Corollary 2

Let E, F be two well-ordered sets. If there exists an isomorphism $f$ of E onto a segment T of F and an isomorphism $g$ of F onto a segment S of E, then we must have $S=E$, $T=F$, and $g,f$ are inverses of each other.

For $g\circ f$ is an isomorphism of E onto the segment $g(T)\subset S$ of E; by Corollary 1 we have $g(T)=S=E$, and $g\circ f$ is the identity mapping of E. Similarly, $f\circ g$ is the identity mapping of F, whence the result.