The lifting property in all four descent cases, the cover criterion, reflection deletion, and directedness

For (1a), assume $\mathrm{\ell }(vs)<\mathrm{\ell }(v)$ and $\mathrm{\ell }(us)>\mathrm{\ell }(u)$. Let $r:=\mathrm{\ell }(vs)=\mathrm{\ell }(v)-1$ and choose a reduced expression $vs=t_1\cdots t_r$, so that $v=(vs)s=t_1\cdots t_{r}s$ is a word of length $r+1=\mathrm{\ell }(v)$, hence a reduced expression of $v$; put $t_{r+1}:=s$. The subword criterion [F1] applied to $u\le v$ gives a reduced subword expression $u=t_{i_1}\cdots t_{i_k}$ with $1\le i_1<\cdots <i_k\le r+1$ of the word $t_1\cdots t_{r}s$. This subword does not retain position $r+1$: if it did, then $k\ge 1$ and $i_k=r+1$, and $u=u^{\mathrm{\prime }}s$ where $u^{\mathrm{\prime }}$ is the product of the letters at $i_1,\dots ,i_{k-1}$, so $\mathrm{\ell }(u^{\mathrm{\prime }})=k-1=\mathrm{\ell }(u)-1$ and $\mathrm{\ell }(us)=\mathrm{\ell }(u^{\mathrm{\prime }})=\mathrm{\ell }(u)-1$, contradicting $\mathrm{\ell }(us)>\mathrm{\ell }(u)$. Hence all retained positions lie in $\{1,\dots ,r\}$, including when $k=0$, so $u$ is a reduced subword of $t_1\cdots t_r$, a reduced expression of $vs$, and $u\le vs$ by [F1]; moreover $us=t_{i_1}\cdots t_{i_k}s$ is the product of the subword at positions $i_1<\cdots <i_k<r+1$ of the reduced word $t_1\cdots t_{r}s$ of $v$, so $us\le v$ by [F1].

For (1b), assume $\mathrm{\ell }(vs)>\mathrm{\ell }(v)$. Then $s_1\cdots s_{q}s$ has length $q+1=\mathrm{\ell }(vs)$, so it is a reduced expression of $vs$. The reduced subword expression of $u$ inside $s_1\cdots s_q$ given by [F1] is also a subword of $s_1\cdots s_{q}s$, so $u\le vs$; and $us$ is the product of the subword at the positions $i_1<\cdots <i_k<q+1$ of the reduced word $s_1\cdots s_{q}s$ of $vs$, so $us\le vs$.

For (1d), assume $\mathrm{\ell }(vs)>\mathrm{\ell }(v)$ and $\mathrm{\ell }(us)<\mathrm{\ell }(u)$. Then $us\to u$ is a Bruhat edge, because $u=(us)s$ with $s\in S\subseteq T$ and $\mathrm{\ell }(u)>\mathrm{\ell }(us)$; likewise $v\to vs$ is an edge. Hence $us\le u\le v\le vs$, which gives both $us\le v$ and $u\le vs$.

For the criterion (2), prove (i) and (ii) equivalent. If (i) holds and $\mathrm{\ell }(v)>\mathrm{\ell }(u)+1$, then [F3] produces $x_1$ with $u<x_1<v$, so $u$ is not covered by $v$; hence (i) implies (ii). Conversely, if $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$ and $u<x<v$, then $\mathrm{\ell }(u)<\mathrm{\ell }(x)<\mathrm{\ell }(v)$ by the strict length increase of [F2], which is impossible; hence (ii) implies (i).

For (1c), assume $\mathrm{\ell }(vs)<\mathrm{\ell }(v)$ and $\mathrm{\ell }(us)<\mathrm{\ell }(u)$. Then $us\to u$ is a Bruhat edge, so $us\le u\le v$; and $s$ is an ascent of $us$, since $\mathrm{\ell }((us)s)=\mathrm{\ell }(uss)=\mathrm{\ell }(u)>\mathrm{\ell }(us)$. Applying (1a), proved in step 1.1, to the pair $us\le v$ (legitimate: $\mathrm{\ell }(vs)<\mathrm{\ell }(v)$ is the hypothesis and $\mathrm{\ell }((us)s)>\mathrm{\ell }(us)$ was just checked) gives $(us)s\le v$, which is $u\le v$, and $us\le vs$; the extra claim $us\le v$ is the already established relation $us\le u\le v$.

For (3), first compute $v{t}_i=(s_1\cdots s_q)(s_q\cdots s_{i+1})s_i(s_{i+1}\cdots s_q)=s_1\cdots s_{i-1}{s}_{i+1}\cdots s_q=v_i$ by cancelling the tail $s_{i+1}\cdots s_q$ against its inverse and $s_i{s}_i=1$. Hence $v_i$ is the product of a word of length $q-1$, so $\mathrm{\ell }(v_i)\le q-1<q=\mathrm{\ell }(v)$ and $v=v_i{t}_i$ exhibits the edge $v_i\to v$, giving $v_i<v$. Applying the criterion of step 1.4 to the pair $v_i\le v$, the element $v_i$ is covered by $v$ if and only if $\mathrm{\ell }(v)=\mathrm{\ell }(v_i)+1$, i.e. if and only if $\mathrm{\ell }(v_i)=q-1$; and $\mathrm{\ell }(v_i)=q-1$ holds if and only if the word $s_1\cdots \widehat{s_i}\cdots s_q$ is reduced, because that word represents $v_i$ and has length $q-1$.

