There's a good function going the other way, \\(f^{\ast}: PY \rightarrow PX\\), the **preimage** function, defined by

$$f^{\ast}(S \in PY) = \\{x \in X: f(x) \in S\\} .$$

Claim: this is right adjoint to the image function \\(f_{\ast}: PX \rightarrow PY\\).

Proof: \\(f_{\ast}(S) \subseteq T\\) means that \\(S\\) maps into \\(T\\), which means that \\(S\\) is included in the preimage of \\(T\\), i.e., \\(S \subseteq f^{\ast}(T)\\).

$$f^{\ast}(S \in PY) = \\{x \in X: f(x) \in S\\} .$$

Claim: this is right adjoint to the image function \\(f_{\ast}: PX \rightarrow PY\\).

Proof: \\(f_{\ast}(S) \subseteq T\\) means that \\(S\\) maps into \\(T\\), which means that \\(S\\) is included in the preimage of \\(T\\), i.e., \\(S \subseteq f^{\ast}(T)\\).