TITLE: A map is continuous if and only if for every set, the image of closure is contained in the closure of image
QUESTION [81 upvotes]: As a part of self study, I am trying to prove the following statement:
Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$. 
Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?

REPLY [3 votes]: The assertion is equivalent to:
$\overline{A}\subseteq f^{-1}(\overline{f(A)})$
So, the assertion follows from:
$\overline{A}\subseteq\overline{f^{-1}(f(A))}\subseteq\overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$

Inclusion: $A\subseteq f^{-1}(f(A)) \Rightarrow \overline{A}\subseteq\overline{f^{-1}(f(A))}$
Inclusion: $f(A)\subseteq\overline{f(A)} \Rightarrow f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)}) \Rightarrow \overline{f^{-1}(f(A))}\subseteq \overline{f^{-1}(\overline{f(A)})}$
Equality: $\overline{f(A)} \text{ closed} \Rightarrow f^{-1}(\overline{f(A)}) \text{ closed} \Rightarrow \overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$

The converse assertion is equivalent to:
$\overline{B}=B \Rightarrow \overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)\subseteq\overline{f^{-1}(B)}\subseteq f^{-1}(f(\overline{f^{-1}(B)}))\subseteq f^{-1}(\overline{f(f^{-1}(B))}) \subseteq f^{-1}(\overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=\overline{f^{-1}(B)}$

Inclusion: $A\subseteq \overline{A} \text{ in general}$
Inclusion: $A\subseteq f^{-1}(f(A)) \text{ in general}$
Inclusion: $f(\overline{A})\subseteq \overline{f(A)} \text{ by assumption}$
Inclusion: $f(f^{-1}(B))\subseteq B \text{ in general} \Rightarrow \overline{f(f^{-1}(B))}\subseteq \overline{B} \Rightarrow f^{-1}(\overline{f(f^{-1}(B))})\subseteq f^{-1}(\overline{B})$
Equality: $\overline{B}=B \Rightarrow f^{-1}(\overline{B})=f^{-1}(B)$