Prove that if $x^*$ is an equilibrium point of $y(n+1)=f^2(y(n))$, then it need not be equilibrium of $x(n+1)=f(x(n))$

Question

This question is related to Problem 13 in section 1.5 in Elaydi's "An Introduction to Difference Equations":

Prove that if $x^*$ is an equilibrium point of $x(n+1)=f(x(n))$, then it is an equilibrium of $y(n+1)=f^2(y(n))$, where $f^2(x)=f(f(x))$. Prove that converse is false in general."

This is the solution:

I don't understand the part where it says: "since $f(x)$ is not monotonic, then there exists $z^*\neq y^*$ such that $f^{-1}(y^*) = z^* \neq y^*$". If $f$ is not monotonic, how do we know that inverse exists and how do we know that there is preimage of $y^*$ other than $y^*$? I cannot figure this out, so I would be grateful for any help. Thanks a lot in advance.

Answer 1

You are correct, this argument is a little strange. If $y^*$ is a fixed point of $g$, then $z^*=f(y^*)$ is already completely determined. There is no way to select a different suitable candidate in the pre-image of $y^*$.

That 2-periodic orbits are possible can be demonstrated just simply by giving an example, like $f(x)=-x^3$ with $y^*=1$ and $z^*=-1$.