Let $A$ be a non-degenerate algebra and let $\Delta: A \to M(A \otimes A)$ be a non-degenerate morphism. We can extend the algebra morphism
$$\iota \otimes \Delta: M(A \otimes A) \to M(A \otimes A \otimes A)$$


Suppose I want to show that ($1$ is the unit of $M(A))$:
$$(\iota \otimes \Delta)(x \otimes 1) = x \otimes 1 \otimes 1, \quad x \in M(A).$$
This looks trivial, but I believe that there is something to prove. To prove the above, we can observe that if $a,b,c,d,e \in A$
$$(\iota \otimes \Delta)(x \otimes 1) (\iota \otimes \Delta)(a \otimes b)(c \otimes d \otimes e) = (\iota \otimes \Delta)(xa \otimes b)(c\otimes d \otimes e)$$
$$= xac \otimes \Delta(b)(d \otimes e)$$


and $$(x \otimes 1 \otimes 1)(\iota \otimes \Delta)(a \otimes b)(c \otimes d \otimes e) = (xac) \otimes \Delta(b)(d \otimes e)$$
and invoke non-degeneracy of $\iota \otimes \Delta$ to conclude that both multipliers must be equal (actually, this shows that both left multipliers must be equal, but since left multipliers determine the right multipliers uniquely this is enough). Note that I used here that if $\phi: A \to M(B)$ is a non-degenerate morphism, then its extension $\overline{\phi}: M(A) \to M(B)$ is given by


$$\overline{\phi}(T) \phi(a)b := \phi(Ta) b, \quad b\phi(a)\overline{\phi}(T) := b\phi(aT)$$
where $a,b \in A$.


Is it correct that there is something to prove here or does this follow from something else I am missing?



About extensions between morphisms on the multiplier algebra
I think there is *something* to prove. But asking "... something to prove here?" is always going to be in the [eye of the beholder](https://www.collinsdictionary.com/dictionary/english/in-the-eye-of-the-beholder). One mathematician's checking of details is another's tedium; one mathematician's concise writing is another's bafflement.


However, in this case my opinion would be that there *is* something to prove; but it's the sort of thing which is to be proved once, and then to be used without proof in the future. (To contrast with things which genuinely are "obvious"; or things which are "routine to the expert").




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I would structure the argument this way.


* If $A$ is non-degenerate, then so is $A\otimes A$. There is an injective algebra homomorphism $M(A)\otimes M(A) \rightarrow M(A\otimes A)$. This is shown in the appendix of [van Daele's original paper](https://www.zbmath.org/?q=an%3A0809.16047).
* So $\iota\otimes\Delta:A\otimes A \rightarrow A\otimes M(A\otimes A) \rightarrow M(A\otimes (A\otimes A))$ makes sense. (*Notice* that *implicitly* I use here, for example, that $\Delta$ is a homomorphism means $\iota\otimes\Delta$ is a homomorphism. This is obvious; but if I were lecturing undergraduates about tensor products of algebras for the first time, this might be an exercise to set them to solve.)
* This homomorphism is non-degenerate. For example,
$$ \big((\iota\otimes\Delta)(a\otimes b)\big) (c\_1\otimes c\_2\otimes c\_3)
= ac\_1 \otimes \Delta(b)(c\_2\otimes c\_3), $$
and so taking linear span gives all of $A\otimes (A\otimes A)$ because $\Delta$ is non-degenerate.
* Thus we can extend $\iota\otimes\Delta$ to a homomorphism $M(A\otimes A)\rightarrow M(A\otimes (A\otimes A))$.
* For $x\in A$ we have that $x\otimes 1\in M(A\otimes A)$. The map $M(A)\otimes M(A)\rightarrow M(A\otimes A)$ does indeed send $1\otimes 1$ to $1 = 1\_{A\otimes A}$. We can of course identify $A\otimes (A\otimes A)$ with $(A\otimes A)\otimes A$ with $A\otimes A\otimes A$.
* So we *do* need to check that the *extension* of $\iota\otimes\Delta$ maps $x\otimes 1\in M(A\otimes A)$ to $x\otimes 1\otimes 1 \in M(A\otimes A\otimes A)$.
* Now proceed as you do.


So actually I have checked a lot more. What I'm trying to do is think a bit more systematically about what is going on, which will hopefully remove your doubts that there is something to check.




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We might alternatively argument thus: extend $\Delta$ to $\overline\Delta:M(A)\rightarrow M(A\otimes A)$; again obviously $\overline\Delta(1)=1\otimes 1$. Thus $\iota\otimes\overline\Delta: A\otimes M(A) \rightarrow A\otimes M(A\otimes A)$ sends $x\otimes 1$ to $x\otimes 1\otimes 1$.


However, how do we associate $\iota\otimes\overline\Delta$ to $\overline{\iota\otimes\Delta}$? Let $B=M(A)$, and consider $A\otimes B$, and thus $M(A\otimes B)$, and try to compare this to $M(A\otimes A)$. I think these can be related, but again there is some work which needs to be done.


