# Weak Multiplier Hopf Algebras II: Source and Target Algebras

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

#### 1.1. Content of the Paper

#### 1.2. Conventions and Notations

#### 1.3. Basic References

## 2. Preliminaries on Weak Multiplier Hopf Algebras

**Lemma**

**1.**

**Proof.**

**Remark**

**1.**

- (i)
- First, rewrite the (images of the) canonical maps ${T}_{1}$ and ${T}_{2}$, and of ${T}_{3}$ and ${T}_{4}$ in the regular case, using the Sweedler notation, as$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \Delta \left(a\right)(1\otimes b)=\sum _{\left(a\right)}{a}_{\left(1\right)}\otimes {a}_{\left(2\right)}b\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}(c\otimes 1)\Delta \left(a\right)=\sum _{\left(a\right)}c{a}_{\left(1\right)}\otimes {a}_{\left(2\right)}\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& (1\otimes b)\Delta \left(a\right)=\sum _{\left(a\right)}{a}_{\left(1\right)}\otimes b{a}_{\left(2\right)}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\Delta \left(a\right)(c\otimes 1)=\sum _{\left(a\right)}{a}_{\left(1\right)}c\otimes {a}_{\left(2\right)}\hfill \end{array}$$where $a,b,c\in A$. In all four expressions, ${a}_{\left(1\right)}$ is covered by c and/or ${a}_{\left(2\right)}$ by b. This is by the assumption put on the coproduct, requiring that the canonical maps have range in $A\otimes A$.
- (ii)
- Next, consider the expressions$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}{a}_{\left(1\right)}\otimes S\left({a}_{\left(2\right)}\right)b\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}cS\left({a}_{\left(1\right)}\right)\otimes {a}_{\left(2\right)}\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}{a}_{\left(1\right)}\otimes bS\left({a}_{\left(2\right)}\right)\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}S\left({a}_{\left(1\right)}\right)c\otimes {a}_{\left(2\right)}\hfill \end{array}$$
- (iii)
- If, on the one hand, we first apply S in the first or the second factor of the expressions in (9) and multiply and if, on the other hand, we simply apply multiplication on the expressions in (11), we get the four elements$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}S\left({a}_{\left(1\right)}\right){a}_{\left(2\right)}b\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}c{a}_{\left(1\right)}S\left({a}_{\left(2\right)}\right)\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}{a}_{\left(1\right)}S\left({a}_{\left(2\right)}\right)b\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}cS\left({a}_{\left(1\right)}\right){a}_{\left(2\right)}\hfill \end{array}$$in A for all $a,b,c\in A$. This is used to define the source and target maps in the next section (see Definition 1 in the next section).
- (iv)
- Now, we combine the coverings obtained in (i) and (ii). Consider, e.g., the two expressions$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}\Delta \left({a}_{\left(1\right)}\right)(1\otimes S\left({a}_{\left(2\right)}\right)b)\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}(cS\left({a}_{\left(1\right)}\right)\otimes 1)\Delta \left({a}_{\left(2\right)}\right)\hfill \end{array}$$where $a,b,c\in A$. The first expression (13) is obtained by applying the canonical map ${T}_{1}$ to the first of the two expressions in (11). Thus, this gives an element in $A\otimes A$ and we know that it is $E(a\otimes b)$ as we can see from Formula (1). Similarly, the second expression (13) is obtained by applying the canonical map ${T}_{2}$ to the second of the two expressions in (11). We know that this is $(b\otimes a)E$, as shown in Formula (2). Note that $E(a\otimes b)$ and $(b\otimes a)E$ belong to $A\otimes A$ because by assumption $E\in M(A\otimes A)$, but that, on the other hand, it is not obvious (as we see from the above arguments) that the expressions that we obtain for these elements belong to $A\otimes A$.
- (v)
- Finally, as a consequence of the above statements, also the four expressions$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}S\left({a}_{\left(1\right)}\right){a}_{\left(2\right)}S\left({a}_{\left(3\right)}\right)b\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}c{a}_{\left(1\right)}S\left({a}_{\left(2\right)}\right){a}_{\left(3\right)}\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \sum _{\left(a\right)}cS\left({a}_{\left(1\right)}\right){a}_{\left(2\right)}S\left({a}_{\left(3\right)}\right)\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}{a}_{\left(1\right)}S\left({a}_{\left(2\right)}\right){a}_{\left(3\right)}b\hfill \end{array}$$are well-defined in A for all $a,b,c\in A$ (also in the non-regular case as $S:A\to M\left(A\right)$). This justifies a statement made earlier about the properties of the antipode.

## 3. The Symmetric Pair of Source and Target Algebras

#### 3.1. The Source and Target Algebras B and C

**Definition**

**1.**

**Proposition**

**1.**

**Proof.**

**Proposition**

**2.**

**Proof.**

**Lemma**

**2.**

- (i)
- $\Delta \left(x\right)=(x\otimes 1)E$; and
- (ii)
- $\Delta \left(x\right)=E(x\otimes 1)$.

