= Solution
\b[The tensor differential.] Use homological grading, so each differential lowers degree by one. The <tensor product of chain complexes> has
$$
(C\otimes_R C')_k=\bigoplus_{p+q=k}C_p\otimes_R C'_q,\qquad
D(x\otimes y)=d_Cx\otimes y+(-1)^p x\otimes d_{C'}y
$$
for $x\in C_p$. This <Koszul sign rule> gives
$$
D^2(x\otimes y)=d_C^2x\otimes y+
\bigl((-1)^{p-1}+(-1)^p\bigr)d_Cx\otimes d_{C'}y+
x\otimes d_{C'}^2y=0.
$$
Thus the graded <tensor product> is a <chain complex>.
\b[The Hom differential.] Write $M_j=\prod_p\operatorname{Hom}_R(C_p,C'_{p+j})$. For a degree-$j$ element of this <graded Hom complex of chain complexes>, use the prescribed differential
$$
d_M f=f d_C+(-1)^{j-1}d_{C'}f.
$$
The next application uses degree $j-1$, so
$$
d_M^2f=f d_C^2+
\bigl((-1)^{j-1}+(-1)^{j-2}\bigr)d_{C'}f d_C+
(-1)^{2j-3}d_{C'}^2f=0.
$$
In degree zero, $d_Mf=f d_C-d_{C'}f$, so its kernel consists exactly of <chain maps>. A degree-one element $h$ has $d_Mh=h d_C+d_{C'}h$, which is exactly the change between two <chain maps> related by a <chain homotopy>. Consequently
$$
\boxed{H_0(M(C,C'))\cong
\{\text{chain maps }C\to C'\}/\text{chain homotopy}.}
$$
This is natural: precomposition and postcomposition by <chain maps> preserve both degree-zero cycles and degree-zero boundaries.
\b[The dual complex and the sign adjustment.] Define the <reversed dual chain complex>
$$
X_i=\operatorname{Hom}_R(C_{-i},R),\qquad
d_X\phi=\phi d_C.
$$
Here $d_X\phi$ is a functional on $C_{1-i}$, so it belongs to $X_{i-1}$, and $d_X^2=0$. The absence of an additional sign in $d_X$ is intentional.
Interpret finite generation of the free <chain complexes> as finiteness of their total graded <free modules>. Then only finitely many degrees occur, and the finite-free evaluation isomorphisms assemble into a graded isomorphism
$$
\psi:X\otimes_R C'\longrightarrow M(C,C'),\qquad
\psi(\phi\otimes y)(c)=\phi(c)y,
$$
where $\phi\in X_i$, $y\in C'_j$, and this component is zero outside $C_{-i}$. The <dual module> construction and evaluation make $\psi$ natural. Finite rank is needed for evaluation to be an isomorphism; finite total support also makes the sums on the tensor side agree with the products on the <Hom functor> side.
Under $\psi$, the <Hom functor> differential has the form
$$
\psi^{-1}d_M\psi(\phi\otimes y)=d_X\phi\otimes y+
(-1)^{i+j-1}\phi\otimes d_{C'}y.
$$
The usual <tensor product of chain complexes> instead has second coefficient $(-1)^i$. Set
$$
\boxed{\rho(j)=\frac{j(j-1)}2,\qquad
A|_{X_i\otimes C'_j}=(-1)^{\rho(j)}\,\mathrm{id}.}
$$
This is integer-valued for every $j\in\mathbb Z$, including negative $j$, and satisfies $\rho(j)-\rho(j-1)=j-1$. Conjugating the usual tensor differential by $A$ leaves the first term unchanged and changes its second coefficient to $(-1)^{i+j-1}$. Hence $d_M=\psi A D A^{-1}\psi^{-1}$, giving the <sign conjugation for the tensor-Hom identification>
$$
\boxed{\Phi=\psi A:X\otimes_R C'\xrightarrow{\;\cong\;}M(C,C'),
\qquad
\Phi(\phi\otimes y_j)(c)=(-1)^{j(j-1)/2}\phi(c)y_j.}
$$
This is an isomorphism of <chain complexes>, not merely of their <homology>. If one instead assumes only degreewise finite rank with unbounded grading, the ordinary tensor need not identify with the product defining $M$; the finite-total convention is essential to this last conclusion.
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