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I m stacked in one prove which dealt with sets and functions. I m concerned to prove that:

                fog ( X ∩ Y) ⊂ fog( X) ∩ fog ( Y)

Assume that g is function from A to B and f is from B to C, Furthermore X ⊂ A and Y ⊂ A.

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Why not prove that $h(X \cap Y) \subset h(X) \cap h(Y)$ first? Suppose $x \in h(X \cap Y)$. Where does that lead you? –  copper.hat Nov 6 '13 at 16:49
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Notice that the composition of functions is not important. If you define $h = f \circ g: A \to C$, then you are trying to show that $h(X \cap Y) \subseteq h(X) \cap h(Y)$. –  Sammy Black Nov 6 '13 at 16:53
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Whoa, @copper.hat. Synchronicity. –  Sammy Black Nov 6 '13 at 16:54
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