homological algebra - Homomorphism between exterior powers of a free module of finite rank

Here is a basis free expression.

Let the rank of $M$ be $r$. Pick an isomorphism $phi: M to M^*=hom_R(M,R)$. Now, if $min M$, define a map $f_m:Lambda^rMtoLambda^{r-1}M$ by contracting with $phi(m)$, so that if $m_1$, $dots$, $m_rin M$, then

$$f_m(m_1wedgecdotswedge m_r)=sum_{i=1}^r;(-1)^i;langlephi(m),m_irangle; m_1wedgecdotswedgehat m_iwedgecdotswedge m_r.$$ Here $langlemathord-,mathord-rangle:M^*times Mto R$ is the evaluation map. It is not hard to show that $min Mmapsto f_minhom(Lambda^rM, Lambda^{r-1}M)$ is an isomorphism

(This isomorphism is not natural, because it depends on the choice of $phi$. Of course, there is a natural isomorphism $Psi_M:M^*tohom(Lambda^rM, Lambda^{r-1}M)$ given by essentially the formula, and there is no natural isomorphism $Mtohom(Lambda^rM, Lambda^{r-1}M)$, because such a thing would, when composed with the inverse of the natural isomorphism $Psi_M$, give a natural isomorphism $Mto M^*$, which does not exist.)

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