testing for continuity via closure operation

Proposition 1.

Let $X\mathrm{,}Y$ be topological spaces, and $f\mathrm{:}X\mathrm{\to }Y$ a function. Then the following are equivalent:

1. $f$ is continuous,

2. for any closed set $D\subseteq Y$, the set $f^{-1}(D)$ is closed in $X$,

3. $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ is the closure of $A$,

4. $\overline{f^{-1}(B)}\subseteq f^{-1}(\overline{B})$,

5. $f^{-1}(C^{\circ })\subseteq f^{-1}{(C)}^{\circ }$, where $C^{\circ }$ is the interior of $C$.

Proof.

• $(1)\iff (2)$. Use the identity $f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$ for any function $f$. Then $f^{-1}(Y-D)=X-f^{-1}(D)$. So if $D$ is closed (or open), $f^{-1}(Y-D)$ is open (or closed), whence $f^{-1}(D)$ is closed (or open).

• $(2)\iff (3)$. Suppose first that $f:X\to Y$ is continuous. Since

  $$\overline{f(A)}=\bigcap \{C\mid C\text{ closed in }Y,\text{ and }f(A)\subseteq C\},$$

  $A\subseteq f^{-1}f(A)\subseteq f^{-1}(C)$, which is closed in $X$. So $\overline{A}\subseteq f^{-1}(C)$, and therefore $f(\overline{A})\subseteq f{f}^{-1}(C)\subseteq C$. As a result,

  $$f(\overline{A})\subseteq \bigcap \{C\mid C\text{ closed in }Y,\text{ and }f(A)\subseteq C\}=\overline{f(A)}.$$

  Conversely, let $V$ be closed in $Y$. Then $\overline{V}=V$. Let $U=f^{-1}(V)$. So $f(U)=V$. Let $W=\overline{U}$. Then $f(W)=f(\overline{U})\subseteq \overline{f(U)}=\overline{V}=V$. So $W\subseteq f^{-1}f(W)\subseteq f^{-1}(V)=U\subseteq \overline{U}=W$. As a result, $U=W$ is closed.

• $(3)\iff (4)$. First, assume $(2)$. Let $B\subseteq Y$ and $A=f^{-1}(B)$. So $f(A)\subseteq B$. Then $f(\overline{A})\subseteq \overline{f(A)}\subseteq \overline{B}$. As a result, $\overline{f^{-1}(B)}=\overline{A}\subseteq f^{-1}f(\overline{A})\subseteq f^{-1}(\overline{B})$.

  Conversely, assume $(3)$. Let $A\subseteq X$ and $B=f(A)$. So $A\subseteq f^{-1}(B)$. Then

  $$f(\overline{A})\subseteq f(\overline{f^{-1}(B)})\subseteq f{f}^{-1}(\overline{B})\subseteq \overline{B}=\overline{f(A)}.$$

• $(4)\iff (5)$. First, assume $(3)$. We use the identity: $C^{\circ }=Y-\overline{Y-C}$. Then

  $f^{-1}(C^{\circ })$$=$$f^{-1}(Y-\overline{Y-C})=f^{-1}(Y)-f^{-1}(\overline{Y-C})\subseteq X-\overline{f^{-1}(Y-C)}$

  $=$$X-\overline{f^{-1}(Y)-f^{-1}(C)}=X-\overline{X-f^{-1}(C)}=f^{-1}{(C)}^{\circ }.$

  Conversely, assume $(4)$. We use the identity $\overline{B}=Y-{(Y-B)}^{\circ }$. Then

  $\overline{f^{-1}(B)}$$=$$X-{(X-f^{-1}(B))}^{\circ }=X-f^{-1}{(Y-B)}^{\circ }$

  $\subseteq$$X-f^{-1}({(Y-B)}^{\circ })=f^{-1}(Y-{(Y-B)}^{\circ })=f^{-1}(\overline{B}).$

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