## Description

Let A and B be nonempty sets, and consider a function f : A → B. For any subset C⊆A,wedefinetheimageofC underf tobetheset

f(C) = {f(x) | x ∈ C} = {y ∈ B | y = f(x) for some x ∈ C}.

Similarly, for any subset D ⊆ B, we define the inverse image or pre-image of D under f to

be the set

f−1(D) = {x ∈ A | f(x) ∈ D}.

(Note that in the definition of f−1(D), we do not have to assume that f is invertible, i.e., that f−1 exists as a function.)

Let A and B be sets, and let f : A → B. Prove the following statements which refer to the above defintions.

(a) If E and F are subsets of A, then

f(E ∪ F) = f(E) ∪ f(F).

(b) If f is injective and E and F are subsets of A, then f(E ∩ F) = f(E) ∩ f(F).

(c) If f is surjective and if D is a subset of B, then f(f−1(D)) = D.

## Explanation & Answer

The proofs are ready and I am sure they satisfy all your requirements. But of course I am ready to give more explanations or make some corrections.docx and pdf file are identical

Let 𝐴 and 𝐵 be sets, and let 𝑓: 𝐴 → 𝐵. Prove the following statements which refer to the

above definitions.

(a) If 𝐸 and 𝐹 are subsets of 𝐴, then 𝑓(𝐸 ∪ 𝐹) = 𝑓(𝐸) ∪ 𝑓(𝐹).

(b) If 𝑓 is injective and 𝐸 and 𝐹 are subsets of 𝐴, then 𝑓(𝐸 ∩ 𝐹) = 𝑓(𝐸) ∩ 𝑓(𝐹).

(c) If 𝑓 is surjective and if 𝐷 is a subset of 𝐵, then 𝑓(𝑓 −1 (𝐷)) = 𝐷.

Proof.

To prove that some sets 𝑃 and 𝑄 are equal, it is sufficient to prove that 𝑃 ⊂ 𝑄 and 𝑄 ⊂ 𝑃,

which in turn means that 𝑥 ∈ 𝑃 ⇒ 𝑥 ∈ 𝑄 and 𝑥 ∈ 𝑄 ⇒ 𝑥 ∈ 𝑃.

(a) Prove that 𝑓(𝐸 ∪ 𝐹) ⊂ 𝑓(𝐸) ∪ 𝑓(𝐹) and 𝑓(𝐸 ∪ 𝐹) ⊃ 𝑓(𝐸) ∪ 𝑓(𝐹).

(i) If 𝑥 ∈ 𝑓(𝐸 ∪ 𝐹) then there is some 𝑦 ∈ 𝐸 ∪ 𝐹: 𝑥 = 𝑓(𝑦).

Because 𝑦 ∈ 𝐸 ∪ 𝐹, 𝑦 ∈ 𝐸 or 𝑦 ∈ 𝐹. In the first case, 𝑥 ∈ 𝑓(𝐸),...