Kernels and cokernels of multicomplex homomorphisms

Question

Let $\mathcal A$ be a (complete and cocomplete) Abelian category.

A multicomplex in $\mathcal A$ is a bigraded object $X^{(\bullet,\bullet)}$ with differentials $$ d^{(i,j)}_r\colon X^{(i,j)}\to X^{(i+r,j-r+1)}, $$ for all $(i,j)\in \mathbb Z\times \mathbb Z$ and $r\in \mathbb N$, with the property that \begin{equation}\label{multidifferential} \sum_{r+s=n}d^{(i+r,j-r+1)}_sd^{(i,j)}_r=0 \end{equation} for all $(i,j)\in \mathbb Z\times \mathbb Z$ and $n\in \mathbb N$ (notice that each of these sums is finite, so it represents a well-defined morphism $X^{(i,j)}\to X^{(i+n,j-n+1)}$ that we want to be trivial).

A morphism of multicomplexes $\phi\colon X\to Y$ between two multicomplexes $X=(X^{(\bullet,\bullet)},(d_r^{(\bullet,\bullet)})_{r\in \mathbb N})$ and $Y=(Y^{(\bullet,\bullet)},(e_s^{(\bullet,\bullet)})_{s\in \mathbb N})$, is a family of morphisms $$ \phi^{(i,j)}_t\colon X^{(i,j)}\to Y^{(i+t,j-t)} $$ which is compatible with differentials in the following sense: $$ \sum_{r+s=n}\phi^{(i+r,j-r+1)}_sd^{(i,j)}_r=\sum_{r+s=n}e^{(i+r,j-r)}_s\phi^{(i,j)}_r $$ for all $(i,j)\in \mathbb N\times \mathbb Z$ and $n\in\mathbb N$ (again these are finite sums identifying morphisms $X^{(i,j)}\to Y^{(i+n,j-n+1)}$). We denote by $\mathbb M(\mathcal A)$ the category of multicomplexes over $\mathcal A$.

I am trying to understand this category but cannot find complete references, just claims here and there, without explicit computations. I have tried to understand for a while if this category has co/kernels but I cannot give a complete construction. More in detail:

Let $\phi=(\phi_i)_{i}\colon A\to B$ be a morphism of multicomplexes, can we construct a multicomplex $K$ and a morphism of multicomplexes $\kappa\colon K\to A$ that is a kernel of $\phi$ in $\mathbb M(\mathcal A)$? It seems natural to take $K^{i,j}:=\mathrm{Ker}(\phi_0^{i,j})$, so that $\kappa^{i,j}_0$ is just the inclusion of $K^{i,j}$ in $A^{i,j}$. Using the universal property of kernels, it is easy to define a $0$-differential for $K$. But at this point I am not able to go on: I do not know how to construct the higher differentials for $K$, nor the higher components of $\kappa$. Maybe I just have a bad definition for the objects $K^{i,j}$, and starting with different objects everything is easy, but I would not know what to choose now.

Could you please give me some good reference where this is explained in detail or give me some (explicit enough) indication on how to go on with the construction?

Answer 1

If I'm not mistaken, multicomplexes in $\mathcal{A}$ are the same as unbounded cochain complexes in another category constructed from $\mathcal{A}$ as follows:

• its objects are $\mathbb{Z}$-graded sequences $X = (X^{(i)})_{i \in \mathbb{Z}}$ of objects of $\mathcal{A}$,

• a morphism $f$ from $(X^{(i)})_{i \in \mathbb{Z}}$ to $(Y^{(i)})_{i \in \mathbb{Z}}$ is given by a family of maps $f^{(i)}_r : X^{(i)} \to Y^{(i+r)}$ for $r \in \mathbb{N}$, $i \in \mathbb{Z}$,

• the composition $f \circ g$ is given by the formula $$(f \circ g)_t^{i} = \sum_{r+s=t} f_s^{(i+r)} g_r^{(i)},$$

• the identity map is the identity for $r = 0$ and zero for $r > 0$.

The correspondence takes a multicomplex $X^{(i,j)}$ to a cochain complex whose $k$th term is the graded object $(X^{(i,k-i)})_{i \in \mathbb{Z}}$. So, it suffices to compute (co)kernels in this latter category.

Let's take $\mathcal{A} = R\textrm{-Mod}$ and let $R[i]$ denote the object which is $R$ in degree $i$ and zero elsewhere. For any $X$ and any element $x$ of $X^{(j)}$ with $j - i \ge 0$ there is a morphism $g_x : R[i] \to X$ with $g_{j-i}^{(i)}$ the map sending $1$ to $x$ and all other $g_*^{(*)}$ zero. If $x$ is in the kernel of $f : X \to Y$, then $f \circ g_x$ should be zero. From the formula for $f \circ g$, this means $f_s^{(j)}(x)$ must be zero for every $s \ge 0$. So we see that the kernel of $f$ should actually consist of those elements of each $X^{(i)}$ on which every $f_r^{(i)}$ vanishes, not just $f_0^{(i)}$. So your guess for $K^{i,j}$ seems to be wrong; we should take the joint kernel of all the $\phi_t^{(i,j)}$. Similarly, I guess that the cokernel of $f : X \to Y$ should be constructed by quotienting out $Y^{(i)}$ by the sum of the images of all the $X^{(i')}$ which can map to it (those with $i' \le i$).