- (i)
- $\Delta \left(y\right)=E(1\otimes y)$; and
- (ii)
- $\Delta \left(y\right)=(1\otimes y)E$.

**Proof.**

**Notation**

**1.**

**Proposition**

**3.**

**Proof.**

**Proposition**

**4.**

**Proof.**

**Proposition**

**5.**

- (i)
- The sets ${\epsilon}_{s}\left(A\right)$ and ${\epsilon}_{t}\left(A\right)$ are subalgebras.
- (ii)
- The algebra ${\epsilon}_{s}\left(A\right)$ is a right ideal of ${A}_{s}$ and ${\epsilon}_{t}\left(A\right)$ is a left ideal of ${A}_{t}$.

**Notation**

**2.**

**Proposition**

**6.**

**Proof.**

**Corollary**

**1.**

**Lemma**

**3.**

**Proof.**

**Theorem**

**1.**

**Proof.**

#### 3.2. The Antipode on the Source and Target Algebras

**Proposition**

**7.**

- (i)
- If $x,y\in M\left(A\right)$ and $(1\otimes x)E=(y\otimes 1)E$, then $x\in {A}_{t}$ and $y\in {A}_{s}$.
- (ii)
- If $x,y\in M\left(A\right)$ and $E(1\otimes x)=E(y\otimes 1)$, then $x\in {A}_{t}$ and $y\in {A}_{s}$.
- (iii)
- If $x\in {A}_{t}$, then $S\left(x\right)\in {A}_{s}$ and $(1\otimes x)E=\left(S\right(x)\otimes 1)E$.
- (iv)
- If $y\in {A}_{s}$, then $S\left(y\right)\in {A}_{t}$ and $E(y\otimes 1)=E(1\otimes S(y\left)\right)$.

**Proof.**

- (i)
- Assume $x,y\in M\left(A\right)$ and that $(1\otimes x)E=(y\otimes 1)E$. If we apply $\iota \otimes \Delta $ to this equation, we find$$\begin{array}{ccc}\hfill (1\otimes \Delta (x\left)\right)(E\otimes 1)& & =(y\otimes 1\otimes 1)(E\otimes 1)(1\otimes E)\hfill \\ & & =(1\otimes x\otimes 1)(E\otimes 1)(1\otimes E)\hfill \\ & & =(1\otimes x\otimes 1)(1\otimes E)(E\otimes 1).\hfill \end{array}$$Now, we use the property that $(1\otimes a)E=0$ implies that $a=0$ (see Lemma 1 in Section 2). This will eventually give $\Delta \left(x\right)=(x\otimes 1)E$. This proves that $x\in {A}_{t}$. If we apply $\Delta \otimes \iota $ instead, we obtain that $y\in {A}_{s}$.
- (ii)
- The second property is proven in completely the same way.
- (iii)
- Let $x\in {A}_{t}$ so that $\Delta \left(x\right)=E(x\otimes 1)$. Then, for all $a\in A$, we have $\Delta \left(ax\right)=\Delta \left(a\right)(x\otimes 1)$ and so$$\begin{array}{cc}\hfill (1\otimes ax)E\phantom{\rule{1.em}{0ex}}& =\sum _{\left(ax\right)}(S\left({\left(ax\right)}_{\left(1\right)}\right)\otimes 1)\Delta \left({\left(ax\right)}_{\left(2\right)}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(a\right)}(S\left({a}_{\left(1\right)}x\right)\otimes 1)\Delta \left({a}_{\left(2\right)}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(a\right)}(S\left(x\right)S\left({a}_{\left(1\right)}\right)\otimes 1)\Delta \left({a}_{\left(2\right)}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\left(S\right(x)\otimes a)E\hfill \end{array}$$This implies $(1\otimes x)E=\left(S\right(x)\otimes 1)E$. It follows from (i) that $S\left(x\right)\in {A}_{s}$.
- (iv)
- Similarly, we get $S\left(y\right)\in {A}_{t}$ when $y\in {A}_{s}$ and $E(y\otimes 1)=E(1\otimes S(y\left)\right)$.

**Proposition**

**8.**

**Proof.**

- (i)
- Take $x\in C$. Then, $x\in {A}_{t}$ and from Proposition 4 we know that ${\epsilon}_{s}\left(ax\right)=S\left(x\right){\epsilon}_{s}\left(a\right)$ for all a. Because now also ${\epsilon}_{s}\left(aS\left(x\right)\right)={\epsilon}_{s}\left(a\right)S\left(x\right)$ for all a, we see that $S\left(x\right)\in M\left(B\right)$. Similarly, $S\left(y\right)\in M\left(C\right)$ when $y\in B$. It follows that ${S}_{C}$ is an anti-homomorphism from C to $M\left(B\right)$ and that ${S}_{B}$ is an anti-homomorphism of B to $M\left(C\right)$.
- (ii)
- As $BA=A$ and ${\epsilon}_{t}\left(ya\right)={\epsilon}_{t}\left(a\right)S\left(y\right)$ for $y\in B$, we see that $CS\left(B\right)=C$. On the other hand, we have$$A=S\left(A\right)A=S\left(AB\right)A=S\left(B\right)S\left(A\right)A=S\left(B\right)A$$Hence, ${S}_{B}:B\to M\left(C\right)$ and ${S}_{C}:C\to M\left(B\right)$ are non-degenerate anti-homomorphisms.

