Solution check for proof of f(A∩f−1(B))⊆f(A)∩Bf(A \cap f^{-1}(B)) \subseteq f(A) \cap B

[Ozma]

Problem. Prove or disprove that for $f: X \to Y$, with $A \subseteq X$ and $B \subseteq Y$, we have $f(A \cap f^{-1}(B)) \subseteq f(A) \cap B$.

Can someone check if the following proof is correct, please?

Proof. Let $y \in f(A \cap f^{-1}(B))$ be arbitrary. Hence, there exists $x \in A \cap f^{-1}(B)$ such that $y=f(x)$. This implies that there exists $x \in A$ such that $y=f(x)$ and there exists $x \in f^{-1}(B)$ such that $y=f(x)$. By definition of image and preimage, this implies that $f(x) \in f(A)$ and that $f(x) \in B$. Since $y=f(x)$, it follows $y \in f(A) \cap B$.

 

[fresh_42]

Problem. Prove or disprove that for $f: X \to Y$, with $A \subseteq X$ and $B \subseteq Y$, we have $f(A \cap f^{-1}(B)) \subseteq f(A) \cap B$.

Can someone check if the following proof is correct, please?

Proof. Let $y \in f(A \cap f^{-1}(B))$ be arbitrary. Hence, there exists $x \in A \cap f^{-1}(B)$ such that $y=f(x)$. This implies that

I would drop all the "exists". You have already one certain $x$ in hand. So I would simply write: This implies that ...

there exists $x \in A$ such that $y=f(x)$ and there exists $x \in f^{-1}(B)$ such that $y=f(x)$.

... $f(x) = y \in f(A)$ and $f(x)=y\in f(f^{-1}(B))=f(\{x\,|\,f(x)\in B\})=B,$ i.e. $y\in f(A)\cap B$ what had to be shown.

By definition of image and preimage, this implies that $f(x) \in f(A)$ and that $f(x) \in B$. Since $y=f(x)$, it follows $y \in f(A) \cap B$.

It is important to concentrate on what is given, how things are defined, and what we aim at, in that order. Scientific sound comes automatically.