= Sign conjugation for the tensor-Hom identification
{title2=$A|_{X_i\otimes C'_j}=(-1)^{j(j-1)/2}$}
For finite free total <chain complexes>, evaluation identifies the underlying graded <tensor product> $X\otimes C'$ 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.
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