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I've been thinking about it and I can't find a reason why it wouldn't be the case.

Even if we have say $f^{-1}(x) = \sqrt x$, the statement is true because $-x$ is undefined for both $f(A)$ and $f(B)$.

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If by "inverse" you mean "preimage," then the statement $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ is always true. – Bey Oct 26 '12 at 16:45
what is the difference between preimage and inverse? – Penny Loren Oct 26 '12 at 16:51
@PennyLoren: Inverse need not exist, it exists if the function is injective. Preimage is the set of "sources". – Asaf Karagila Oct 26 '12 at 16:52
For a concrete example, $f(x)=x^2$; what is $f^{-1}(2)$? Is it $+\sqrt2$ or $-\sqrt2$? – Asaf Karagila Oct 26 '12 at 17:04
up vote 1 down vote accepted

Assuming that $f^{-1}(A)=\{x\mid f(x)\in A\}$, let us see that this is true.

Instead of guessing, one can (and should) just go ahead and prove:

Let $x\in f^{-1}(A\cup B)$, then $f(x)\in A\cup B$ therefore either $f(x)\in A$ or $f(x)\in B$. If $f(x)\in B$ then $x\in f^{-1}(B)$, and therefore $x\in f^{-1}(A)\cup f^{-1}(B)$. Similarly if $x\in f^{-1}(A)$ the same argument applies.

Therefore we have proved that $f^{-1}(A\cup B)\subseteq f^{-1}(A)\cup f^{-1}(B)$.

Let $x\in f^{-1}(A)\cup f^{-1}(B)$, then either $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$. If $x\in f^{-1}(A)$ then $f(x)\in A$ and therefore in $A\cup B$; and if $x\in f^{-1}(B)$ we have that $f(x)\in B$ and therefore $f(x)\in A\cup B$.

Either way we have that $f(x)\in A\cup B$ and therefore $x\in f^{-1}(A\cup B)$. Therefore $f^{-1}(A\cup B)\supseteq f^{-1}(A)\cup f^{-1}(B)$.

We have proved a two-sided inclusion and therefore equality as wanted.

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