About Fixed Point Sets

Question

Let $f,g:X\to X$, and denote by $Fix(g)$ the set of all fixed points of $g$, i.e., $$Fix(g)=\{x\in X: x=g(x)\}.$$ Now suppose $f$ and $g$ commute in the sense that $f\circ g = g \circ f$. Is it true that $$f(Fix(g))=Fix(g) ?$$ Clearly $f(Fix(g)) \subseteq Fix(g)$, but does the reverse inclusion hold, or do we need more assumptions on how $f$ and $g$ are defined?

Answer 1

What you are asking can be reformulated as follows:

Consider pairs $(X,g)$ where $g\colon X\to X$ is some map. Between such pairs, consider maps $f\colon (X,g)\to (X', g')$, i.e. maps $f\colon X\to X'$ such that $f\circ g= g'\circ f$.

To each such pair $(X,g)$ we can associate the set $Fix(g)\subseteq X$ of fixed points. Each map $f\colon (X,g)\to (X',g')$ between such pairs gives rise to a restricted map $f_*\colon Fix(g)\to Fix(g')$ which maps $x\mapsto f(x)$ (this is well defined because if $x\in Fix(g)$, then $g'(f(x))=f(g(x))=f(x)$, i.e. $f(x)\in Fix(g')$).

What you are asking is: Given a map $f\colon (X,g)\to (X,g)$ from a pair $(X,g)$ to itself, when is $f_*$ surjective?

Note, that if we have maps $(X,g)\xrightarrow{f^1}(X',g')\xrightarrow{f^2}(X'',g'')$, then by definition $(f^2\circ f^1)_* = f^2_*\circ f^1_*$ and if we consider $id\colon (X,g)\to (X,g)$ then $id_* = id$. This makes the assignment $(X,g)\mapsto Fix(g)$, $f\mapsto f_*$ into what is called a functor.

It is useful to be able to recognize such functors because they have some nice properties: For instance, if $f\colon (X,g)\to (X',g')$ is an isomorphism, i.e. if there is a map $f^{-1}\colon (X',g')\to (X,g)$ such that $f\circ f^{-1}=id$ and $f^{-1}\circ f=id$; then also $f_*$ is an isomorphism (with $(f_*)^{-1}= (f^{-1})_*$).

In your example, this means that if $f\colon (X,g)\to (X,g)$ is invertible (such as the case of fractional linear transformations), then $f_*$ will also be invertible, hence in particular surjective.

More generally, assume you could find a map $f'\colon (X,g)\to (X,g)$ such that $f\circ f'=id$ (remember that the notation $f'\colon (X,g)\to (X,g)$ means that $f'\colon X\to X$ and $f'\circ g=g\circ f'$). Then we would obtain $f_*\circ f'_*=(f\circ f')_*=id_*=id$ which implies that $f_*$ is surjective (because for every x\in Fix(g) we would have $f_*(f'_*(x))=x$).

It is NOT enough to require that $f\colon X\to X$ is surjective, as following example shows: Let $X=\mathbb N\times \mathbb N$ and $g\colon X\to X$ given by $g(u,x)= (-u,x)$). Now define $f\colon X\to X$ by mapping $(u,x)$ to $(p(u),x)$ if $u\neq 0$ and to $(0, x+1)$ if $u=0$ (where $p(u)=u-1$ if $u>0$ and $p(u)=u+1$ if $u<0$). The function $f\colon X\to X$ is surjective and satisfies $f\circ g=g\circ f$ but the restriction of $f$ to $Fix(g)= \{0\}\times \mathbb N$ is $(0,x)\mapsto (0,x+1)$ which is not surjective.