$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proposition 2.5.1.12. Let $(C_{\ast }, \partial )$ and $(D_{\ast }, \partial )$ be chain complexes. Then there is a unique homomorphism of graded abelian groups
\[ \partial : (C \boxtimes D)_{\ast } \rightarrow (C \boxtimes D)_{\ast - 1} \]
satisfying the identity
\[ \partial ( x \boxtimes y) = (\partial (x) \boxtimes y) + (-1)^{m} (x \boxtimes \partial (y)) \]
for $x \in C_{m}$ and $y \in D_{n}$. Moreover, this homomorphism satisfies $\partial ^2 = 0$, so we can regard the pair $( (C \boxtimes D)_{\ast }, \partial )$ as a chain complex.
Proof.
For every pair of integers $m,n \in \operatorname{\mathbf{Z}}$, the construction
\[ (x,y) \mapsto (\partial x \boxtimes y) + (-1)^{m} (x \boxtimes \partial y) \]
determines a bilinear map $C_ m \times D_{n} \rightarrow (C \boxtimes D)_{m+n-1}$. Invoking the universal property of tensor products and direct sums, we deduce that there is a unique map $\partial : (C \boxtimes D)_{\ast } \rightarrow (C \boxtimes D)_{\ast -1}$ with the desired properties. The identity $\partial ^2 = 0$ follows from the calculation
\begin{eqnarray*} \partial ^2(x \boxtimes y) & = & \partial ( (\partial x \boxtimes y) + (-1)^{m} (x \boxtimes \partial y) ) \\ & = & (\partial ^2 x \boxtimes y) + (-1)^{m-1} (\partial x \boxtimes \partial y) + (-1)^{m} ( \partial x \boxtimes \partial y) + (-1)^{2m} ( x \boxtimes \partial ^2 y) \\ & = & 0. \end{eqnarray*}
$\square$