Block interchanges transport arbitrary braid boxes

Strand placements and empty blocks. The assignments ${\sigma }_i\mapsto {\sigma }_i$ from $B_p$ and ${\sigma }_j\mapsto {\sigma }_{p+j}$ from $B_q$ preserve every defining relation, so they define homomorphisms into $B_{p+q}$. Their images commute: the closest possible generator indices are $p-1$ and $p+1$, whose difference is two. This defines $\alpha \otimes \beta$ independently of word representatives. If $p=0$ or $q=0$, every row of $Q_{p,q}$ is empty or there are no rows; $Q_{p,q}=1$, and both identities reduce to the same braid on the nonempty block. Hence assume $p,q\ge 1$.

The descending-row identity. Put $D_{n,l}={\sigma }_n{\sigma }_{n-1}\cdots {\sigma }_l$. For $l\le k<n$, commute a leading ${\sigma }_k$ past ${\sigma }_n,\dots ,{\sigma }_{k+2}$, replace ${\sigma }_k{\sigma }_{k+1}{\sigma }_k$ by ${\sigma }_{k+1}{\sigma }_k{\sigma }_{k+1}$, and commute the last ${\sigma }_{k+1}$ past ${\sigma }_{k-1},\dots ,{\sigma }_l$. Every latter index differs from $k+1$ by at least two. The resulting word is $D_{n,l}{\sigma }_{k+1}$. Thus ${\sigma }_k{D}_{n,l}=D_{n,l}{\sigma }_{k+1}$. When $n=k+1$ the initial commuting segment is empty; when $k=l$ the final commuting segment is empty. Both endpoint cases therefore use the same braid relation.

Generators of the first block. For $1\le i<p$, push ${\sigma }_i$ through the rows of $Q_{p,q}$ using step 1.2. At row $j$ its index is $k=i+j-1$, with $j\le k<p+j-1$, exactly the required range for $D_{p+j-1,j}$; after that row the index is $i+j$. After all $q$ rows, ${\sigma }_i{Q}_{p,q}=Q_{p,q}{\sigma }_{q+i}$. Multiplying this equality by the appropriate inverses gives ${\sigma }_i^{-1}{Q}_{p,q}=Q_{p,q}{\sigma }_{q+i}^{-1}$ as well. For $p=1$ there are no first-block generators to check.

Generators of the second block. In $B_N$, $N=p+q$, index reflection ${\sigma }_i\mapsto {\sigma }_{N-i}$ preserves the Artin relations. It sends $Q_{p,q}$ to $Q_{q,p}$: the reflected word is the product of the grid entries ${\sigma }_{q-i+j}$ first in increasing $i=1,\dots ,q$, then increasing $j=1,\dots ,p$, whereas $Q_{q,p}$ orders the same grid first by $j$, then by $i$. To transpose these orders, only pairs with $i<i^{\mathrm{\prime }}$ and $j>j^{\mathrm{\prime }}$ must change order. Their indices differ by $(i^{\mathrm{\prime }}-i)+(j-j^{\mathrm{\prime }})\ge 2$, so every such swap is a far commutation. Apply step 2.1 to $Q_{q,p}$ and its first-block generator ${\sigma }_{q-j}$, $1\le j<q$, then reflect back: this gives ${\sigma }_{p+j}{Q}_{p,q}=Q_{p,q}{\sigma }_j$, and the same formula for inverse generators. For $q=1$ this verification is vacuous.

Arbitrary boxes and inverse crossings. Apply steps 2.1 and 3.1 successively to any words for $\alpha$ and $\beta$, including inverse letters. They give $(\alpha \otimes \beta )Q_{p,q}=Q_{p,q}(\beta \otimes \alpha )$ with the indicated shifted embeddings. Multiplying by $Q_{p,q}^{-1}$ on both sides gives the asserted inverse identity. The displayed equations use algebraic word order: the rightmost factor runs first geometrically. Thus $Q_{p,q}$ physically takes ordered input blocks $(q,p)$ to output blocks $(p,q)$, and $Q_{p,q}^{-1}$ takes physical input $(p,q)$ to output $(q,p)$. Each strand of one block crosses each of the other once with uniform sign, and order inside either block is preserved. A chronological record is obtained by reversing the actual word; when chronological input is written $(p,q)$, the negative chronological interchange is $Q_{q,p}^{-1}$, whose actual geometric word is its reversal. These are distinct reading conventions, not a reflection or change of generator sign. The two algebraic identities already proved transport the boxes at their specified input/output frames. Thus both signs transport arbitrary boxes, with all zero-width and one-width cases covered in steps 1.1, 2.1 and 3.1. No closure equivalence or Markov theorem is used.