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:


 * $\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$