The preïmage of the image and image of the preïmage are not necessarily the identity

Given an arbitrary function $𝑓 : 𝑋 \to 𝑌$ , we understand the image $𝑓 ⋆ : P (𝑋) \to P (𝑌)$ and the preïmage $𝑓 ⋆ : P (𝑌) \to P (𝑋)$ . The result of composing these functions together is not necessarily the identity, but rather has the following properties: #m/thm/general

𝐴 \subseteq 𝑓 ⋆ (𝑓 ⋆ (𝐴)) 𝐵 \supseteq 𝑓 ⋆ (𝑓 ⋆ (𝐵))

where $𝐴 \subseteq 𝑋$ and $𝐵 \subseteq 𝑌$ .

Proof

Let $𝑎 \in 𝐴$ . Then $𝑓 (𝑎) \in 𝑓 ⋆ (𝐴)$ , and thus $𝑎 \in 𝑓 ⋆ (𝑓 ⋆ (𝐴))$ . Therefore $𝐴 \subseteq 𝑓 ⋆ (𝑓 ⋆ (𝐴))$ .

Similarly, let $𝑏 \in 𝑓 ⋆ (𝑓 ⋆ (𝐵))$ . It follows that there exists $𝑥 \in 𝑓 ⋆ (𝐵)$ such that $𝑓 (𝑥) = 𝑏$ , whence $𝑏 = 𝑓 (𝑥) \in 𝐵$ .

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