Proof with functions and inverse - Spivak 



How does he know that $f^{-1}$ is one-one? Doesn't he have to prove that? Or is he applying his first theorem in the chapter to $f$? That is $f$ is a function if and only if $f^{-1}$ is one-one?
EDIT: Probably a very stupid question, but can someone actually show me the case for $-f$? Or is Spivak simply saying that we can take it for granted (and I believe this because it make sense) and not actually try to prove $-f$ is increasing?
 A: It seems to me you may retrieve the one-to-one property from a similar reasoning as the one used in the proof.
Suppose that $f^{-1}$ is not one-to-one. This means there exist $a < b$ such that $f^{-1}(a) = f^{-1}(b) = k$. Hence $f(k) = f(f^{-1}(a)) = a$ and $f(k) = f(f^{-1}(b)) = b$, that is $a = b$. But this can't be because we supposed $a < b$.
Hence $f^{-1}(a) > f^{-1}(b)$ or $f^{-1}(a) < f^{-1}(b)$, which one? See Spivak's proof.
A: To answer the additional question in your edit: if $f$ is decreasing then $f(b)<f(a)$ whenever $b>a.$  Therefore $-f(b)>-f(a)$ whenever $b>a.$  Hence $-f$ is increasing.  By the previous result, the inverse of $-f$ is therefore increasing.  This implies that $f^{-1}$ is decreasing.
To understand the final step, we need to ask $``$what is the inverse of $-f?"$  If $\eta$ is the negation map, $x\mapsto -x,$ then $-f=\eta\circ f$ and so the inverse of $-f$ is $f^{-1}\circ\eta^{-1}=f^{-1}\circ\eta$ since the negation map is an involution.  So $f^{-1}\circ\eta$ is increasing.  This means that $f^{-1}(\eta(b))>f^{-1}(\eta(a))$ whenever $b>a.$  But this means that $f^{-1}(-b)>f^{-1}(-a)$ whenever $b>a$ and, letting $b'=-b$ and $a'=-a,$ that $f^{-1}(b')>f^{-1}(a')$ whenever $-b'>-a'.$  Consequently $f^{-1}(a')<f^{-1}(b')$ whenever $a'>b'.$  Hence $f^{-1}$ is decreasing.
