Kernels and cokernels of multicomplex homomorphisms











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



A multicomplex in $mathcal A$ is a bigraded object $X^{(bullet,bullet)}$ with differentials
$$
d^{(i,j)}_rcolon X^{(i,j)}to X^{(i+r,j-r+1)},
$$

for all $(i,j)in mathbb Ztimes mathbb Z$ and $rin mathbb N$, 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 $(i,j)in mathbb Ztimes mathbb Z$ and $nin mathbb N$ (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 $phicolon Xto Y$ between two multicomplexes $X=(X^{(bullet,bullet)},(d_r^{(bullet,bullet)})_{rin mathbb N})$ and $Y=(Y^{(bullet,bullet)},(e_s^{(bullet,bullet)})_{sin mathbb N})$, is a family of morphisms
$$
phi^{(i,j)}_tcolon X^{(i,j)}to Y^{(i+t,j-t)}
$$

which is compatible with differentials in the following sense:
$$
sum_{r+s=n}phi^{(i+r,j-r+1)}_sd^{(i,j)}_r=sum_{r+s=n}e^{(i+r,j-r)}_sphi^{(i,j)}_r
$$

for all $(i,j)in mathbb Ntimes mathbb Z$ and $ninmathbb N$ (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 $phi=(phi_i)_{i}colon Ato B$ be a morphism of multicomplexes, can we construct a multicomplex $K$ and a morphism of multicomplexes $kappacolon Kto A$ 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?










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    up vote
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    favorite
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    Let $mathcal A$ be a (complete and cocomplete) Abelian category.



    A multicomplex in $mathcal A$ is a bigraded object $X^{(bullet,bullet)}$ with differentials
    $$
    d^{(i,j)}_rcolon X^{(i,j)}to X^{(i+r,j-r+1)},
    $$

    for all $(i,j)in mathbb Ztimes mathbb Z$ and $rin mathbb N$, 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 $(i,j)in mathbb Ztimes mathbb Z$ and $nin mathbb N$ (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 $phicolon Xto Y$ between two multicomplexes $X=(X^{(bullet,bullet)},(d_r^{(bullet,bullet)})_{rin mathbb N})$ and $Y=(Y^{(bullet,bullet)},(e_s^{(bullet,bullet)})_{sin mathbb N})$, is a family of morphisms
    $$
    phi^{(i,j)}_tcolon X^{(i,j)}to Y^{(i+t,j-t)}
    $$

    which is compatible with differentials in the following sense:
    $$
    sum_{r+s=n}phi^{(i+r,j-r+1)}_sd^{(i,j)}_r=sum_{r+s=n}e^{(i+r,j-r)}_sphi^{(i,j)}_r
    $$

    for all $(i,j)in mathbb Ntimes mathbb Z$ and $ninmathbb N$ (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 $phi=(phi_i)_{i}colon Ato B$ be a morphism of multicomplexes, can we construct a multicomplex $K$ and a morphism of multicomplexes $kappacolon Kto A$ 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?










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      up vote
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      up vote
      4
      down vote

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



      A multicomplex in $mathcal A$ is a bigraded object $X^{(bullet,bullet)}$ with differentials
      $$
      d^{(i,j)}_rcolon X^{(i,j)}to X^{(i+r,j-r+1)},
      $$

      for all $(i,j)in mathbb Ztimes mathbb Z$ and $rin mathbb N$, 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 $(i,j)in mathbb Ztimes mathbb Z$ and $nin mathbb N$ (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 $phicolon Xto Y$ between two multicomplexes $X=(X^{(bullet,bullet)},(d_r^{(bullet,bullet)})_{rin mathbb N})$ and $Y=(Y^{(bullet,bullet)},(e_s^{(bullet,bullet)})_{sin mathbb N})$, is a family of morphisms
      $$
      phi^{(i,j)}_tcolon X^{(i,j)}to Y^{(i+t,j-t)}
      $$

      which is compatible with differentials in the following sense:
      $$
      sum_{r+s=n}phi^{(i+r,j-r+1)}_sd^{(i,j)}_r=sum_{r+s=n}e^{(i+r,j-r)}_sphi^{(i,j)}_r
      $$

      for all $(i,j)in mathbb Ntimes mathbb Z$ and $ninmathbb N$ (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 $phi=(phi_i)_{i}colon Ato B$ be a morphism of multicomplexes, can we construct a multicomplex $K$ and a morphism of multicomplexes $kappacolon Kto A$ 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?










