I am reading https://ocw.mit.edu/courses/mathematics/18-905-algebraic-topology-i-fall-2016/lecture-notes/MIT18_905F16_lec28.pdf .

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?


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