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This question already has an answer here:

Given $A$ and $B$ is the subset of $C$ and $f:C\mapsto D$, $$f(A\cap B)\subseteq f(A) \cap f(B)$$ and the equality holds if the function is injective.

But why for the inverse, suppose that $E$ and $F$ is the subset of $D$, $$f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$$ without saying that the inverse function is injective. So if $$x\in f^{-1}(E) \cap f^{-1}(F)$$ $$x\in f^{-1}(E) \text{ and } x\in f^{-1}(F)$$ This means that there exists elements $y_1 \in E$ and $y_2 \in F$. So here how do we know that these two elements are equal.

I am independent learner so I hope I can get an explaination in more details.

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marked as duplicate by Community Mar 11 '17 at 2:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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proof: $$x\in f^{-1}(E \cap F)\Leftrightarrow f(x)\in E\cap F\\\Leftrightarrow f(x)\in E \ and \ f(x)\in F\\\Leftrightarrow x\in f^{-1}(E) \ and \ x\in f^{-1}(F) \\ \Leftrightarrow x\in f^{-1}(E)\cap f^{-1}(F)\\$$ so the equality holds.

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They don't mean the inverse by $f^{-1}$ but the pre-image under f. Maybe once you have studied that definition, it will become more evident!

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  • $\begingroup$ ya i know here pre-image is not inverse. But how we know the two images are the same? $\endgroup$ – Alan Wang Oct 21 '14 at 8:07
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The sentence "This means there exists elements $y_1\in E$ and $y_2\in F$" is incomplete. The mere fact that $y_1,y_2$ exist is not very useful. Say what property those elements have, and you will see why they are equal.

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  • $\begingroup$ Do you mean that $y$ must be inside intersection of $E$ and $F$ in order for the previous statement exists? $\endgroup$ – Alan Wang Oct 21 '14 at 8:22
  • $\begingroup$ I mean that opening the sentence with "This means" indicates a relation with the previous statement, which is "$x\in f^{-1}(E)$ and $x\in f^{-1}(F)$". So there should be some relation between $y_1,y_2$ and the $x$ of that statement. Write down that relation and you're almost done. $\endgroup$ – Marc van Leeuwen Oct 21 '14 at 8:54

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