TITLE: Is tensor of chain complex commutative?
QUESTION [5 upvotes]: Let $B_*$ and $C_*$ be chain complexes (say of $R$-modules). Then is $B_*\otimes_R C_*$ isomorphic as a chain complex to $C_*\otimes B_*$? 

There are lots of signs involved, and I am not sure if it can be possible to arrange them such that the two chain complexes are isomorphic.

REPLY [5 votes]: Yes, they are naturally isomorphic.  Define $f:B_*\otimes C_*\to C_*\otimes B_*$ by $f(b\otimes c)=(-1)^{|b||c|}c\otimes b$ when $b$ and $c$ are homogeneous.  We can compute $$f(d(b\otimes c))=(-1)^{|db||c|}c\otimes db+(-1)^{|b|}(-1)^{|b||dc|}dc\otimes b=(-1)^{(|b|-1)|c|}c\otimes db+(-1)^{|b||c|}dc\otimes b$$
and $$d(f(b\otimes c))=(-1)^{|b||c|}dc\otimes b+(-1)^{|b||c|}(-1)^{|c|}c\otimes db=(-1)^{|b||c|}dc\otimes b+(-1)^{(|b|+1)|c|}c\otimes db.$$
These are equal, since $|b|+1$ and $|b|-1$ have the same parity, so $f$ commutes with $d$.