**Theorem**

**2.**

**Proof.**

#### 3.3. The Canonical Idempotent E as a Separability Idempotent in $M(B\otimes C)$

**Lemma**

**4.**

**Proof.**

**Theorem**

**3.**

**Proof.**

- (i)
- By the lemma, we find that $E(1\otimes a)$ belongs to $B\otimes A$. We therefore can apply ${\epsilon}_{t}$ on the second leg of this expression. We know that the second leg of E belongs to ${\epsilon}_{t}\left(A\right)$ and this is a subalgebra of ${A}_{t}$. In Proposition 4, we show that ${\epsilon}_{t}\left(xa\right)=x{\epsilon}_{t}\left(a\right)$ for all $x\in {A}_{t}$. Therefore, $(\iota \otimes {\epsilon}_{t})\left(E(1\otimes a)\right)=E(1\otimes {\epsilon}_{t}\left(a\right))$. We conclude that $E(1\otimes {\epsilon}_{t}\left(a\right))\in B\otimes C$ for all a and so $E(1\otimes C)\subseteq B\otimes C$.
- (ii)
- We now show that E is full in the sense of Definition 1.1 of [4]. For this, assume that V is a subspace of B so that $E(1\otimes x)\subseteq V\otimes C$ for all $x\in C$. Then, $(1\otimes b)E(1\otimes xa)\in V\otimes A$ for all $a,b\in A$ and $x\in C$. In Proposition 6, we show that $CA=A$ and in Proposition 1 that B is spanned by elements of the form $(\iota \otimes \omega (a\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}}b\left)\right)E$ where $a,b\in A$ and $\omega $ is a linear functional on A. Then, we must have $V=B$ proving that the left leg of E (as an idempotent in $M(B\otimes C)$) is still all of B, and similarly for the right leg. Hence, E is full.
- (iii)
- Finally, we know already from Proposition 8 that the antipode is a non-degenerate anti-homomorphism from B to $M\left(C\right)$ as well as a non-degenerate anti-homomorphism from C to $M\left(B\right)$. As in Proposition 7, they satisfy$$(1\otimes x)E=\left(S\right(x)\otimes 1)E\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{and}\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}E(y\otimes 1)=E(1\otimes S(y\left)\right)$$

**Proposition**

**9.**

**Proof.**

**Proposition**

**10.**

**Proof.**

#### 3.4. Existence of Local Units

**Proposition**

**11.**

**Proof.**

**Remark**

**2.**

- (i)
- As we see from the proof of Lemma 4 and from earlier arguments, we find that $(\iota \otimes \epsilon )\left((1\otimes a)E\right)={\epsilon}_{s}\left(a\right)$ when $a\in A$. The formula makes sense as an equality of left multipliers of A. Note that we do not expect $(1\otimes a)E$ to belong to $B\otimes A$. Similarly, we find $(\epsilon \otimes \iota )\left(E(a\otimes 1)\right)={\epsilon}_{t}\left(a\right)$ for a in A, now as right multipliers of A. Again, we do not expect $E(A\otimes 1)\subseteq A\otimes C$.
- (ii)
- On the other hand, we do have $E(1\otimes A)\subseteq B\otimes A$ and $(A\otimes 1)E\subseteq A\otimes C$, as shown in the lemma. As shown above, if we apply ε on the second leg in the first case and on the first leg in the second case, we get$$(\iota \otimes \epsilon )\left(E(1\otimes a)\right)={\epsilon}_{s}^{\prime}\left(a\right)\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}(\epsilon \otimes \iota )\left((a\otimes 1)E\right)={\epsilon}_{t}^{\prime}\left(a\right)$$
- (iii)
- From the proof of the lemma, we see that the range of ${\epsilon}_{s}^{\prime}$ is the same as the range of ${\epsilon}_{s}$, namely B. Indeed, we have$$\sum _{\left(c\right)}{\epsilon}_{s}\left({c}_{\left(1\right)}\right)\otimes {c}_{\left(2\right)}a{a}^{\prime}=\sum _{\left(a\right)}{\epsilon}_{s}^{\prime}\left({a}_{\left(1\right)}\right)b\otimes {a}_{\left(2\right)}{a}^{\prime}$$
- (iv)
- In the regular case, we get$${\epsilon}_{s}^{\prime}\left(a\right)=\sum _{\left(a\right)}{a}_{\left(2\right)}{S}^{-1}\left({a}_{\left(1\right)}\right)\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}{\epsilon}_{t}^{\prime}\left(a\right)=\sum _{\left(a\right)}{S}^{-1}\left({a}_{\left(2\right)}\right){a}_{\left(1\right)}$$$$S\left({\epsilon}_{t}^{\prime}\left(a\right)\right)={\epsilon}_{t}\left(a\right)\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}and\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}S\left({\epsilon}_{t}^{\prime}\left(a\right)\right){\epsilon}_{s}\left(a\right)$$

