I am given this equation: $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

I want to prove it: what i did is

I take any $a \in f^{-1}(B_1 \cap B_2)$, then there is $b \in (B_1 \cap B_2)$ so that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.

this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.

is it ok?


2 Answers 2


Here is essentially the first answer, written out a bit more formally.

Let's start with the basic property of $\;f^{-1}[\cdot]\;$: $$ a \in f^{-1}[B] \;\equiv\; f(a) \in B $$ for any $\;a,B\;$. Using this, we can simply calculate \begin{align} & a \in f^{-1}[B_1 \cap B_2] \\ \equiv & \;\;\;\;\;\text{"the above basic property"} \\ & f(a) \in B_1 \cap B_2 \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$"} \\ & f(a) \in B_1 \;\land\; f(a) \in B_2 \\ \equiv & \;\;\;\;\;\text{"the above basic property, twice"} \\ & a \in f^{-1}[B_1] \;\land\; a \in f^{-1}[B_2] \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;\cap\;$ using its definition"} \\ & a \in f^{-1}[B_1] \cap f^{-1}[B_2] \\ \end{align} By set extensionality, the statement in question follows.

  • $\begingroup$ ...and this would even be simpler if the proof was done in the other direction, starting with the most complex side of the equation. $\endgroup$ Commented Jun 11, 2022 at 12:24

Yeah...this can be actually written in this way;

$a\in f^{-1}(B_1\cap B_2)$, means $f(a)\in B_1\cap B_2$ and so $f(a)\in B_1$ and $f(a)\in B_2$. Hence, $a\in f^{-1}(B_1)$ and $a\in f^{-1}(B_2)$

  • $\begingroup$ this also means that $B_1=B_2$ is either the same or it's interesection is empty $\endgroup$
    – Rainb
    Commented May 7, 2023 at 10:40

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