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)\$$.