## 4. Examples and Special Cases

#### 4.1. The Groupoid Examples

**Example**

**1.**

- (i)
- Consider a groupoid G. First, there is the algebra A, defined as the space $K\left(G\right)$ of complex functions on G with finite support and pointwise product. Recall that the coproduct Δ on $K\left(G\right)$ is defined by$$\Delta \left(f\right)(p,q)=\left\{\begin{array}{cc}f\left(pq\right)\hfill & if\phantom{\rule{4.pt}{0ex}}pq\phantom{\rule{4.pt}{0ex}}is\phantom{\rule{4.pt}{0ex}}defined,\hfill \\ 0\hfill & otherwise.\hfill \end{array}\right.$$The pair $(A,\Delta )$ is a regular weak multiplier Hopf algebra (in the sense of Definitions 1.14 and 4.1 in [6]). The canonical idempotent E in $M(A\otimes A)$ is given by the function on pairs $(p,q)$ in $G\times G$ that is 1 if $pq$ is defined and 0 if this is not the case. The antipode S is defined by $\left(S\left(f\right)\right)\left(p\right)=f\left({p}^{-1}\right)$ whenever $f\in K\left(G\right)$ and $p\in G$.In this example, the algebra ${A}_{s}$ is the algebra of all complex functions on G so that $f\left(p\right)=f\left(q\right)$ whenever $p,q\in G$ satisfy $s\left(p\right)=s\left(q\right)$. It is naturally identified with the algebra of all complex functions on the set ${G}_{0}$ of units in G. The source map ${\epsilon}_{s}$ from A to ${A}_{s}$ is defined by $\left({\epsilon}_{s}\left(f\right)\right)\left(p\right)=f\left({p}^{-1}p\right)$ whenever $p\in G$ and $f\in K\left(G\right)$. The image of the source map ${\epsilon}_{s}\left(A\right)$, what we called in this paper the source algebra, is identified with the algebra of complex functions with finite support on the units. Symmetrically, the algebra ${A}_{t}$ consists of functions f on G so that $f\left(p\right)=f\left(q\right)$ if $t\left(p\right)=t\left(q\right)$ for $p,q\in G$. It is also identified with the space of all complex functions on the units. The target map ${\epsilon}_{t}$ from A to ${A}_{t}$ is defined by $\left({\epsilon}_{t}\left(f\right)\right)\left(p\right)=f\left(p{p}^{-1}\right)$ for all p and $f\in K\left(G\right)$. The target algebra, i.e., the image ${\epsilon}_{t}\left(A\right)$ of the target map, is again identified with the space of functions with finite support on the units. Recall that these two algebras are subalgebras of the multiplier algebra $M\left(A\right)$ (here, the algebra of all complex functions on G). Observe also that the source and target algebras, ${\epsilon}_{s}\left(A\right)$ and ${\epsilon}_{t}\left(A\right)$, can be strictly smaller than the algebras ${A}_{s}$ and ${A}_{t}$, respectively. This happens when the set of units is infinite. In that case, we see that ${A}_{s}$ is indeed the multiplier algebra $M\left({\epsilon}_{s}\left(A\right)\right)$ of ${\epsilon}_{s}\left(A\right)$ and similarly for the target map.
- (ii)
- For the second case, we take the groupoid algebra $\mathbb{C}G$ of G. If we use $p\mapsto {\lambda}_{p}$ for the canonical embedding of G in $\mathbb{C}G$, then, if $p,q\in G$, we have ${\lambda}_{p}{\lambda}_{q}={\lambda}_{pq}$ if $pq$ is defined and 0 otherwise. The coproduct on $\mathbb{C}G$ is given by $\Delta \left({\lambda}_{p}\right)={\lambda}_{p}\otimes {\lambda}_{p}$ for all $p\in G$. The idempotent E is $\sum {\lambda}_{e}\otimes {\lambda}_{e}$ where the sum is only taken over the units e of G. The antipode is given by $S\left({\lambda}_{p}\right)={\lambda}_{{p}^{-1}}$ for all $p\in G$.The symmetric pair of source and target maps is given by ${\epsilon}_{s}\left({\lambda}_{p}\right)={\lambda}_{e}$ where $e=s\left(p\right)$ and ${\epsilon}_{t}\left({\lambda}_{p}\right)={\lambda}_{e}$ where now $e=t\left(p\right)$ for $p\in G$. Here, the source and target algebras coincide and it is the algebra of the span of elements of the form ${\lambda}_{e}$ where e is a unit of G. In addition, here the source and target algebras need not be unital and so can be strictly smaller then their multiplier algebras.

