diagonal map on exterior algebra Let $F$ be a free module of rank $2m$ over a commutative ring $R$. In Buchsbaum-Eisenbud's paper from 1977 about structure theorems for free resolutions of ideals of codimension 3, they give a proof that the determinant of a $2m \times 2m$ alternating matrix is some polynomial squared. That polynomial is called the pfaffian. Their proof uses multilinear algebra, and I cannot figure it out, and I think my problem comes down to their use of the action of $\wedge F$ on $\wedge F^*$, where $F^* = \text{Hom}_R(F,R)$ is the dual of $F$. 
They start the appendix by saying there is a map $\Delta: \wedge F \to \wedge F \otimes \wedge F$ which is induced by the map $\Delta(a) = a\otimes 1 + 1\otimes a \in \wedge F \otimes \wedge F$, for $a\in F$, and such that $\Delta$ is an algebra homomorphism. 
My question is, how does this induce the desired $\Delta$? That is, what is $\Delta(a\wedge b)$ for $a,b\in F$? And if $\alpha\in \wedge^{2m} F$, then what is the component of $\Delta(\alpha)$ in $\wedge^{2m} F \otimes \wedge^0 F$? (I need this last part to calculate the divided power).
 A: As discussed in the comments, the relevant tensor product here is the super tensor product; if $A$ and $B$ are two superalgebras then their super tensor product has multiplication given on homogeneous elements by
$$(a_1 \otimes b_1)(a_2 \otimes b_2) = (-1)^{|b_1| |a_2|} a_1 a_2 \otimes b_1 b_2$$
which has a sign owing to the need to swap $b_1$ and $a_2$. One desirable feature of this definition is that it gives a natural isomorphism $\Lambda(V \oplus W) \cong \Lambda(V) \otimes \Lambda(W)$.
As you say, $\Delta$ is defined by the condition that it is an algebra homomorphism, and consequently (I will omit the wedges for readability)
$$\Delta(ab) = \Delta(a) \Delta(b) = (a \otimes 1)(b \otimes 1) + (a \otimes 1)(1 \otimes b) + (1 \otimes a)(b \otimes 1) + (1 \otimes a)(1 \otimes b)$$
(the point of writing all those out was to be very careful about signs), which simplifies to
$$\Delta(ab) = ab \otimes 1 + a \otimes b - b \otimes a + 1 \otimes ab.$$
If we write $\alpha \in \Lambda^{2m} F$ as a product $e_1 \dots e_{2m}$ where $e_i$ is a basis of $F$ (again omitting wedges), then $\Delta(\alpha)$ is a product of terms of the form $(e_i \otimes 1 + 1 \otimes e_i)$, and consequently the component in $\Delta^{2m} F \otimes \Delta^0 F$ is just $\alpha$ again. 
