subseteq X$, so $f(A) \subseteq f(\overline A)$ implies $\overline{f(A)} \subseteq f(\overline A)$.

($\impliedby$): Suppose $A \subseteq X$ is closed. Then $\overline{f(A)} \subseteq f(\overline A) = f(A)$, so $f(A) = \overline{f(A)}$. So $f(A)$ is closed whenever $A$ is closed; thus $f$ is closed.

Proposition 1.3: is fairly similar to 1.2.

Proposition 1.4: ($\implies$) Suppose $f$ is open. Then $A$ open implies $f(A)$ open, so $f( {\rm int} \, f^{-1}(B)) \subseteq {\rm int} \, B$ (why?) and ${\rm int} \, f^{-1}(B) \subseteq f^{-1}({\rm int} \, B)$.

($\impliedby$) Suppose $A$ is open and $\mbox{int} f^{-1}(B) \subseteq f^{-1}(\mbox{int} \, B)$ for all $B \subseteq Y$. Then $$A = {\rm int}\, A \subseteq {\rm int} \, f^{-1}( f(A)) \subseteq f^{-1}({\rm int} \, f(A)),$$ so $f(A) \subseteq f(f^{-1}({\rm int} \, f(A))) \subseteq {\rm int} \, f(A)$. So $f(A)$ is open.

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