Let $f: X \rightarrow Y$ be a function and $ A \subseteq Y$ and $B \subseteq Y$. Prove that $f^{-1}(A)\setminus f^{-1}(B)= f^{-1}(A\setminus B)$.

Question

Let $f: X \rightarrow Y$ be a function and $ A \subseteq Y$ and $B \subseteq Y$. Prove that $f^{-1}(A)\setminus f^{-1}(B)= f^{-1}(A \setminus B)$.

My defintion of inverse image is Let $f: X \rightarrow Y$ be a function and let $V \subseteq Y$. The inverse image of V is the set $f^{-1}(V)$ $=$ $\{x\in X$ such that $f(x) \in V \}$

I know I need to prove the RHS is an element in the LHS and vice versa. Where should I start?

See also: Proof of $f^{-1}(B_{1}\setminus B_{2}) = f^{-1}(B_{1})\setminus f^{-1}(B_{2})$ — Martin Sleziak, Sep 13 '19 at 12:42

score 1 · Accepted Answer · answered Mar 06 '15 at 23:26

Here's one direction:

Let $x\in f^{-1}(A)\setminus f^{-1}(B)$. Then $x\in f^{-1}(A)$ and $x\not\in f^{-1}(B)$. Therefore, $f(x)\in A$ and $f(x)\not\in B$. Therefore, $f(x)\in A\setminus B$. Since $f(x)\in A\setminus B$, $x\in f^{-1}(A\setminus B)$.

The other direction just follows this proof backwards.

Let $f: X \rightarrow Y$ be a function and $ A \subseteq Y$ and $B \subseteq Y$. Prove that $f^{-1}(A)\setminus f^{-1}(B)= f^{-1}(A\setminus B)$.

1 Answers1

Linked