Let $A$ be a non-degenerate algebra with multiplier algebra $M(A)$. Let $S: A \to M(A)$ be an antimultiplicative linear map, i.e.
$$S(ab) = S(b)S(a).$$
Consider the mapping $$\iota \otimes S: A \otimes A \to A \otimes M(A) \subseteq M(A \otimes A).$$
Does it extend to a map $\iota \otimes S: M(A \otimes A) \to M(A \otimes A)$?
The results that I know for extending linear maps to the multiplier algebra require (anti)multiplicativity to make sure the extension is well-defined. Here however, $\iota \otimes S$ is neither multiplicative nor antimultiplicative, but I guess there should be some way around this problem.
To give some context, in Van Daele's paper "An algebraic framework on group duality", in proposition 3.11, he considers an expression $(\iota \otimes S)(\Delta(q))$ where $S: A \to A$ is the antipode on the regular multiplier Hopf algebra $(A, \Delta)$. Since $\Delta(q) \in M(A \otimes A)$, we need to be able to make sense of $\iota \otimes S$ on the multiplier algebra $M(A \otimes A)$. Here is a screenshot: