# Be f : A → B a function and X,Y ⊂ B. Show that: any thoughts? thanks

## a) f^-1(X∩Y) = f^-1(X) ∩ f^-1(Y) b) If $f$ is surjective, then $f \left({f}^{-} 1 \left(X\right)\right) = X$

Dec 31, 2017

To show two sets are equal, show that each is a subset of the other (see below).

#### Explanation:

a) To show that ${f}^{- 1} \left(X \cap Y\right) \subseteq {f}^{- 1} \left(X\right) \cap {f}^{- 1} \left(Y\right)$, assume that $a \in {f}^{- 1} \left(X \cap Y\right)$. This means that $f \left(a\right) \in X \cap Y$ so that $f \left(a\right) \in X$ and $f \left(a\right) \in Y$. But this implies that $a \in {f}^{- 1} \left(X\right)$ and $a \in {f}^{- 1} \left(Y\right)$, leading to the conclusion that $a \in {f}^{- 1} \left(X\right) \cap {f}^{- 1} \left(Y\right)$.

The previous logic is reversible to show that ${f}^{- 1} \left(X\right) \cap {f}^{- 1} \left(Y\right) \subseteq {f}^{- 1} \left(X \cap Y\right)$. For if $a \in {f}^{- 1} \left(X\right) \cap {f}^{- 1} \left(Y\right)$, then $a \in {f}^{- 1} \left(X\right)$ and $a \in {f}^{- 1} \left(Y\right)$, implying that $f \left(a\right) \in X$ and $f \left(a\right) \in Y$. But this means that $f \left(a\right) \in X \cap Y$ so that $a \in {f}^{- 1} \left(X \cap Y\right)$.

The last two paragraphs lead to the conclusion that ${f}^{- 1} \left(X \cap Y\right) = {f}^{- 1} \left(X\right) \cap {f}^{- 1} \left(Y\right)$.

b) Assume that $f$ is surjective (onto).

To show that $f \left({f}^{- 1} \left(X\right)\right) \subseteq X$, suppose that $b \in f \left({f}^{- 1} \left(X\right)\right)$. This means that there exists $a \in {f}^{- 1} \left(X\right)$ such that $f \left(a\right) = b$. But $a \in {f}^{- 1} \left(X\right)$ implies that $b = f \left(a\right) \in X$.

The logic of the previous paragraph is not completely reversible since we did not need the fact that $f$ is surjective. To reverse the logic, we need that assumption. To show that $X \subseteq f \left({f}^{- 1} \left(X\right)\right)$, suppose that $b \in X$. Since $f$ is surjective (onto), there exists an $a \in A$ such that $f \left(a\right) = b$, which also means that $a \in {f}^{- 1} \left(X\right)$ (since $b \in X$). But this then means that $b \in f \left({f}^{- 1} \left(X\right)\right)$.

The last two paragraphs lead to the conclusion that $f \left({f}^{- 1} \left(X\right)\right) = X$ when $f$ is surjective.