The image map of a bijection is a bijection

Given a bijection $𝑓 : 𝑋 \to 𝑌$ , the image map $𝑓 ⋆ : P (𝑋) \to P (𝑌)$ is also a bijection. #m/thm/general Moreover:

(𝑓 ⋆) - 1 = 𝑓 ⋆ = (𝑓 - 1) ⋆

Proof

Let $𝑓 : 𝑋 \to 𝑌$ be a bijection between arbitrary sets, It follows that for a given subset $𝐴 \subseteq 𝑋$

𝑥 \in 𝑓 ⋆ (𝑓 ⋆ (𝐴)) ⟺ 𝑓 (𝑥) \in 𝑓 ⋆ (𝐴) ⟺ \exists 𝑎 \in 𝐴 : 𝑓 (𝑎) = 𝑓 (𝑥) ⟺ \exists 𝑎 \in 𝐴 : 𝑎 = 𝑥 ⟺ 𝑥 \in 𝐴

thus $𝐴 = 𝑓 ⋆ (𝑓 ⋆ (𝐴))$ . Likewise, for a given subset $𝐵 \subseteq 𝐵$

𝑦 \in 𝑓 ⋆ (𝑓 ⋆ (𝐴)) ⟺ \exists 𝑥 \in 𝑓 ⋆ (𝐴) : 𝑓 (𝑥) = 𝑦 ⟺ \exists 𝑥 \in 𝑓 ⋆ (𝐴) : 𝑥 = 𝑓 - 1 (𝑦) ⟺ 𝑓 - 1 (𝑦) \in 𝑓 ⋆ (𝐴) ⟺ 𝑓 (𝑓 - 1 (𝑦)) \in 𝐴 ⟺ 𝑦 \in 𝐴

thus $𝐴 = 𝑓 ⋆ (𝑓 ⋆ (𝐴))$ . Therefore $𝑓 ⋆ \circ 𝑓 ⋆ = 𝑓 ⋆ \circ 𝑓 ⋆ = i d$ , hence $𝑓 ⋆$ is a bijection with inverse $𝑓 ⋆$ .

#state/tidy | #lang/en | #SemBr