Sign conjugation for the tensor-Hom identification ($A{|}_{X_i\otimes C_j^{\prime }}=(-1{)}^{j(j-1)/2}$)

For finite free total chain complexes, evaluation identifies the underlying graded tensor product $X\otimes C^{\prime }$ with the graded Hom complex of chain complexes, where $X$ is the reversed dual chain complex. Under the precomposition-first Hom differential, conjugation by the sign $(-1{)}^{j(j-1)/2}$ makes this a chain isomorphism. The identity $\rho (j)-\rho (j-1)=j-1$ works for every integer degree. If the grading is unbounded and only degreewise finite, products on the Hom side need not equal direct sums on the tensor side.