Bijection has Left and Right Inverse/Proof 2

Proof
Suppose $f$ is a bijection.

From Bijection iff Inverse is Bijection and Composite of Bijection with Inverse is Identity Mapping, it is shown that the inverse mapping $f^{-1}$ such that:
 * $f^{-1} \circ f = I_S$
 * $f \circ f^{-1} = I_T$

is a bijection.