Subwords, reflection deletions and the covers of the longest element in S4

Multiplying $s_1{s}_2{s}_3{s}_1{s}_2{s}_1$ by the position-swapping rule [F2] gives $4321$, whose six pairs are all inversions; thus the word is reduced of length $6$ by [F1] and [F3]. For (i), the word $s_1{s}_3$ is a subword of $s_1{s}_2{s}_3{s}_1{s}_2{s}_1$ at positions $1,3$, and it is reduced because $s_1$ and $s_3$ act on disjoint pairs of positions, so its value $s_1{s}_3$ has one-line form $2143$ and inversion number $2$, equal to its length of $2$ by [F1] and [F2]. The subword criterion [F4] applied to $x=u$ and $v=w_0$ gives $u\le w_0$, which is (i).

For (ii), first note that $S_4=[1,w_0]$: $S_4$ has $24$ elements and is finite, the Bruhat order on it is directed (The lifting property in all four descent cases, the cover criterion, reflection deletion, and directedness (4)), so finitely many pairwise upper bounds combine to a greatest element $z$ with $1\le x\le z$ for every $x\in S_4$; by the strict length increase of [F7] the element $z$ must have maximal length, and the maximum of $\mathrm{inv}$ on $S_4$ is $6$, attained only by the reverse permutation $4321=w_0$; hence $z=w_0$ and every element of $S_4$ satisfies $x\le w_0$, while conversely $x\le w_0$ means $x\in S_4$.

For (iii), multiplying out each deletion and reading the inversion numbers with [F1], [F2] and [F3]: deletion of position $1$, $4$ or $6$ gives $4312$, $4231$ or $3421$, each of length $5$; deletion of position $2$ or $5$ gives $4123$ or $2341$, each of length $3$; and deletion of position $3$ gives $1324$ of length $1$. By [F5] each value equals $w_0{\tau }_i$ for the reflection ${\tau }_i$ conjugated to the letter at position $i$, and each is strictly below $w_0$. Since $\mathrm{\ell }(w_0)=6$, the cover criterion [F5] says a deletion value is covered by $w_0$ exactly when its length is $5$: this holds for $i=1,4,6$ (the deletion words of length $5$ are reduced, since their length equals the inversion number of their value) and fails for $i=2,5,3$, where the deletion words of length $5$ are not reduced. Finally, the elements of $S_4$ of length $5$ are exactly $4312$, $4231$ and $3421$: a permutation has inversion number $5$ if and only if exactly one of the six pairs $i<j$ satisfies $\sigma (i)<\sigma (j)$, and the enumeration of the $24$ permutations confirms that this holds only for those three; hence the covers of $w_0$ are precisely the three elements of length $5$, and no element of length $<5$ can be covered by $w_0$ by [F5].

For (ii), every element of $S_4$ lies below $w_0$ by step 1.2 and therefore occurs as a subword value by [F4]; conversely every subword value belongs to $S_4$. Thus the $64$ position subsets realize exactly $S_4=[1,w_0]$, a set of $24$ elements. By the position-swapping rule [F2], the seven listed position sets evaluate respectively to $2134$, $1243$, $2143$, $1342$, $2341$, $2431$ and $4321$. Evaluating all $64$ subsets also gives exactly the six listed position sets for $2134$: the singletons $\{1\}$, $\{4\}$ and $\{6\}$ carry $s_1$, and $\{1,2,5\}$, $\{1,4,6\}$ and $\{2,5,6\}$ evaluate to $s_1{s}_2{s}_2=s_1$, $s_1^3=s_1$ and $s_2{s}_2{s}_1=s_1$, respectively.

Collecting: (i) is a direct instance of the subword criterion; (ii) shows that the $64$ subwords of one fixed reduced expression of $w_0$ realize exactly the $24$ elements of $S_4=[1,w_0]$, so subwords of one expression may repeat values, and the element $s_1$ arises from six different position sets; (iii) shows that the single-letter deletions of a reduced expression of $w_0$ realize three covers and three shorter elements, realizing the general cover criterion and reflection-deletion statements. All computations are finite and use no choice principle.