Fixed points and iterates of an invertible function

Question

Suppose that $g : [0,1] \rightarrow [0,1]$ is a continuous and strictly increasing function such that $g(0)=0$ and $g(1)=1$. Under these hypotheses $g(x)$ has an inverse function $g^{-1} :[0,1] \to [0,1]$ such that $g^{-1}(g(x)) =x$ and $g(g^{-1}(x)) =x$ for all $x \in[0,1]$. How do I show that the fixed point of $g(x)$ is also a fixed point of $g^{-1}(x)$. Also would $g^{-1}(x)$ have the same iterates of $g(x)$?

Answer 1

Suppose $a$ is a fixed point of $g(x)$. Then $g(a)=a$. Applying $g^{-1}$ to both sides we get $g^{-1}(g(a))=g^{-1}(a)$, but the LHS is $a$, so $a$ is a fixed point of $g^{-1}(x)$.

Answer 2

If $g(x)=x$, then $g^{-1}(x)=g^{-1}(g(x))=x$. If $g^{-1}(x)=x$ then $g(x)=g(g^{-1}(x))=x$.

Answer 3

The inverse function $g^{-1}$ "undoes" the action of $g$. A fixed point of $g$ is a point to which $g$ doesn't do anything. How do you undo not-doing-anything? By not doing anything! So fixed points of $g$ must be fixed points of $g^{-1}$.

In general, if $x_0$ is sent to $x_1$ by $g$ then $g^{-1}$ sends $x_1$ back to $x_0$.

Another way to think about this: in this particular case, you can draw the graph of $g$ as a line from one corner of a square to the other that is always going upwards. Then the graph of $g^{-1}$ is what you get when you reflect this square in the diagonal line $y = x$ (basically because if $(x,y)$ is a point on the graph of $g$ then $(y,x)$ is a point on the graph of $g^{-1}$). The points that stay the same are exactly the points that are already on the diagonal, which is exactly the fixed points.