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$f(S \cap T) \subseteq f(S) \cap f(T)$

Suppose there is a $x$ that is in $S$, but not in $T$, then there is a value $y$ such that $f(x) = x$, that is in $f(S)$, but not in $f(S \cap T)$. Suppose there is a $x$ that is in $T$, but not in $S$, then there is a value $y$ such that $f(x) = x$ that is in $f(T)$, but is not in $f(S \cap T)$.

*correction made

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    No. If $x$ is in $S$ but not $T$, and $y$ is in $T$ but not in $S$, it's possible there are $z$ and $w$ in $S\cap T$ such that $f(z)=f(y)$ and $f(w)=f(x)$.2012-11-04
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    You might want to begin your proof with this: Let $y \in f(S \cap T)$ so $y = f(x)$ for some $x \in $ ...2012-11-04
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    ahh im too confused now2012-11-04
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    Please, try to make the title of your question more informative. E.g., *Why does $a imply $a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader.*2012-11-04

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