Let $f:X \to Y$ be a map between two topological spaces. Prove that the following statements are equivalent.

Question

I'm doing this exercise and I got stuck on the $2 \implies 3$ part. Here's my try:

Let $\ f:X \to Y$ be a map between two topological spaces. Prove that the following statements are equivalent.

$1)$ $\ f$ is continuous.

$2)$ $\ f^{-1}(Int(B))\subset Int(f^{-1}(B))$, $\forall B\subset Y.$

$3)$ $\ f(Cl(A))\subset Cl(f(A))$, $\forall A\subset X.$

I did this so far:

$1 \implies 2$

Suppose $f$ continuous. We have

$Int(B)\subset B \iff \ f^{-1}(Int(B))\subset f^{-1}(B) \iff \\ \ Int(f^{-1}(Int(B)))\subset Int(f^{-1}(B)) \iff f^{-1}(Int(B))\subset Int(f^{-1}(B))$

I can't see how to start attacking the $2 \implies 3$ part. I'd appreciate any hint.

Thanks for your time.

Answer 1

Suppose $y \in f[\operatorname{Cl}(A)]$, so $ y = f(x)$ with $x \in \operatorname{Cl}(A)$. We want to show that $y \in \operatorname{Cl}(f[A])$, so let $O$ be an open neighbourhood of $y$, and we must show $O$ intersects $f[A]$. As $O$ is open, $\operatorname{Int}(O) = O$ and so by (2) we have $f^{-1}[O] \subseteq \operatorname{Int}(f^{-1}[O])$, which shows $f^{-1}[O]$ is open and as it contains $x$ (as $f(x) = y \in O$), $f^{-1}[O]$ intersects $A$ (as $x \in \operatorname{Cl}(A)$), say in $x' \in A \cap f^{-1}[O]$, but then $f(x') \in O \cap f[A]\neq \emptyset$ and we are done.

Answer 2

Hint:

Note that $$ f(Cl(A)) =f(X-Int (A^\complement))$$

where $A^\complement$ is the complement of $A$

Answer 3

(3) $\Rightarrow$ (2):

Observe that $\overline{A^C} \dot{\cup}Int A = X \therefore \overline{A} \dot{\cup}Int A^C = X$, where $\dot \cup$ means disjoint union. We will show that $(Int F^{-1}(B))^C \subset (F^{-1}(Int B))^C$. $$ p\notin Int F^{-1}(B) \Leftrightarrow p \in \overline{F^{-1}(B)^C} = (Int F^{-1}(B))^C \Rightarrow F(p)\in F \left(\overline{F^{-1}(B)^C} \right) \subset \overline{ F \left(F^{-1}(B)^C \right) } \\ \Rightarrow F(p) \notin Int(F \left(F^{-1}(B)^C \right))^C \supset Int B \Rightarrow p \notin F^{-1}(Int B), $$ since $F \left(F^{-1}(B)^C \right) \subset B^C \therefore \left( F \left(F^{-1}(B)^C \right) \right)^C \supset B$.

(2) $\Rightarrow$ (1) is fine.

(1) $\Rightarrow$ (3):

Given $p\in F(\overline A)$, $p = f(\alpha)$ for some $\alpha \in \overline A$. Let $U\subset Y$ be a neighborhood of $p$. Since $F$ is continuous, $F^{-1}(U) \subset X$ is a neighborhood of $\alpha$. There exists $a\in A\cap F^{-1}(U) \Rightarrow f(a) \in F(A)\cap U \Rightarrow p \in \overline{F(A)}$, since $U \supset F(F^{-1}(U))$.

Answer 4

Starting with

$\begin{align}\small f^{-1}(Int(Y-f(A)))&\subset Int(f^{-1}(Y-f(A)))\\ &\subset lnt(X-A), \end{align}$ by taking complements and evaluating by function we have

$\begin{array} .f(Cl(A))&\subset f(X-f^{-1}(Int(Y-f(A))))\\ &\subset Cl(f(A)). \end{array}$