I am trying to understand the commutativity of diagram (2) of Proposition 3.11 from this paper. It should be a simple computation but I don't understand how they do it and when I try to do it myself I don't achieve commutativity.
Context
Let $A=\bigoplus A_i^j$ be a $(\Bbb{Z},\Bbb{Z})$-bigraded module. We say that $a\in A^i_j$ has bidegree $(i,j)$. For $x,y$ of bidegrees $(x_1,x_2)$, $(y_1,y_2)$, let $\langle x,y\rangle=x_1y_1+x_2y_2$. The tensor product of maps is defined according to the sign convention $(f\otimes g)(a\otimes b)=(-1)^{\langle g,a\rangle}f(a)\otimes g(b)$.
The totalization of $A$ is defined to be the graded module with degree components $\DeclareMathOperator{\Tot}{Tot} \Tot(A)^n=\prod_{i\leq 0}A^{n+i}_i\oplus\bigoplus_{i>0}A^{n+i}_i$. There is a definition of $\Tot$ applied to morphisms of twisted complexes, but it is claimed to be defined in general bigraded modules, so I understand that the general definition should be similar. In the case of twisted complexes we have that a morphism $f:A\to B$ is a family of maps $f_m:A\to B$ of bidegree $(-m,-m+1)$ for $m\geq 0$ and its totalization is defined to be
$(\Tot(f)(a))_j=\sum_{m\geq 0}(-1)^{mn}f_m(a_{j+m})$
for $a=(a_i)_{i\in\Bbb{Z}}\in\Tot(A)^n$, where $a_i\in A^{n+i}_i$ denotes the $i$-th component of $a$.
Therefore, $\Tot$ defines a functor from bigraded modules (or twisted complexes) to graded modules. Proposition 3.11 claims that this functor is lax monoidal with a structure map $\mu_{A,B}:\Tot(A)\otimes \Tot(B)\to \Tot(A\otimes B)$ given by
$(\mu_{A,B}(a\otimes b))_k=\sum_{k_1+k_2=k}(-1)^{k_1n_2}a_{k_1}\otimes b_{k_2}$.
for $a=(a_i)_i\in \Tot(A)^{n_1}$ and $b=(b_j)_j\in \Tot(B)^{n_2}$
Problem
I want to show that $\mu$ is natural, which means that for $f:A\to A'$ and $g:B\to B'$ we have $\Tot(f\otimes g)\circ \mu_{A,B}=\mu_{A',B'}\circ \Tot(f)\otimes \Tot(g)$ (diagram (2) of Proposition 3.11). I don't understand how it is done in the paper so I am going to show my attempt and I hope someone can help me find the mistakes.
Attempt
- Let me compute $((\Tot(f\otimes g)\circ \mu_{A,B})(a\otimes b))_j$ on homogeneous elements $a\in\Tot(A)^{n_1}$ and $b\in\Tot(B)^{n_2}$, so let $n=n_1+n_2$. In the paper they seem to compute $\Tot$ first, but I think it makes more sense to do things in order, so I am going to first compute
$$(\mu_{A,B}(a\otimes b))_{j+m}=\sum_{k_1+k_2=j+m}(-1)^{k_1n_2}a_{k_1}\otimes b_{k_2}.$$
For convenience and similarity of notation with the paper let $m=m_1+m_2$ and rewrite the above sum as
$$\underset{ m_1,m_2\geq 0}{\sum_{k_1+k_2=j}}(-1)^{(k_1+m_1)n_2}a_{k_1+m_1}\otimes b_{k_2+m_2}.$$ Now let me compute $(\Tot(f\otimes g)\circ \mu_{A,B})(a\otimes b))_j$ by taking the $m$-th component of $f\otimes g$ on each summand:
$$((\Tot(f\otimes g)\circ \mu_{A,B})(a\otimes b))_j=\Tot(f\otimes g)(\underset{ m_1,m_2\geq 0}{\sum_{k_1+k_2=k}}(-1)^{(k_1+m_1)n_2}a_{k_1+m_1}\otimes b_{k_2+m_2})=$$
$$\underset{ m_1,m_2\geq 0}{\sum_{k_1+k_2=k}}(-1)^{(k_1+m_1)n_2}\Tot(f\otimes g)(a_{k_1+m_1}\otimes b_{k_2+m_2})=\underset{ m_1,m_2\geq 0}{\sum_{k_1+k_2=k}}(-1)^{(k_1+m_1)n_2}\sum_{m\geq 0}(-1)^{mn}(f\otimes g)_m(a_{k_1+m_1}\otimes b_{k_2+m_2})$$
Using that $m=m_1+m_2$ I can write this sum as
$$\underset{ m_1,m_2\geq 0}{\sum_{k_1+k_2=k}}(-1)^{(k_1+m_1)n_2+mn}(f_{m_1}\otimes g_{m_2})(a_{k_1+m_1}\otimes b_{k_2+m_2})$$
Now, to evaluate $(f_{m_1}\otimes g_{m_2})(a_{k_1+m_1}\otimes b_{k_2+m_2})$ they use the sign convention I specified above, giving $(-1)^{m_2n_1}$ as a result. However, since at this point we are in the totalization, I would instead use the sign convention of graded modules with respect to the total degree (the degree component in totalization), which just gives $(-1)^{n_1}$ since the total degree of $a$ is $n_1$ and that of $g$ is $1$ (in the case of twisted complexes, unspecified in general, the sign would be $(-1)^{|g|n_1}$). Why should one use the bigraded sign convention here?
- Now let us compute the other side of the commutativity equation, $((\mu_{A',B'}\circ\Tot(f)\otimes\Tot(g))(a\otimes b))_j$. Again I would first evaluate $\Tot(f)\otimes\Tot(g))(a\otimes b)$ which should involve again total degree conventions, so that the evaluation results in $(-1)^{|g|n_1}\Tot(f)(a)\otimes\Tot(g)(b)$. Then I apply $\mu_{A',B'}$ to obtain
$$\mu_{A',B'}((-1)^{|g|n_1}\Tot(f)(a)\otimes\Tot(g)(b))_j=\sum_{k_1+k_2=j}(-1)^{|g|n_1+k_1(n_2+|g|)}(\Tot(f)(a))_{k_1}\otimes (\Tot(g)(b))_{k_2}$$
Here I write $n_2+|g|$ because by definition of $\mu$ I need the total degree of $\Tot(g)(b)$. In the case of a twisted complex it would be $n_2+1$. In any case it looks different to what they have, which is $k_1n_2$.
Now we can compute each $\Tot$, giving
$$\underset{m_1,m_2\geq 0}{\sum_{k_1+k_2=j}}(-1)^{|g|n_1+k_1(n_2+|g|)+m_1n_1+m_2n_2}f_{m_1}(a_{m_1+k_1})\otimes g_{m_2}(b_{m_2+k_2})$$
Summary
The evaluation signs would cancel and I get on the left hand side $(-1)^{(k_1+m_1)n_2+(m_1+m_2)(n_1+n_2)}$ and on the right hand side I get $(-1)^{k_1(n_2+|g|)+m_1n_1+m_2n_2}$, which are not equal and therefore $\mu$ is not natural. What mistakes did I make here? There are mainly two points where I am not sure that I highlighted, could someone explain me why I shouldn't proceed the way I did?
Edit
I have spotted a mistake, maps of twisted complexes are of bidegree $(-m,-m)$ so the total degree is $0$, so I do get $k_1n_2$ on the right hand side, and in bigraded modules I should impose that morphisms have total degree $0$. However, even with this change, there is a $(-1)^{m_2n_1}$ factor that obstructs commutativity.