Definition of multiplier bialgebra Consider the following fragments from "An invitation to quantum groups and duality" by Timmerman:

Question: In remark 2.1.6 (ii), it is stated that the homomorphism $\Delta\otimes \text{id}: A \otimes A \to A \otimes A \otimes A$ is extended to a homomorphism $M(A \otimes A) \to M(A \otimes A \otimes A)$. However, don't we start from a homomorphism
$$\Delta \otimes \text{id}: A \otimes A \to M(A \otimes A) \otimes A$$
? If so, how does this extension work? One possibility I see is to embed $$M(A \otimes A)\otimes A \subseteq M(A \otimes A) \otimes M(A) \subseteq M(A \otimes A \otimes A)$$ and we then obtain a homomorphism
$$\Delta\otimes \text{id}: A \otimes A \to M(A \otimes A \otimes A)$$ which (probably) can be extended to a homomorphism
$$\Delta\otimes \text{id}: M(A \otimes A) \to M(A \otimes A \otimes A)$$
Are the observations I make in this post correct? Or am I missing something?
 A: $\newcommand{\bC}{\mathbb{C}}\newcommand{\finite}{\text{fin}}$The answer is yes to both; all references are to Timmermann.

*

*The map $M(A \otimes A) \otimes A \to M(A \otimes A \otimes A)$ is the composition of obvious inclusion $M(A \otimes A) \otimes A \hookrightarrow M(A \otimes A) \otimes M(A)$ with the natural embedding $$M(A \otimes A) \otimes M(A) \hookrightarrow M((A \otimes A) \otimes A) = M(A \otimes A \otimes A)$$ of Proposition 2.1.4, part v.

*Once you’re satisfied that the composition
$$
 A \otimes A \xrightarrow{\Delta \otimes \operatorname{id}} M(A \otimes A) \otimes A \hookrightarrow M(A \otimes A) \otimes M(A) \hookrightarrow M(A \otimes A \otimes A)
$$
is non-degenerate, it extends uniquely to a homomorphism $A \otimes A \to M(A \otimes A \otimes A)$ by Proposition 2.1.4, part vi.


Addendum on motivation:
It’s worth remembering just why multiplier algebras arise in noncommutative geometry—you can see why even at the most elementary, purely algebraic level.
Given a set $X$, let $\bC_\finite(X)$ denote the $\ast$-algebra of finitely supported functions $X \to \mathbb{C}$ with the usual pointwise operations. Let $\phi : X \to Y$ be a function. You’d like to define a $\ast$-homomorphism $\phi^t: \bC_\finite(Y) \to \bC_\finite(X)$ by
$$
 \forall a  \in \bC_\finite(Y), \quad \phi^t(a) := a \circ \phi.
$$
The moment that there exists $y \in Y$ such that $\phi^{-1}(y)$ is infinite, your would-be $\ast$-homomorphism $\phi^t$ fails to have the correct codomain, since any function $a \in \bC_\finite(Y)$ whose support contains $y$ will yield a function $\phi^t(a) : X \to \mathbb{C}$ with infinite support. However, even in this worst case scenario, $\phi^t(a)$ will still yield a multiplier of $\bC_\finite(X)$ acting via pointwise multiplication. Hence, no matter how badly behaved $\phi$ may be, we can always make sense of $\phi^t$ as a non-degenerate $\ast$-homomorphism
$$
 \phi^t: \bC_\finite(Y) \to \bC(X) \cong M(\bC_\finite(X)).
$$
In short, if you want $\mathbb{C}_{\mathrm{fin}}$ to define a contravariant functor from the category of sets and functions to some category of (not-necessarily unital) $\ast$-algebras, you need the codomain to be the category where objects are $\ast$-algebras and where a morphism from $A$ to $B$ is a non-degenerate $\ast$-homomorphism $f: A \to M(B)$; in the $C^\ast$-algebraic context, this insight is apparently due to Woronowicz (see this older MathOverflow question).
So, what, then is the remark saying? It’s simply saying that the given definition of multiplier bialgebra is the obvious result of taking the definition of bialgebra from the category of unital $\ast$-algebras and unital $\ast$-homomorphisms and adapting it systematically to the above category of (not-necessarily unital) $\ast$-algebras by replacing any unital $\ast$-homomorphism $A \to B$ by a non-degenerate $\ast$-homomorphism $A \to M(B)$.
