Sign convention for total complex

Question

Let $C^{\bullet,\bullet}$ be a double complex with differentials $d,e$ both of degree $+1$.. I use the convention that the squares commute. I am mainly thinking of $C^{i,j}=A^i\otimes B^j$ or $C^{i,j}=hom(A^{-i},B^j)$ for cochain complexes $(A^\bullet,d), (B^\bullet,e)$.

Then we can define the sum total complex $$ Tot^\oplus(C)^n=\bigoplus_{i+j=n} C^{i,j} $$ and the product total complex $$ Tot^\Pi(C)^n=\prod_{i+j=n} C^{i,j} $$ It is my understanding that the differentials in both cases would be induced by $d+(-1)^ie$ for a summand / factor $C^{i,n-i}$ of $Tot^?(C)^n$.

(EDIT: I switched up $d$ and $e$ in the convention for the internal hom. What I want is as in the nlab) However this does not agree with the usual convention for the internal hom of chain complexes, which would have us take $e-(-1)^nd$. And if one writes down the condition of the adjunction $Hom(A^\bullet\otimes B^\bullet, C^\bullet)\cong Hom(A^\bullet, hom(B^\bullet,C^\bullet))$ the additional sign appears as clear as day.

Now my questions are:

1. Are the two ways to define differentials on $hom(A^\bullet, B^\bullet)$ isomorphic?
2. Given an adjunction of $2$ (or $n$) with left adjoint $F:\mathcal{A}\times\mathcal{B}\to \mathcal{C}$, with right adjoints $G^{1,2}$, how do we define an induced adjunction $$ Ch(F):Ch(\mathcal{A})\times Ch(\mathcal{B})\to Ch(\mathcal{C}) $$
3. Where can I find such discussions in the literature?

I think the answer to 1 is no, since I tried several canidates but could not find an isomorphism, which is of course not a proof. For 2 I think, one can just generalize the formula $d-(-1)^ne$, but this seems just inelegant to me, and there should be a clearer more formal / category-theoretical way, e.g. through a middle adjoint $$ Tot^\oplus \dashv \;? \dashv Tot^\Pi $$

Answer 1

There's an isomorphism between the two versions of the total complex induced by $\text{id}:C^{i,j}\to C^{i,j}$ when

• $i\equiv 0\pmod{4}$, or
• $i\equiv 1\pmod{4}$ and $j$ is odd, or
• $i\equiv 3\pmod{4}$ and $j$ is even

and by $-\text{id}:C^{i,j}\to C^{i,j}$ otherwise.

[In the original version of the question, the two proposed differentials were $d+(-1)^ie$ and $d-(-1)^ne$ (rather than $d+(-1)^ie$ and $e-(-1)^nd$. For that case, an isomorphism between the two versions of the total complex is induced by $\text{id}:C^{i,j}\to C^{i,j}$ for $j\equiv 0,3\pmod{4}$ and $-\text{id}:C^{i,j}\to C^{i,j}$ for $j\equiv 1,2\pmod{4}$.]

In fact, for any two reasonable sign conventions (by which I mean that some arrows in the double complex are assigned minus signs in such a way that every $1\times 1$ square has an odd number of minus signs on its edges), there will be an isomorphism given by $\pm\text{id}$ on each $C^{i,j}$. To decide whether to use $+\text{id}$ or $-\text{id}$, use $+\text{id}$ on $C^{0,0}$, count how many minus signs are assigned to the arrows on a path in the double complex from $(0,0)$ to $(i,j)$. The parity of the difference between the two numbers obtained from the two sign conventions will be independent of the path. Use $+\text{id}$ if this parity is even, and $-\text{id}$ if it's odd.