Preimage of Subset under Composite Mapping/Proof 1

Proof
A mapping is a specific kind of relation.

Hence, Inverse of Composite Relation applies, and it follows that:


 * $\left({g \circ f}\right)^{-1} \left[{S_3'}\right] = \left({f^{-1} \circ g^{-1}}\right) \left[{S_3'}\right]$