      share|cite|improve this 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
      $$
      d^{(i,j)}_rcolon X^{(i,j)}to X^{(i+r,j-r+1)},
      $$

      for all $(i,j)in mathbb Ztimes mathbb Z$ and $rin mathbb N$, 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 $(i,j)in mathbb Ztimes mathbb Z$ and $nin mathbb N$ (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 $phicolon Xto Y$ between two multicomplexes $X=(X^{(bullet,bullet)},(d_r^{(bullet,bullet)})_{rin mathbb N})$ and $Y=(Y^{(bullet,bullet)},(e_s^{(bullet,bullet)})_{sin mathbb N})$, is a family of morphisms
      $$
      phi^{(i,j)}_tcolon X^{(i,j)}to Y^{(i+t,j-t)}
      $$

      which is compatible with differentials in the following sense:
      $$
      sum_{r+s=n}phi^{(i+r,j-r+1)}_sd^{(i,j)}_r=sum_{r+s=n}e^{(i+r,j-r)}_sphi^{(i,j)}_r
      $$

      for all $(i,j)in mathbb Ntimes mathbb Z$ and $ninmathbb N$ (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 $phi=(phi_i)_{i}colon Ato B$ be a morphism of multicomplexes, can we construct a multicomplex $K$ and a morphism of multicomplexes $kappacolon Kto A$ 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?







      homological-algebra spectral-sequences abelian-categories






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      edited Nov 29 at 17:03

























      asked Nov 29 at 16:25









      Simone Virili

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          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 $mathcal{A} = Rtextrm{-Mod}$ 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 $i' le i$).






          share|cite|improve this answer





















          • Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
            – Simone Virili
            7 hours ago










          • Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
            – Simone Virili
            7 hours ago










          • Because of simple examples like the one above I am starting to think that this category is not Abelian
            – Simone Virili
            7 hours ago











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          up vote
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          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 $mathcal{A} = Rtextrm{-Mod}$ 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 $i' le i$).






          share|cite|improve this answer





















          • Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
            – Simone Virili
            7 hours ago










          • Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
            – Simone Virili
            7 hours ago










          • Because of simple examples like the one above I am starting to think that this category is not Abelian
            – Simone Virili
            7 hours ago















          up vote
          6
          down vote













          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 $mathcal{A} = Rtextrm{-Mod}$ 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 $i' le i$).






          share|cite|improve this answer





















          • Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
            – Simone Virili
            7 hours ago










          • Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
            – Simone Virili
            7 hours ago










          • Because of simple examples like the one above I am starting to think that this category is not Abelian
            – Simone Virili
            7 hours ago













          up vote
          6
          down vote










          up vote
          6
          down vote









          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 $mathcal{A} = Rtextrm{-Mod}$ 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 $i' le i$).






          share|cite|improve this answer












          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 $mathcal{A} = Rtextrm{-Mod}$ 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 $i' le i$).







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          answered Nov 29 at 17:37









          Reid Barton

          18.1k150103




          18.1k150103












          • Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
            – Simone Virili
            7 hours ago










          • Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
            – Simone Virili
            7 hours ago










          • Because of simple examples like the one above I am starting to think that this category is not Abelian
            – Simone Virili
            7 hours ago


















          • Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
            – Simone Virili
            7 hours ago










          • Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
            – Simone Virili
            7 hours ago










          • Because of simple examples like the one above I am starting to think that this category is not Abelian
            – Simone Virili
            7 hours ago
















          Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
          – Simone Virili
          7 hours ago




          Thank you for the nice point of view on the category of multicomplexes! It seems that your observation is correct and that multicomplexes can be seen as complexes over a category of, say, "multiobjects". That said, I have the same difficulties to show that the category of multiobjects is Abelian, even starting with a category of modules, your example shows that my guess was wrong but I do not think that the intersection of the kernels of all degrees is correct either.
          – Simone Virili
          7 hours ago












          Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
          – Simone Virili
          7 hours ago




          Consider the following example in Abelian groups: $A=(mathbb Z,mathbb Z, 0,0,dots)$, $B=(mathbb Z,mathbb Z, 0,0,dots)$ and the map $phicolon Ato B$ defined by $phi_0^0=0$, $phi_1^0=- mathrm{id}$ and $phi_0^1=mathrm{id}$ (the other components are necessarily trivial). Then the intersection of the kernels of all degrees is trivial, but the kernel of $phi$ is not the inclusion $0to A$ because there is a map $psicolon Cto A$ that composes to zero but it is not trivial: $C=(mathbb Z,0,0,dots)$ and $psi_0^0=mathrm{id}$, $psi_{1}^0=mathrm{id}$.
          – Simone Virili
          7 hours ago












          Because of simple examples like the one above I am starting to think that this category is not Abelian
          – Simone Virili
          7 hours ago




          Because of simple examples like the one above I am starting to think that this category is not Abelian
          – Simone Virili
          7 hours ago


















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