Determinant of exact sequence

Denote $p := \text{dim}\ A$, $q := \text{dim}\ C$ and $n := p+q = \text{dim}\ B$. Then I claim the following makes sense$$(\ddagger)\quad {\bigwedge}^p A\otimes {\bigwedge}^q C\to{\bigwedge}^n B,\quad (a_1\wedge ...\wedge a_p)\otimes (c_1\wedge ...\wedge c_q)\longmapsto a_1\wedge ...\wedge a_p\wedge \tilde{c}_1\wedge ...\wedge \tilde{c}_q$$ for arbitrary preimages $\tilde{c}_i$ of the $c_i$ under $B\twoheadrightarrow C$.

First, for fixed $a_i$ and $c_j$ the r.h.s. is independent of the choice of the $\tilde{c}_i$:

• If the $a_i$ do not form a basis of $A$, the right hand side vanishes independently of the choice of $\tilde{c}_i$.
• If the $a_i$ do form a basis of $A$, any other choice $\tilde{c}_j^{\prime}$ of preimage for $c_j$ differs from $\tilde{c}_j$ by a linear combination of the $a_i$, which cancels in the r.h.s. of $(\ddagger)$.

Hence $(\ddagger)$ makes sense as a map $A^p\times C^q\to {\bigwedge}^n B$, and since it is antisymmetric and multilinear in the $A$- and $C$-variables, it descends to a morphism as indicated.

Finally, if $(a_1,...,a_p)$ is a basis of $A$ and $(c_1,...,c_q)$ is a basis of $B$, the morphism $(\ddagger)$ sends the generator $(a_1\wedge ...\wedge a_p)\otimes (c_1\wedge ...\wedge c_q)$ of the l.h.s. to the generator $a_1\wedge ...\wedge a_p\wedge \tilde{c}_1\wedge ...\wedge \tilde{c}_q$ of the r.h.s. corresponding to the basis $(a_1,...,a_p,\tilde{c}_1,...,\tilde{c}_q)$ of $B$, hence is an isomorphism.

Daniel Murfet's notes, Tensor, Symmetric and exterior algebras, Section 3.3.