Inverse function being onto.

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Please refer to this follow up on a question.

As you can see from that link:

$\Phi=f^{-1}\circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ now in my textbook the next theorem following what I posted in the link, is as follows:

Theorem 2 Let $f:[a,b]\subseteq R\to R^n$ and $h:[c,d]\subseteq R\to R^n$ be 1-1 parametrizations of a simple arc C in $R^n$ Then there exists a unique function $\Phi$ from [a,b] onto [c,d] such that $h=f\circ\Phi$. Moreover, $\Phi$ is continuous and strictly monotonic.

Now I am getting confused as the proof provided for theorem 2 is the theorem provided in the link but in the follow up question $\Phi$ maps [c,d] onto [a,b] and not opposite. What am I missing?

To be more specific the proof states:

By Theorem 1 (link) the function $\Phi$ is 1-1, continuous and strictly monotonic.