Why the $(-1)^{pq}$ in tensor of cochain (whitney alexander map)?

We defined the Alexander Whitney map $$C_*( X\times Y) \to C_*(X) \otimes C_*(Y)$$, and would like now like to take $$Hom(*,R)$$ in order to get a cup product in the cohomology.

So given two chain complexes $$C_*, D_*$$, we'd like a map

$$F:Hom(C_*,R) \otimes Hom(D_*,R) \to Hom(C_* \otimes D_*,R)$$

(The convention is that when you tensor, the differential gets minus at vertical maps with odd $$x$$ coordinate).

The map $$F$$ (up to signs) does the obvious- given $$f\otimes g$$ on $$x\otimes y$$ return $$f(x)g(y)$$. My question is why do we need to add a sign to $$F$$? The author adds the sign $$(-1)^{pq}$$, in fact if we add the sign, $$F$$ doesn't commute with $$d_h$$ and $$d_v$$ I think. What am I missing?