Let $A,B$ be two graded algebras. We know that $$f:A \otimes B \to B \otimes A$$ defined by $f(a\otimes b)= (-1)^{\deg a.\deg b}(b \otimes a)$ is isomorphism.
My question is
(i) Why $f(a\otimes b)= (-1)^{\deg a.\deg b}(b \otimes a)$ ? instead $f(a\otimes b)= b \otimes a$ ?
(ii) If $A,B,C$ are three graded algebras then so is $A \otimes B \otimes C.$ determined graded algebra structure over $A \otimes B \otimes C.$?
I assume that $A,B$ are graded commutative. Then the graded product on $B \otimes A$ is defined on homogeneous tensors by
$$ (b \otimes a) \cdot (b' \otimes a') = (-1)^{\deg b' \deg a} (b \cdot b') \otimes (a \cdot a') $$
where the sign is there so that the product will be commutative in the graded sense and similarly for $A \otimes B$. Then the sign factor in the definition of $f$ is there in order to make $f$ a homomorphism:
$$ f((a \otimes b) \cdot (a' \otimes b')) = f((-1)^{\deg b \deg a'} (a \cdot a') \otimes (b \cdot b')) = (-1)^{\deg b \deg a'}(-1)^{\deg(a \cdot a') \deg(b \cdot b')} (b \cdot b') \otimes (a \cdot a') = (-1)^{\deg b \deg a' + \deg a \deg b + \deg a \deg b' + \deg a' \deg b + \deg b \deg b'} (b \cdot b') \otimes (a \cdot a') \\ = (-1)^{\deg a \deg b'} (-1)^{\deg a \deg b} (b \otimes a) \cdot (-1)^{\deg a' \deg b'} (b' \otimes a') = f( a \otimes b) \cdot f(a' \otimes b').$$