Hello! The following are the latex commands of a note. Can U prove that for a given f, if there is some t>0 such that Z^t(f)=f, then t=2?The commands are:\documentclass[leqno]{article}
\usepackage{amsmath,amssymb,amsthm,mathtools}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}

\title{On Counting and Involutions on Posets}
\author{Evelyn Chieh}
\date{\today}

\begin{document}
\maketitle

Let $P$ be a finite poset of $n$ elements. 
For each $a\in P$, denote by $\iota_P(a)$ the number of elements in $P$ that are less than $a$.
When the contex is clear, we will simply write $\iota$ instead of $\iota_P$. 
For each order-preserving map $f$ from $P$ to the chain $[n]\coloneq\{0,1,\ldots,n\}$, 
let $Zf(a)$ be the number of elements in $P$ whose values under $f$ are less than or equal to $\iota(a)$.  
Then
\begin{equation}
    Zf(a)=\sum_{i\leq\iota(a)}\left\vert f^{-1}(i)\right\vert,
\end{equation}
where $\vert f^{-1}(i)\vert$ is the size of the preimage of $i$ under $f$.    

For each order-preserving map $f$, denote by $\partial f$ the M\"{o}bius inversion of $f$. 
Since $Zf$ is also an order-preserving map, for all $a\in P$ we have
\begin{equation}
    Zf(a)=\sum_{b\leq a}\partial Zf(b).
\end{equation}

Let $C_n$ denote a chain of $n$ elements. 
Since $\iota$ is an isomorphism between $C_n$ and $[n-1]$, for each $a\in C_n$, we have the following results.

\begin{lemma}
    For every order-preserving map $f:C_n\to [n]$,
    $$
    \partial Zf(a)=\left\vert f^{-1}(\iota(a))\right\vert
    $$
    for all $a\in C_n$.
\end{lemma}

We now study the structure of $\mathrm{Hom}_{\mathsf{Pos}}\left(P,[n]\right)$ the set of all order-preserving maps from $P$ to $[n]$.

\begin{lemma}
   Let $P$ be a finite poset of $n$ elements. 
   For $f$, $g\in\mathrm{Hom}_{\mathsf{Pos}}\left(P,[n]\right)$, 
   define $f\leq g$ if $f(a)\leq g(a)$ for all $a\in P$. 
   If $f\leq g$ then $Zf\geq Zg$. 
\end{lemma}

\begin{proof}
    Suppose that $f\leq g$. 
    If $g(x)=k\in [n]$, then $f(x)\leq k$. 
    Which means that $g^{-1}(k)\subset\bigcup_{i\leq k}f^{-1}(i)$. 
    Thus
    \begin{align*}
        Zg(a)&=\sum_{i\leq\iota(a)}\left\vert g^{-1}(i)\right\vert \\
        &=\left\vert\bigcup_{i\leq\iota(a)}g^{-1}(i)\right\vert \\
        &\leq\left\vert\bigcup_{i\leq\iota(a)}\bigcup_{j\leq i}f^{-1}(i)\right\vert \\
        &=\left\vert\bigcup_{i\leq\iota(a)}f^{-1}(i)\right\vert \\
        &=Zf(a)
    \end{align*}
    for all $a\in P$. 
\end{proof}

Consider the set $\underline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right)$ consist of all order-preserving maps from $P$ to $[n]$ whose preimages of $0$ are all nonempty, 
and let $\overline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right)$ be the complement of it in $\mathrm{Hom}_{\mathsf{Pos}}\left(P,[n]\right)$. 
Its not hard to show that 
$$
Z\left(\underline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right)\right)\subset\overline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right)
$$
and
$$
Z\left(\overline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right)\right)\subset\underline{\mathrm{Hom}}_{\mathsf{Pos}}\left(P,[n]\right).
$$

Thus for any $f\in\mathrm{Hom}_{\mathsf{Pos}}\left(P,[n]\right)$, 
if $t>0$ is the least integer such that $Z^{t}(f)=f$, then $t$ must be even. 

We conjecture that for any finite poset $P$, there exists an $N\in\mathbb{N}$ such that $Z^{r+2}(f)=Z^r(f)$ whenever $r>N$.



\end{document}