For the converse part of (3), let $x$ be covered by $v$. By step 1.4, $\mathrm{\ell }(x)=\mathrm{\ell }(v)-1=q-1$, and [F1] exhibits $x$ as the product of a reduced subword of $s_1\cdots s_q$ of length $q-1$, which omits exactly one position $i$; then $x=v_i$ by the definition of $v_i$. For uniqueness, suppose $v_i=v_j$ with $i<j$ and put $W:=s_j{s}_{j-1}\cdots s_{i+1}=(s_{i+1}\cdots s_j{)}^{-1}$ and $Y_j:=s_q\cdots s_{j+1}$, so that $Y_i:=s_q\cdots s_{i+1}=Y_{j}W$; from $v{t}_i=v_i=v_j=v{t}_j$ we get $t_i=t_j$, that is, $Y_{j}W{s}_i{W}^{-1}{Y}_j^{-1}=Y_j{s}_j{Y}_j^{-1}$, hence $W{s}_i{W}^{-1}=s_j$, i.e. $s_i{s}_{i+1}\cdots s_j=s_{i+1}\cdots s_j{s}_j=s_{i+1}\cdots s_{j-1}$. The word $s_i{s}_{i+1}\cdots s_j$ is a subword of the reduced word $s_1\cdots s_q$, hence is reduced of length $j-i+1$: if it admitted a shorter expression, substituting that expression into $s_1\cdots s_q$ would produce a word of length $<q$ for $v$, contradicting $\mathrm{\ell }(v)=q$. But the same element is represented by the word $s_{i+1}\cdots s_{j-1}$ of length $j-i-1$, and $j-i-1<j-i+1$ contradicts the minimality of the length. Hence $i=j$ and the position is unique. This completes (3).

For (4), induct on the natural number $\mathrm{\ell }(u)+\mathrm{\ell }(w)$. If $\mathrm{\ell }(u)=0$ then $u=1\le w$ by [F2], and $z:=w$ satisfies $u\le z$ and $w\le z$. Otherwise $u\mathrm{\ne }1$, and if $u=s_1\cdots s_m$ is a reduced expression with $m\ge 1$ then $s:=s_m$ satisfies $\mathrm{\ell }(us)<\mathrm{\ell }(u)$, since $us=s_1\cdots s_{m-1}$ is represented by a word of length $m-1$. By the induction hypothesis applied to the pair $(us,w)$ there is $x\in W$ with $us\le x$ and $w\le x$. If $\mathrm{\ell }(xs)<\mathrm{\ell }(x)$, apply (1a), proved in step 1.1, to the pair $us\le x$ (legal since $s$ is a descent of $x$ and an ascent of $us$: $\mathrm{\ell }((us)s)=\mathrm{\ell }(uss)=\mathrm{\ell }(u)>\mathrm{\ell }(us)$): it gives $(us)s\le x$, i.e. $u\le x$, so $z:=x$ works with $w\le x$. If $\mathrm{\ell }(xs)>\mathrm{\ell }(x)$, apply (1b), proved in step 1.2, to the pair $us\le x$: it gives $u=(us)s\le xs$; and $x\le xs$ because $x\to xs$ is an edge with $\mathrm{\ell }(xs)>\mathrm{\ell }(x)$, so $w\le x\le xs$ and $z:=xs$ works.

For the criterion (2), prove (i) equivalent to (iii). If (i) holds then by step 1.4 $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$, and writing $q:=\mathrm{\ell }(v)$ and fixing a reduced expression $v=s_1\cdots s_q$, the subword criterion [F1] gives a reduced subword expression of $u$ of length $q-1$, i.e. a single-letter deletion; the element is $v_i$ for the omitted position $i$, and by step 2.2 $v_i=v{t}_i$ with $\mathrm{\ell }(v{t}_i)=\mathrm{\ell }(v_i)=q-1$, which is (iii). Conversely, if $u=vt$ with $t\in T$ and $\mathrm{\ell }(vt)=\mathrm{\ell }(v)-1$, then $v=ut$ with $t\in T$ and $\mathrm{\ell }(v)>\mathrm{\ell }(u)$, so $u\to v$ is an edge and $u<v$ with $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$; by step 1.4 this is (i).

For the criterion (2), combine steps 1.4, 3.1 and 2.2: (i) implies (ii) by step 1.4, (ii) implies (i) by step 1.4, (i) implies (iii) and (iii) implies (i) by step 3.1, and step 2.2 identifies the elements of (iii) with the single-letter deletions of a fixed reduced expression; so (i), (ii) and (iii) are equivalent, which is (2).

Collecting: (1) holds by steps 1.1, 1.2, 1.3, 2.1 and 2.2; (2) holds by steps 1.4 and 4.1, which establish both directions of each equivalence; (3) holds by steps 2.2 and 2.3; and (4) holds by step 2.4. No use of the Axiom of Choice is made: all selections are among finitely many positions of a fixed word or are the unique deleted index of strong exchange, and the induction of step 2.4 runs on natural numbers.