Solution check for proof of [imath]f(A \cap f^{-1}(B)) \subseteq f(A) \cap B[/imath]

[Ozma]

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

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

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

 

[fresh_42]

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

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

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

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

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

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

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

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.