In the the accepted answer of this question, Stefan Hamcke used this functor:
$$ Id×(−⊗−): C \times C \to C \times C $$
It maps $(a,(a,a))$ to $(a, a \times a)$. But I wonder how this functor operates in other cases.
A functor has to map every object to a new object. But what would it do for $(a,b)$, or what would it do for $((a,b),c)$?
EDIT:
Okay I understood the statement in the original answer by Stefan Hamcke wrong. Now I understand that $Id(-\otimes -): C \times C \times C \to C$, however in that case:
What kind of triple in $C \times C \times C$ shows the associativity between $(z,(a,(b,c)))$ and $(z,((a,b),c))$?
I could imagine two new functors and their natural isomorphism:
$$ z \otimes (Id \times (- \otimes -)): \mathcal{C} \times \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\ z \otimes((- \otimes -) \times Id): \mathcal{C} \times \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\ \alpha_2: z \otimes (Id \times (- \otimes -)) \cong z \otimes((- \otimes -) \times Id) $$
Is that the case? However in this case, the associator would need to a family of natural morphisms? Or is it okay, because $\alpha_2$ can be modelled as $z \otimes \alpha$?