the relationship between $f^{-1}(x)$ and $x$ A elementary question here. If I have 
$$f(x)<x\quad\forall\,\,x\in (0,1).$$
Can I deduce $f^{-1}(x)>x$ for $x\ne 0,x\ne 1$?
If yes, could anyone show me how?
 A: Apply your condition $f(y) < y$ to $y:=f^{-1}(x)$, and you immediately get 
$$x = f(f^{-1}(x))<f^{-1}(x)$$
A: Think in terms of the graph of $f$. If you know that $f(x)<x$ for all $x$, then the graph of $f$ lies below the line $y=x$ (because that line is the graph of $g(x)=x$, and you are told that $f(x)<g(x)$).
But the graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$. That means the graph of $f^{-1}$ lies above the line $y=x$, which is equivalent (much as before) to the statement that $f^{-1}(x)>x$.
A: Think of $f$ as a sender. $f(x)$ is the element that this $f$  sends $x$ to. 
Your hypothesis says that the sender $f$ has the shrinking property: it sends any $x$ to something smaller than it. Now $f^{-1}$ is a function that reverse-engineers $f$. Given a $y$ it treats it as an element sent by $f$ and tries to figure out which one was sent to it by $f$. So it hs to be bigger than the $y$: that is $f^{-1}(y)>y$. (Moral: sometimes it is better to use different symbols).
A: the domain of $f$ is $(0,1)$ and the range of $f$ is a subset of $(0,1)$ because of the constraint $f(x) < x \text{ for } 0 < x < 1.$  let $y$ be a point in the range of $f$  and $f(x) = y, x = f^{-1}(y).$ 
now, $$ f(x)< x $$ is equivalent to $$ y < f^{-1}(y) \text{ or } f^{-1}(y) > y.$$  what we have is $$f^{-1}(y) > y \quad \forall y \in range(f)  $$ and $range(f)$ is not necessarily $(0,1).$
