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Given $f:X\rightarrow Y$ as a function, the image of $x$ if $f(x)$. The preimage of $y$ is $f^{-1}(y)=\{x\ |\ f(x)=y\}$, with the symbol PreIm$(Y)$

Given the definition, could you prove the following statement? Thank you.

Given that $f:X\rightarrow Y$ is a function, and that $A, B \subseteq Z$.

PreIm$(A\cup B)$ = PreIm$(A) \cup$ PreIm$(B)$

Asaf Karagila
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user12488
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    Try doing a search; you should be able to find a variety of proofs of this statement already on math.stackexchange. – goblin GONE Apr 03 '14 at 05:07
  • Note the discrepancy in your notation, you define preimage for elements of $y$ but then PreIm is defined for sets. – Asaf Karagila Apr 03 '14 at 05:13
  • @AsafKaragila: Of course, you are familiar with the standard abuse of notation identifying $f^{-1}(y)$ and $f^{-1}({y})$, I'm sure – MPW Apr 03 '14 at 05:35
  • @MPW The definition using "${ x\mid f(x)=y}$" doesn't apply for subsets of $Y$ though, that's what OP is suggesting. – Christoph Apr 03 '14 at 07:09
  • @AsafKaragila: And, in fact, I suppose it is the very result OP wants to prove that allows the extension from points to sets. Point taken. – MPW Apr 03 '14 at 11:11

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First, please dispense with the silly notation "PreIm()". The standard notation is $f^{-1}(\cdot)$.

That said, one notes that $f^{-1}$ is generally well-behaved.

If $x\in f^{-1}(A\cup B)$, then $f(x)\in A\cup B$, so either $f(x)\in A$ or $f(x)\in B$, which means $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$, hence $x\in f^{-1}(A)\cup f^{-1}(B)$. We have shown $$f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B).$$

If $x\in f^{-1}(A)\cup f^{-1}(B)$, then either $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$, so either $f(x)\in A$ or $f(x)\in B$, so $f(x)\in A\cup B$, hence $x\in f^{-1}(A\cup B)$. We have shown that $$f^{-1}(A)\cup f^{-1}(B)\subset f^{-1}(A\cup B).$$

Thus $$ \boxed{f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)} $$ as desired.

MPW
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