#### 4.2. Examples Associated with Separability Idempotents

**Theorem**

**4.**

**Proof.**

- (i)
- The algebra P is non-degenerate and idempotent because this is true for its components B and C.
- (ii)
- Because $E\in M(B\otimes C)$, we have that ${\Delta}_{P}(c\otimes b)$, defined as $c\otimes E\otimes b$, belongs to $M(P\otimes P)$. Because ${E}^{2}=E$, it is clear that ${\Delta}_{P}$ is a homomorphism. By assumption, we have that $E(1\otimes C)$ and $(B\otimes 1)E$ are subsets of $B\otimes C$. Therefore,$${\Delta}_{P}\left(P\right)({1}_{P}\otimes P)\subseteq P\otimes P\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{and}\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}(P\otimes {1}_{P}){\Delta}_{P}\left(P\right)\subseteq P\otimes P.$$The coproduct ${\Delta}_{P}$ is coassociative and $({\Delta}_{P}\otimes {\iota}_{P}){\Delta}_{P}(c\otimes b)=c\otimes E\otimes E\otimes b$ for all $b\in B$ and $c\in C$. This coproduct is full because E is assumed to be full (as in Definition 1.1 of [4]).
- (iii)
- Now, we prove that there is a counit ${\epsilon}_{P}$ on $(P,{\Delta}_{P})$. First, define ${\epsilon}_{P}(c\otimes b)={\phi}_{C}\left(c{S}_{B}\left(b\right)\right)$. For all $b\in B$ and $c\in C$, we have that$$\begin{array}{cc}\hfill ({\iota}_{P}\otimes {\epsilon}_{P}){\Delta}_{P}(c\otimes b)\phantom{\rule{1.em}{0ex}}& =({\iota}_{P}\otimes {\epsilon}_{P})(c\otimes E\otimes b)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =({\iota}_{P}\otimes {\phi}_{C})(c\otimes E(1\otimes {S}_{B}\left(b\right)))\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =({\iota}_{P}\otimes {\phi}_{C})(c\otimes E(b\otimes 1))=c\otimes b.\hfill \end{array}$$On the other hand, if we define ${\epsilon}_{P}^{\prime}(c\otimes b)={\phi}_{B}\left({S}_{C}\left(c\right)b\right)$, we find similarly$$({\epsilon}_{P}^{\prime}\otimes \iota ){\Delta}_{P}(c\otimes b)=c\otimes b$$
- (iv)
- Take any elements $b,{b}^{\prime}\in B$ and $c,{c}^{\prime}\in C$. Then,$${\Delta}_{P}(c\otimes b)(1\otimes 1\otimes {c}^{\prime}\otimes {b}^{\prime})=(1\otimes E\otimes 1)(c\otimes 1\otimes {c}^{\prime}\otimes b{b}^{\prime}).$$If we replace ${c}^{\prime}$ by elements of the form ${S}_{B}\left({b}^{\prime \prime}\right){c}^{\prime \prime}$, the right hand side will be$$(1\otimes E\otimes 1)(c\otimes {b}^{\prime \prime}\otimes {c}^{\prime \prime}\otimes b{b}^{\prime}).$$Next, we use that B is idempotent and that the map ${S}_{B}$ is non-degenerate. Then, we can conclude from this that ${\Delta}_{P}\left(P\right)({1}_{P}\otimes P)={E}_{P}(P\otimes P)$ with ${E}_{P}=1\otimes E\otimes 1$. Similarly, we find $(P\otimes {1}_{P}){\Delta}_{P}\left(P\right)=(P\otimes P){E}_{P}$ and it follows that ${E}_{P}$ is the canonical idempotent for $(P,{\Delta}_{P})$.It is straightforward to verify that the legs of ${E}_{P}$ commute. Moreover,$$({\iota}_{P}\otimes {\Delta}_{P})\left({E}_{P}\right)=1\otimes E\otimes E\otimes 1$$
- (v)
