TITLE: Do we have always $f(A \cap B) = f(A) \cap f(B)$?
QUESTION [10 upvotes]: Suppose $A$ and $B$ are subsets of a topological space and $f$ is any function from $X$ to another topological space $Y$. Do we have always $f(A \cap B) = f(A) \cap f(B)$?
Thanks in advance

REPLY [6 votes]: I don't know about topological spaces. But given $f:X\rightarrow Y$ and $A,B \subseteq X$ here's an attempt for proving that $$ f \: \text{injective} \rightarrow f(A \cap B) = f(A) \cap f(B),$$ where 
$$ (f(A \cap B) = f(A) \cap f(B)) 
\leftrightarrow 
(
\underbrace{f(A \cap B) \subseteq f(A) \cap f(B)}_{(i)}
) 
\wedge 
(
\underbrace{f(A \cap B) \supseteq f(A) \cap f(B)}_{(ii)}
). 
$$
(i) (contrapositive) $y \notin f(A) \cap f(B)$ is equivalent to saying that $y \notin f(A)$ and $y \notin f(B)$. Given $y \notin f(A)$ it is also not in $f(A \cap B)$, since $f(A) \supseteq f(A \cap B).$ Accordingly, $y \notin f(B)$ requires that $y \notin f(A \cap B).$ 
(ii) $y \in f(A)$ and $y \in f(B)$ implies, since $(\forall x,z \in X) f(x) = f(z) \rightarrow x = z$, that $y \in \{ f(x) : x \in A \wedge x \in B \}$ , or $y \in f(A \cap B)$. In short
$$ (y \in f(A) \wedge y \in f(B) \rightarrow y \in f(A \cap B)) \leftrightarrow f(A \cap B) \supseteq f(A) \cap f(B). $$ $\Box$ 


See also this page.