How can I show that the unique fixed point of a composition of continuous functions implies another unique fixed point?

Question

Consider the continuous functions $f : \mathbb R \to \mathbb R$ and $g : \mathbb R \to \mathbb R$. Is it possible to show that, if there exists a unique $x_0 \in \mathbb R$ such that $x_0 = f(g(x_0))$, then there exists a unique $y_0 \in \mathbb R$ such that $y_0 = g(f(y_0))$?

The proof of the existence of a fixed point $y_0$ of $g \circ f$ is straightforward. Because $x_0 = f(g(x_0))$, then $g(x_0) = g(f(g(x_0))$. Then, we let $y_0 = g(x_0)$, such that $y_0 = g(f(y_0))$, as desired. However, I can't yet prove the uniqueness of $y_0$, so I would appreciate any suggestions.

Answer 1

This is basically what John Hughes said in the comments. If $y_0, y_1$ were two fixed points of $g\circ f$ then your argument would give two fixed points of $f \circ g$, given by $$ x_0 = f(y_0), x_1 = f(y_1). $$ If $f \circ g$ has a unique fixed point then $x_0 = x_1$ but then $$ y_0 = gf(y_0) = gf(y_1) = y_1 $$

Answer 2

To complement @Holmes' helpful answer, here is a complete proof that generalizes the statement in the question.

Consider the (not necessarily continuous) functions $f : X_f \to Y_f$ and $g : X_g \to Y_g$, where $X_f,X_g,Y_f,$ and $Y_g$ are arbitrary sets. Consider the function composition $f \circ g : X_g \to Y_f$, and suppose that there exists a unique fixed point $x_0 \in X_g$ of $f \circ g$, such that $x_0 = f(g(x_0))$. We want to show that there exists a unique fixed point $y_0 \in X_f$ of the function composition $g \circ f : X_f \to Y_g$, such that $y_0 = g(f(y_0))$.

First, we show that $y_0$ exists. Because $x_0 = f(g(x_0))$, then $g(x_0) = g(f(g(x_0)))$ (this stems from the univalence (a.k.a "right-uniqueness") property of functions). Then, let $y_0 = g(x_0)$, such that $y_0 = g(f(y_0))$, as desired.

Next, we show that $y_0$ is unique. Suppose that there exists $y_1 \in X_f$ such that $y_1 \neq y_0$ and $y_1 = g(f(y_1))$. Because $y_1 = g(f(y_1))$, then $f(y_1) = f(g(f(y_1)))$, such that $f(y_1)$ is a fixed point of $f \circ g$. Similarly, because $y_0 = g(f(y_0))$, then $f(y_0)$ is also a fixed point of $f \circ g$. However, by assumption, $f \circ g$ has a unique fixed point $x_0$, and so it must be the case that $f(y_1) = f(y_0)$, which implies that $g(f(y_1)) = g(f(y_0))$. Finally, because $y_0$ and $y_1$ are both fixed points of $g \circ f$, then $y_1 = y_0$, as desired.

Additionally, by symmetry, we can repeat a variation of the proof above to prove the converse statement: if there exists a unique fixed point $y_0 \in X_f$ of $g \circ f$, such that $y_0 = g(f(y_0))$, then there exists a unique fixed point $x_0 \in X_g$ of $f \circ g$, such that $x_0 = f(g(x_0))$.

In conclusion, there exists a unique fixed point $x_0 \in X_g$ of $f \circ g$ if and only if there exists a unique fixed point $y_0 \in X_f$ of $g \circ f$.