The Chung–Feller theorem: for each $k$ with $0\le k\le n$, exactly $C_n$ of the diagonal paths from $(0,0)$ to $(2n,0)$ have exactly $2k$ steps lying above level $0$

A step $i$ of a diagonal path joins the two heights $h(i-1)$ and $h(i)$, which differ by exactly $1$; writing $c$ for the smaller of them, the step lies above level $0$ exactly when $c\ge 0$, hence below level $0$ exactly when $c\le -1$. For an up step $c=h(i-1)$ and for a down step $c=h(i)$, so the steps below level $0$ are the up steps starting at a height $\le -1$ together with the down steps ending at a height $\le -1$, and these two families are disjoint because a step cannot be both up and down.

Let $\beta (q)$ be the number of up steps of $q\in \mathcal{W}((0,0),(2n,0))$ starting at a height $\le -1$. The map $\mathrm{\Delta }$ sending $w\in U$ to the diagonal path of length $2n$ from $(0,0)$ whose step word is $w_1{w}_2\cdots w_{2n}$, with $1$ read as an up step and $-1$ as a down step, is a bijection $U\to \mathcal{W}((0,0),(2n,0))$: the inverse prepends the entry $1$, and by [F1] and [L8] a word of length $2n$ over the two letters with exactly $n$ up letters is the step word of exactly one diagonal path of length $2n$ from $(0,0)$, whose height at the last index is $0$. Moreover $\kappa (w)=1+\beta (\mathrm{\Delta }(w))$: writing $H(i)={\sum }_{i^{\mathrm{\prime }}<i}{w}_{i^{\mathrm{\prime }}}$, the path $q=\mathrm{\Delta }(w)$ has $h(i)=H(i+1)-1$ for $0\le i\le 2n$ and its step $i$ carries the letter $w_i$, so the position $i=0$ always contributes to $\kappa (w)$ because $w_0=1$ and $H(0)=0$, while a position $i\ge 1$ with $w_i=1$ contributes exactly when $H(i)\le 0$, that is exactly when $h(i-1)\le -1$. Finally $U$ is finite with $\mid U\mid =\left(\genfrac{}{}{0px}{}{2n}{n}\right)$, since deleting the entry at the position $0$ is a bijection from $U$ onto the words of length $2n$ over $\{1,-1\}$ with exactly $n$ entries $-1$, and those correspond by [L7], [L9] and [L12] to the $n$-element subsets of a $2n$-element set.

Fix $a\in W$ and let $t_0<t_1<\cdots <t_n$ be the positions of $a$ carrying the entry $1$, extended to all integers by $t_{l+n+1}:=t_l+(2n+1)$. Put $b_l:=S_a(t_{l+1})-S_a(t_l)$. Then $b_{l+n+1}=b_l$, because $S_a(t_l+2n+1)=S_a(t_l)+\parallel a\parallel$ by [F3], so $b$ is determined by $b_0,\dots ,b_n$ and is a word of length $n+1$ of integers; its weight is $\parallel b\parallel =S_a(t_{n+1})-S_a(t_0)=\parallel a\parallel$, which is $1$ because $a$ has $n+1$ entries $1$ and $n$ entries $-1$. The one-step difference identity of [F3] for $b$ reads $S_b(l)-S_b(l-1)=b_{l-1}$, so induction on $r$ gives $S_b(l+r)-S_b(l)=S_a(t_{l+r})-S_a(t_l)$ for every $l\in \mathbb{Z}$ and every $r\in \mathbb{N}$.

For $q\in \mathcal{W}((0,0),(2n,0))$ the steps below level $0$ number $2\beta (q)$, so the steps above level $0$ number $2n-2\beta (q)$ with $0\le \beta (q)\le n$. Send a down step $i$ with $c:=h(i)\le -1$ to $\psi (i):=$ the least index $i^{\mathrm{\prime }}>i$ with $h(i^{\mathrm{\prime }})\ge c+1$, which exists by [L15] because $h(2n)=0\ge c+1$; then $h(\psi (i)-1)\le c$ and $h(\psi (i))\ge c+1$ differ by $1$, so $h(\psi (i)-1)=c$ and $h(\psi (i))=c+1$ and $\psi (i)$ is an up step starting at height $c\le -1$. The map is injective: if $\psi (i_1)=\psi (i_2)=i^{\mathrm{\prime }}$ then $h(i_1)=h(i^{\mathrm{\prime }}-1)=h(i_2)$, and if moreover $i_1<i_2$ then $i_2-1>i_1$ with $h(i_2-1)=c+1$, so $\psi (i_1)\le i_2-1<\psi (i_2)$, a contradiction. It is surjective: given an up step $i^{\mathrm{\prime }}$ with $c:=h(i^{\mathrm{\prime }}-1)\le -1$, the set of $j\le i^{\mathrm{\prime }}-1$ with $h(j)\ge c+1$ contains $0$ and is bounded above, so by [L14] it has a greatest element $j_0$; then $h(j_0)=c+1$ and $h(j_0+1)=c$, so $i:=j_0+1$ is a down step with $h(i)=c\le -1$, every index strictly between $i$ and $i^{\mathrm{\prime }}$ has height $\le c$, and $\psi (i)=i^{\mathrm{\prime }}$. So the two families of step 1.1 are equinumerous by [L12], and [L10] adds them; the number of up steps of $q$ is $n$ by [F1] since $h(2n)=0$, so $\beta (q)\le n$.

