Parts (a) and (c) are standard. See [GHJ1989, Proposition 2.8.1] and [Bou1968, Chapter IV, Section 1.3, Example 2], respectively. Part (b) is a consequence of (a) and the monoid isomorphism in (1.5).

(d) The right way to think of this is to realize that $A_k$ is defined as a presentation by the generators $d\in A_k$ and the relations which specify the composition of diagrams. To prove the presentation in the statement of the theorem we need to establish that the generators and relations in each of these two presentations can be derived from each other. Thus it is sufficient to show that

1. The generators in (1.10) satisfy the relations in Theorem 1.11.
2. Every set partition $d\in A_k$ can be written as a product of the generators in (1.10).
3. Any product $d_1\circ d_2$ can be computed using the relations in Theorem 1.11.

(1) is established by a direct check using the definition of the multiplication of diagrams. (2) follows from (b) and (c) and the fact (1.6) that $A_k=S_k{P}_k{S}_k\text{.}$ The bulk of the work is in proving (3).

Step 1. First note that the relations in (a)–(d) imply the following relations:

Step 2. Analyze how elements of $P_k$ can be efficiently expressed in terms of the generators.

Let $t\in P_k\text{.}$ The blocks of $t$ partition $\{1,\dots ,k\}$ into top blocks and partition $\{1\prime ,\dots ,k\prime \}$ into bottom blocks. In $t,$ some top blocks are connected to bottom blocks by an edge, but no top block is connected to two bottom blocks, for then by transitivity the two bottom blocks are actually a single block. Draw the diagram of $t,$ such that if a top block connects to a bottom block, then it connects with a single edge joining the leftmost vertices in each block. The element $t\in P_k$ can be decomposed in block form as

with $\tau \in S_k,$ $i_1<i_2<\dots <i_r,$ $j_1<j_2<\dots <j_s,$ ${\ell }_1<{\ell }_2<\dots <{\ell }_m,$ and $r_1<r_2<\dots <r_n\text{.}$ The left product of $p_{i}s$ corresponds to the top blocks of $t,$ the right product of $p_{i}s$ corresponds to the bottom blocks of $t,$ and the permutation $\tau$ corresponds to the propagation pattern of the edges connecting top blocks of $t$ to bottom blocks of $t\text{.}$ For example,

The dashed edges of $\tau$ are “non-propagating” edges, and they may be chosen so that they do not cross each other. The propagating edges of $\tau$ do not cross, since $t$ is planar.

Using the relations (f1) and (f2), the non-propagating edges of $\tau$ can be “removed”, leaving a planar diagram which is written in terms of the generators $p_i$ and $p_{i+\frac{1}{2}}\text{.}$ In our example, this process will replace $\tau$ by $p_{2\frac{1}{2}}{p}_2{p}_{3\frac{1}{2}}{p}_3{p}_{5\frac{1}{2}}{p}_5{p}_{4\frac{1}{2}}{p}_4,$ so that

Step 3. If $t\in P_k$ and ${\sigma }_1\in S_k$ which permutes the top blocks of the planar diagram $t,$ then there is a permutation ${\sigma }_2$ of the bottom blocks of $t$ such that ${\sigma }_{1}t{\sigma }_2$ is planar. Furthermore, this can be accomplished using the relations. For example, suppose

is a planar diagram with top blocks $T_1$ and $T_2$ connected respectively to bottom blocks $B_1$ and $B_2$ and

then transposition of $B_1$ and $B_2$ can be accomplished with the permutation

which is planar. It is possible to accomplish these products using the relations from the statement of the theorem. In our example, with ${\sigma }_1=s_2{s}_1{s}_3{s}_2{s}_4{s}_3{s}_5{s}_2{s}_4{s}_6{s}_1{s}_3{s}_5{s}_2{s}_4{s}_3$ and with ${\sigma }_2=s_4{s}_5{s}_6{s}_3{s}_4{s}_5{s}_2{s}_3{s}_4{s}_1{s}_2{s}_3,$

where $T_2^{\prime }{T}_1^{\prime }=\left(p_{1\frac{1}{2}}{p}_2\right)\left(p_{5\frac{1}{2}}{p}_{6\frac{1}{2}}{p}_6{p}_7\right)$ and $B_2^{\prime }{B}_1^{\prime }=\left(p_2{p}_3{p}_{1\frac{1}{2}}{p}_{2\frac{1}{2}}\right)\left(p_5{p}_6{p}_7{p}_{4\frac{1}{2}}{p}_{5\frac{1}{2}}{p}_{6\frac{1}{2}}\right)\text{.}$

Step 4. Let $t,b\in P_k$ and let $\pi \in S_k\text{.}$ Then $t\pi b=tx\sigma$ where $x\in P_k$ and $\sigma \in S_k,$ and this transformation can be accomplished using the relations in (b), (c), and (d).

Suppose $T$ is a block of bottom dots of $t$ containing more than one dot and which is connected, by edges of $\pi ,$ to two top blocks $B_1$ and $B_2$ of $b\text{.}$ Using Step 3 find permutations ${\gamma }_1,{\gamma }_2\in S_k$ and ${\sigma }_1,{\sigma }_2\in S_k$ such that

are planar diagrams with $T$ as the leftmost bottom block of $t\prime$ and $B_1$ and $B_2$ as the two leftmost top blocks of $b\prime \text{.}$ Then

where $b\prime \prime$ is a planar diagram with fewer top blocks than $b$ has. This is best seen from the following picture, where $t\pi b$ equals

and the last equality uses the relations $p_{i+\frac{1}{2}}=p_{i+\frac{1}{2}}^2$ and the fourth relation in (d) (multiple times). Then $t\pi b={\gamma }_1^{-1}t\prime {\gamma }_2\pi {\sigma }_1^{-1}b\prime \prime {\sigma }_2^{-1}=t\pi \prime b\prime \prime {\sigma }_2^{-1},$ where $\pi \prime =\pi {\sigma }_1^{-1}\text{.}$

By iteration of this process it is sufficient to assume that in proving Step 4 we are analyzing $t\pi b$ where each bottom block of $t$ connects to a single top block of $b\text{.}$ Then, since $\pi$ is a permutation, the bottom blocks of $t$ must have the same sizes as the top blocks of $b$ and $\pi$ is the permutation that permutes the bottom blocks of $t$ to the top blocks of $b\text{.}$ Thus, by Step 1, there is $\sigma \in S_k$ such that $x=\pi b{\sigma }^{-1}$ is planar and

Completion of the proof. If $d_1,d_2\in A_k$ then use the decomposition $A_k=S_k{P}_k{S}_k$ (from (1.6)) to write $d_1$ and $d_2$ in the form

and use (b) and (c) to write these products in terms of the generators. Let $\pi ={\pi }_2{\sigma }_1\text{.}$ Then Step 4 tells us that the relations give $\sigma \in S_k$ and $x\in P_k$ such that

Using Step 2 and that $A_k=S_k{P}_k{S}_k,$ this product can be identified with the product diagram $d_1{d}_2\text{.}$ Thus, the relations are sufficient to compose any two elements of $A_k\text{.}$

$\square$