There's a mapping $f:X\rightarrow Y$.

1.for all $A,B\subset X$, $f(A\cap B)=f(A)\cap f(B)$, prove $f$ is injective.

2.for all $A\subset X$, $f(A^{c})=[f(A)]^{c}$, prove $f$ is bijective.

  • $\begingroup$ Try to show us what you have tried yourself, and where you got stuck. This way we can help you more effectively! You will also learn a lot more this way. $\endgroup$ – sxd Sep 9 '12 at 15:05

For 1 you want to show that $f(x) = f(y)$ implies $x = y$. So let $f(x) = f(y)$. Then $f(\{x\}) = f(\{y\})$ and hence $f(\{x\}) \cap f(\{y\}) = f(\{x\}) = f(\{y\}) = f(\{x\} \cap \{y\})$. Hence $x=y$.

  • $\begingroup$ Now try to do 2 in a similar way. $\endgroup$ – Rudy the Reindeer Sep 9 '12 at 15:06

(1) Suppose $f$ is not injective. That is, there exists $a \neq b$ such that $f(a) = f(b)$. Consider the sets $\{a\}$ and $\{b\}$. Then $f(\{a\} \cap \{b\}) = f(\emptyset) = \emptyset \neq f(\{a\}) \cap f(\{b\})$.

(2) Suppose $f$ is not injective. Then there exists an $x$ such that $f^{-1}(x)$ contains more than a single element. Let $a \in f^{-1}(x)$. Let $A = f^{-1}(x) - \{a\}$. Then $x \in f(A^c)$ but $x \notin f(A)^c$. So $f(A^c) \neq f(A)^c$.

Suppose that $f$ is not surjective. Then there exists a $y \in Y$ such that $y \notin f(X)$. Let $A = X$. Then $f(A^c) = f(\emptyset) = \emptyset$. However $y \in f(A)^c$. Hence $f(A^c) \neq f(A)^c$.

Thus, it has been shown that $f$ is injective and surjective.


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