The members of the orbit of $a$ that lie in $U$ are exactly the $n+1$ pairwise distinct words ${\sigma }^{t_l}a$ with $0\le l\le n$, and $\kappa ({\sigma }^{t_l}a)$ takes each of the values $1,\dots ,n+1$ for exactly one such $l$. Since $\parallel a\parallel =1$ and $1$ is coprime to $2n+1$ by [L18], [L2] makes the stabiliser trivial, so the $2n+1$ words ${\sigma }^{j}a$ with $0\le j\le 2n$ are pairwise distinct and form the orbit; the entry of ${\sigma }^{j}a$ at the position $0$ is $a_{j<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>(2n+1)}$, which is $1$ exactly for $j\in \{t_0,\dots ,t_n\}$. For such a $j=t_l$, [L5] gives ${\sum }_{i^{\mathrm{\prime }}<i}({\sigma }^{t_l}a{)}_{i^{\mathrm{\prime }}}=S_a(t_l+i)-S_a(t_l)$, and the positions $i$ with $0\le i\le 2n$ at which ${\sigma }^{t_l}a$ carries the entry $1$ are exactly those with $t_l+i=t_{l+r}$ for some $r$ with $0\le r\le n$, because $t_l<t_{l+1}<\cdots <t_{l+n}<t_l+2n+1$ and the integers whose residue carries the entry $1$ are exactly the $t_j$. Hence $\kappa ({\sigma }^{t_l}a)=\mid \{r:0\le r<n+1,S_a(t_{l+r})\le S_a(t_l)\}\mid$, which by step 1.3 is $\mid \{r:0\le r<n+1,S_b(l+r)\le S_b(l)\}\mid$; and [L1] applied to $b$, of length $n+1$ and weight $1$, says that this is a bijection from $\{0,\dots ,n\}$ onto $\{1,\dots ,n+1\}$.

For $1\le k^{\mathrm{\prime }}\le n+1$ put $I_{k^{\mathrm{\prime }}}:=\{w\in U:\kappa (w)=k^{\mathrm{\prime }}\}$. By step 2.2, $\kappa$ takes values in $\{1,\dots ,n+1\}$ on $U$ and each orbit meets each $I_{k^{\mathrm{\prime }}}$ in exactly one word, so $U$ is the union of the pairwise disjoint sets $I_1,\dots ,I_{n+1}$, and for each pair $k^{\mathrm{\prime }},k^{\mathrm{\prime }\mathrm{\prime }}$ the rule sending $w\in I_{k^{\mathrm{\prime }}}$ to the unique member of its orbit lying in $I_{k^{\mathrm{\prime }\mathrm{\prime }}}$ is a bijection, the orbits being the classes of an equivalence relation by [L4]. Hence all $n+1$ sets have the same cardinality, and by [L10], [L11], [L12] and [L13] we get $(n+1)\mid I_1\mid =\mid U\mid =\left(\genfrac{}{}{0px}{}{2n}{n}\right)$, which is $(n+1)C_n$ by [L6]; cancelling the nonzero factor $n+1$ by [L17] gives $\mid I_{k^{\mathrm{\prime }}}\mid =C_n$ for every $k^{\mathrm{\prime }}$.

By step 2.1 a path $q\in \mathcal{W}((0,0),(2n,0))$ has $2n-2\beta (q)$ steps above level $0$, an even number, and $0\le \beta (q)\le n$, so the count is $2k$ for exactly one $k$ with $0\le k\le n$, namely $k=n-\beta (q)$. By step 1.2 the bijection $\mathrm{\Delta }$ carries $\{w\in U:\kappa (w)=n-k+1\}$ onto ${\mathcal{F}}_{n,k}$, since $\kappa (w)=1+\beta (\mathrm{\Delta }(w))$ and $\beta =n-k$; and by step 3.1 that set has exactly $C_n$ elements, so $\mid {\mathcal{F}}_{n,k}\mid =C_n$ by [L12]. At $n=1$ the two paths from $(0,0)$ to $(2,0)$ have height sequences $0,1,0$ and $0,-1,0$, with two steps above level $0$ and none respectively, so each of $k=1$ and $k=0$ is realised once, and $C_1=1$.