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Let $A$ and $B$ be sets and let $f$ be a function from $A$ to $B$. If $X\subseteq A$ and $Y\subseteq A$, then $f(X\cap Y)=f(X)\cap f(Y)$.

Do I have to use a image and pre-image to prove? I'm honestly confused where to start.

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  • $\begingroup$ Necessarily $f(X\cap Y)\subset f(X)\cap f(Y)$, but you do not necessarily have $f(X\cap Y)=f(X)\cap f(Y)$. $\endgroup$ – neth Apr 19 '16 at 21:10
  • $\begingroup$ Choose some x, x is either in X and not Y, or Y and not X or both or neither .. what sets does f(x) belong to for each case. $\endgroup$ – Doug M Apr 19 '16 at 21:11
  • $\begingroup$ Example: $A=\{1,2,3\}=B$, $X=\{1\}$, $Y=\{2\}$ and $f:A\to A$ given by $f(a)=1\;\forall a\in A$. $\endgroup$ – lulu Apr 19 '16 at 21:13
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Take $z \in f(X \cap Y)$. Then there is $x \in X\cap Y$ such that $f(x) = z$. But since $x \in X$, this means $z = f(x) \in f(X)$ and likewise, since $x \in Y$, we have $z = f(x) \in f(Y)$. Thus $z \in f(X) \cap f(Y).$ This shows that $f(X \cap Y) \subseteq f(X) \cap f(Y)$.

As has been pointed out in the comments, the other direction is not necessarily true. Take the function $f(x) = x^2$ for $x \in \mathbb R$ and let $A = [-2,2]$, $X = [-2, -1]$ and $Y = [1,2]$. Then $X,Y \subseteq A$. Also $X\cap Y = \varnothing$ so $f(X\cap Y) = \varnothing$. However, $f(X) = [1,4] = f(Y)$, so $f(X) \cap f(Y) = [1,4] \neq f(X \cap Y).$

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