- We now define ${S}_{P}(c\otimes b)={S}_{B}\left(b\right)\otimes {S}_{C}\left(c\right)$ for all b and c and we show that all the conditions of Theorem 2.9 of [6] are fulfilled. This will complete the proof.We consider the candidate for the generalized inverse ${R}_{1}$ of the canonical map ${T}_{1}$ using this expression for ${S}_{P}$. We get, using formally ${E}_{\left(1\right)}\otimes {E}_{\left(2\right)}$ for E, that$$\begin{array}{cc}\hfill {R}_{1}(c\otimes b\otimes {c}^{\prime}\otimes {b}^{\prime})& =\left(({\iota}_{P}\otimes {S}_{P})(c\otimes E\otimes b)\right)(1\otimes 1\otimes {c}^{\prime}\otimes {b}^{\prime})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =c\otimes {E}_{\left(1\right)}\otimes {S}_{B}\left(b\right){c}^{\prime}\otimes {S}_{C}\left({E}_{\left(2\right)}\right){b}^{\prime}.\hfill \end{array}$$That this maps $P\otimes P$ to itself is a consequence of the property, obtained in Proposition 1.9 of [4], saying that ${E}_{\left(1\right)}\otimes {S}_{C}\left({E}_{\left(2\right)}\right){b}^{\prime}$ is in $B\otimes B$.Using this candidate for the antipode, we can calculate the candidates for the source and target maps ${\epsilon}_{s}^{P}$ and ${\epsilon}_{t}^{P}$. We find$$\begin{array}{cc}\hfill {\epsilon}_{t}^{P}(c\otimes b)& =(c\otimes {E}_{\left(1\right)})({S}_{B}\left(b\right)\otimes {S}_{C}\left({E}_{\left(2\right)}\right))=c{S}_{B}\left(b\right)\otimes 1\hfill \\ \hfill {\epsilon}_{s}^{P}(c\otimes b)& =({S}_{B}\left({E}_{\left(1\right)}\right)\otimes {S}_{C}\left(c\right))({E}_{\left(2\right)}\otimes b)=1\otimes {S}_{C}\left(c\right)b\hfill \end{array}$$Finally, we have to show that$$\sum _{\left(a\right)}{\epsilon}_{t}^{P}\left({a}_{\left(1\right)}\right){a}_{\left(2\right)}=a\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{and}\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(a\right)}{\epsilon}_{s}^{P}\left({a}_{\left(1\right)}\right){S}_{P}\left({a}_{\left(2\right)}\right)={S}_{P}\left(a\right)$$$${\epsilon}_{t}^{P}(c\otimes {E}_{\left(1\right)})({E}_{\left(2\right)}\otimes b)=c{S}_{B}\left({E}_{\left(1\right)}\right){E}_{\left(2\right)}\otimes b=c\otimes b$$$${\epsilon}_{s}^{P}(c\otimes {E}_{\left(1\right)}){S}_{P}({E}_{\left(2\right)}\otimes b)=(1\otimes {S}_{C}\left(c\right){E}_{\left(1\right)})({S}_{B}\left(b\right)\otimes {S}_{C}\left({E}_{2}\right))={S}_{B}\left(b\right)\otimes {S}_{C}\left(c\right).$$Finally, we have to show that ${T}_{1}{R}_{1}(p\otimes {p}^{\prime})={E}_{P}(p\otimes {p}^{\prime}$ for all $p,{p}^{\prime}\in P$ where ${T}_{1}$ is the canonical map $p\otimes {p}^{\prime}\mapsto {\Delta}_{P}\left(p\right)(1\otimes {p}^{\prime})$ and where ${R}_{1}$ it its generalized inverse constructed with the antipode ${S}_{P}$ as above. With $p=c\otimes b$ and ${p}^{\prime}={c}^{\prime}\otimes {b}^{\prime}$ we find$$\begin{array}{cc}\hfill {T}_{1}{R}_{1}(p\otimes {p}^{\prime})& =(c\otimes {E}_{\left(1\right)}\otimes {E}_{\left(2\right)}{S}_{B}\left(b\right)\otimes 1)(1\otimes 1\otimes {c}^{\prime}\otimes {b}^{\prime})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =(c\otimes {E}_{\left(1\right)}b\otimes {E}_{\left(2\right)}\otimes 1)(1\otimes 1\otimes {c}^{\prime}\otimes {b}^{\prime})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =(1\otimes E\otimes 1)(c\otimes b\otimes {c}^{\prime}\otimes {b}^{\prime})\hfill \end{array}$$This proves that the candidate for the antipode satisfies all the requirements needed for Theorem 2.9 of [6] and it completes the proof.

