# Inverse image of disjoint is disjoint?

If I have two sets that are disjoint i.e. $A\cap B=\emptyset$, and $\varphi \in C^1(U,\mathbb{R}^N)$, then are the inverse images (i.e. $\varphi^{-1}(A), \varphi^{-1}(B)$) also disjoint?

My logic supporting this assertion would be that from the definition of well defined: $$a=b \implies f(a)=f(b)$$

Therefore, taking the contrapositive, the disjointness seems to follow, but I'm not certain.

Does this seem reasonable?

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## 1 Answer

$$x\in \phi^{-1}(A)\cap\phi^{-1}(B)\Longrightarrow \phi(x)\in A\cap B...\text{contradiction...?}$$

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It always holds. It's the definition of the inverse image: $$x\in\phi^{-1}(A)\Longleftrightarrow \phi(x)\in A$$ –  DonAntonio Oct 8 '12 at 10:42
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