**Theorem**

**5.**

**Proof.**

**Proposition**

**12.**

**Proof.**

#### 4.3. Discrete Quantum Groups

**Proposition**

**13.**

#### 4.4. A Quantization of the Groupoid Associated with a Group Action

**Proposition**

**14.**

- (i)
- B and C commute;
- (ii)
- $bq={\sum}_{\left(q\right)}{q}_{\left(1\right)}(b\u25c3{q}_{\left(2\right)})$ for all $b\in b$ and $q\in Q$; and
- (iii)
- $qc={\sum}_{\left(q\right)}({q}_{\left(1\right)}\u25b9c){q}_{\left(2\right)}$ for all $c\in C$ and $q\in Q$.

**Proposition**

**15.**

**Proof.**

- (i)
- First, it is not hard to show that E and $\Delta \left(q\right)$ for all $q\in Q$ are elements of $M(P\otimes P)$. This is a consequence of the fact that the multiplier algebras of B, C and Q all sit in $M\left(P\right)$ and similarly for tensor products.
- (ii)
- We now show that E and $\Delta \left(q\right)$ commute in $M(P\otimes P)$. Using the Sweedler notation, both for E as before and for $\Delta \left(q\right)$, we get$$\begin{array}{cc}\hfill E\Delta \left(q\right)\phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{E}_{\left(1\right)}{q}_{\left(1\right)}\otimes {E}_{\left(2\right)}{q}_{\left(2\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{q}_{\left(1\right)}({E}_{\left(1\right)}\u25c3{q}_{\left(2\right)})\otimes {E}_{\left(2\right)}{q}_{\left(3\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{q}_{\left(1\right)}{E}_{\left(1\right)}\otimes ({q}_{\left(2\right)}\u25b9{E}_{\left(2\right)}){q}_{\left(3\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{q}_{\left(1\right)}{E}_{\left(1\right)}\otimes {q}_{\left(2\right)}{E}_{\left(2\right)}=\Delta \left(q\right)E.\hfill \end{array}$$In the above calculation, we first have used the commutation rule between B and Q (as the first leg of E is in B), then the relation of the actions of Q on E as in Formula (27) and finally the commutation rule between C and Q (as the second leg of E is in C). Of course, to make things precise, we need to cover at the right places with the right elements. This can be done if we multiply from the left in the first factor with $bp$ and from the right in the second factor with $rc$, where $b\in B$, $c\in C$ and $p,r\in Q$.Then, we can define ${\Delta}_{P}$ on P by Formula (29) in the formulation of the proposition. Using the commutation rules, namely that E is an idempotent, it commutes with elements $\Delta \left(q\right)$ and $\Delta $ is a coproduct on Q, it can be shown that ${\Delta}_{P}$ is a coproduct on P. It is full.It is also clear that E, as sitting in $M(P\otimes P)$, has to be the canonical idempotent for ${\Delta}_{P}$.
- (iii)
- We now prove that there is a counit and that it is given by the formulas in the formulation of the proposition.First, define ${\epsilon}_{P}$ on P by ${\epsilon}_{P}\left(qcb\right)=\epsilon \left(q\right){\phi}_{C}\left(c{S}_{B}\left(b\right)\right)$ for $b,c,q$ in $B,C,Q$, respectively. Observe that we use a different order of the elements $b,c,q$ in this definition. Then, we get for all $b,c,q$ that$$\begin{array}{cc}\hfill ({\iota}_{P}\otimes {\epsilon}_{P}){\Delta}_{P}\left(cqb\right)\phantom{\rule{1.em}{0ex}}& =({\iota}_{P}\otimes {\epsilon}_{P})((c\otimes 1)\Delta \left(q\right)E(1\otimes b))\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}{E}_{\left(1\right)}{\epsilon}_{P}\left({q}_{\left(2\right)}{E}_{\left(2\right)}b\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}{E}_{\left(1\right)}\epsilon \left({q}_{\left(2\right)}\right){\phi}_{C}\left({E}_{\left(2\right)}{S}_{B}\left(b\right)\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =cq{E}_{\left(1\right)}b{\phi}_{C}\left({E}_{\left(2\right)}\right)=cqb.\hfill \end{array}$$If, on the other hand, we define ${\epsilon}_{P}^{\prime}$ on P by the formula ${\epsilon}_{P}^{\prime}\left(cbq\right)={\phi}_{B}\left({S}_{C}\left(c\right)b\right))\epsilon \left(q\right)$, a similar calculation will give then that$$({\epsilon}_{P}^{\prime}\otimes {\iota}_{P}){\Delta}_{P}\left(cqb\right)=cqb$$It then follows from the general theory that ${\epsilon}_{P}^{\prime}={\epsilon}_{P}$ and that this is the counit.In the regular case, we consider after the proof of this proposition, we can give a direct argument for the equality of these two expressions for the counit, as done in the simpler case in Theorem 4 (see the remark after the proof of Theorem 5).This takes care of the counit.
- (iv)
- Let us now look at the antipode and the source and target maps. It is expected that the antipode ${S}_{P}$ must coincide with ${S}_{B},{S}_{C},{S}_{Q}$ on $B,C,Q$, respectively.It can be verified that ${S}_{P}$ defined in this way is an anti-homomorphism from P to $M\left(P\right)$. For this, one has to argue that the definition is compatible with the commutation rules between the component $B,C,Q$. We need to use this further in our calculations.To use Theorem 2.9 of [6] again to prove that $(P,{\Delta}_{P})$ is a weak multiplier Hopf algebra, we first must show that the candidates for the maps ${R}_{1}$ and ${R}_{2}$, constructed with the candidate for the antipode map, $P\otimes P$ to itself. We do this for ${R}_{1}$.We have$$\begin{array}{cc}\hfill {R}_{1}(cqb\otimes {c}^{\prime}{q}^{\prime}{b}^{\prime})\phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{E}_{\left(1\right)}{q}_{\left(1\right)}\otimes {S}_{P}\left({E}_{\left(2\right)}{q}_{\left(2\right)}b\right){c}^{\prime}{q}^{\prime}{b}^{\prime}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{E}_{\left(1\right)}{q}_{\left(1\right)}\otimes {S}_{B}\left(b\right)S\left({q}_{\left(2\right)}\right){S}_{C}\left({E}_{\left(2\right)}\right){c}^{\prime}{q}^{\prime}{b}^{\prime}\hfill \end{array}$$To prove the next conditions, we first calculate the candidates for the counital maps ${\epsilon}_{s}^{P}$ and ${\epsilon}_{S}^{P}$. For all $b,c,q$ we find$$\begin{array}{cc}\hfill {\epsilon}_{s}^{P}\left(cqb\right)\phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{S}_{P}\left(c{E}_{\left(1\right)}{q}_{\left(1\right)}\right){E}_{\left(2\right)}{q}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{S}_{P}\left({E}_{\left(1\right)}c{q}_{\left(1\right)}\right){E}_{\left(2\right)}{q}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}S\left({q}_{\left(1\right)}\right){S}_{C}\left(c\right){S}_{B}\left({E}_{\left(1\right)}\right){E}_{\left(2\right)}{q}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}S\left({q}_{\left(1\right)}\right){S}_{C}\left(c\right){q}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}S\left({q}_{\left(1\right)}\right){q}_{\left(2\right)}({S}_{C}\left(c\right)\u25c3{q}_{\left(3\right)})b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =({S}_{C}\left(c\right)\u25c3q)b.\hfill \end{array}$$In a similar way, we find$${\epsilon}_{t}^{P}\left(cqb\right)=c(q\u25b9{S}_{B}\left(b\right))$$Next, we verify that ${T}_{1}{R}_{1}$ is given by left multiplication by E. For this, it is enough to verify that $E(cqb\otimes 1)=(\iota \otimes {\epsilon}_{t}^{P}){\Delta}_{P}\left(cqb\right)$ for all $b,c,q$. For the left hand side, we have$$\begin{array}{cc}\hfill E(cqb\otimes 1)\phantom{\rule{1.em}{0ex}}& =c{E}_{\left(1\right)}qb\otimes {E}_{\left(2\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}({E}_{\left(1\right)}\u25c3{q}_{\left(2\right)})b\otimes {E}_{\left(2\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}{E}_{\left(1\right)}b\otimes {q}_{\left(2\right)}\u25b9{E}_{\left(2\right)}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}{E}_{\left(1\right)}\otimes {q}_{\left(2\right)}\u25b9\left({E}_{\left(2\right)}{S}_{B}\left(b\right)\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}{E}_{\left(1\right)}\otimes ({q}_{\left(2\right)}\u25b9{E}_{\left(2\right)})({q}_{\left(3\right)}\u25b9{S}_{B}\left(b\right))\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}({E}_{\left(1\right)}\u25c3{q}_{\left(2\right)})\otimes {E}_{\left(2\right)}({q}_{\left(3\right)}\u25b9{S}_{B}\left(b\right))\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{E}_{\left(1\right)}{q}_{\left(1\right)}\otimes {E}_{\left(2\right)}({q}_{\left(2\right)}\u25b9{S}_{B}\left(b\right))\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{E}_{\left(1\right)}c{q}_{\left(1\right)}\otimes {E}_{\left(2\right)}({q}_{\left(2\right)}\u25b9{S}_{B}\left(b\right)).\hfill \end{array}$$We find precisely $(\iota \otimes {\epsilon}_{t}^{P}){\Delta}_{P}\left(cqb\right)$. In a similar way, we find that ${T}_{2}{R}_{2}$ is given by right multiplication with E.
- (v)
- Finally, the only thing left is to show that$$\sum _{\left(p\right)}{p}_{\left(1\right)}{S}_{P}\left({p}_{\left(2\right)}\right){p}_{\left(3\right)}=p\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{and}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\sum _{\left(p\right)}{S}_{P}\left({p}_{\left(1\right)}\right){p}_{\left(2\right)}{S}_{P}\left({p}_{\left(3\right)}\right)={S}_{P}\left(p\right)$$$$\sum _{\left(p\right)}{p}_{\left(1\right)}{S}_{P}\left({p}_{\left(2\right)}\right){p}_{\left(3\right)}=\sum _{\left(p\right)}{\epsilon}_{t}\left({p}_{\left(1\right)}\right){S}_{P}\left({p}_{\left(2\right)}\right).$$Now, if $p=cqb$, we get using the Sweedler notation for E that$$\begin{array}{cc}\hfill \sum _{\left(p\right)}{\epsilon}_{t}^{P}\left({p}_{\left(1\right)}\right){S}_{P}\left({p}_{\left(2\right)}\right)\phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}{\epsilon}_{t}^{P}\left(c{q}_{\left(1\right)}{E}_{\left(1\right)}\right){q}_{\left(2\right)}{E}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}c{q}_{\left(1\right)}\u25b9\left({S}_{B}\left({E}_{\left(1\right)}\right)\right){q}_{\left(2\right)}{E}_{\left(2\right)}b\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\sum _{\left(q\right)}cq{S}_{B}\left({E}_{\left(1\right)}\right){E}_{\left(2\right)}b=cqb.\hfill \end{array}$$The other formula is proven in a similar way. This completes the proof.

**Proposition**

**16.**

**Proof.**

## 5. Conclusions and Further Research

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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Daele, A.V.; Wang, S.
Weak Multiplier Hopf Algebras II: Source and Target Algebras. *Symmetry* **2020**, *12*, 1975.
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Daele AV, Wang S.
Weak Multiplier Hopf Algebras II: Source and Target Algebras. *Symmetry*. 2020; 12(12):1975.
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Daele, Alfons Van, and Shuanhong Wang.
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