Construct additive inverse in an abelian category

abelian-categoriesabelian-groupscategory-theory

A (locally small) abelian category has a canonical addition structure on its hom sets. In fact, this structure forms an abelian group.

Let $A$ be an abelian category. Here, an abelian category is a category with a $0$ object, binary products and coproducts, and kernels and cokernels, such that all epis are conormal and all monos are normal.

I have constructed the sum of morphisms $f, g : A \rightarrow B$ in the natural way. I have also found what the zero object should be. But I cannot manage to construct the additive inverse of a morphism. Can someone help me on this matter?

Best Answer

Suppose $f\colon A\to B$. You get a map $M=\begin{pmatrix} 1 & f \\ 0 & 1\end{pmatrix}$ on $A\oplus B$. Let $(a,b)\colon K\to A\oplus B$ be the kernel. Thus the composite $(a,af+b)\colon K\to A\oplus B$ is the zero morphism, so $a=0$, and consequently $b=0$. Thus $M$ has zero kernel, so is a monomorphism.

Dually, let $\begin{pmatrix} a \\ b\end{pmatrix}\colon A\oplus B\to C$ be the cokernel of $M$. Hence the composite $\begin{pmatrix} a+fb \\ b\end{pmatrix}$ is the zero morphism, so $b=0$, and consequently $a=0$. So $M$ is an epimorphism, and hence an isomorphism since we are in an abelian category.

Let $M^{-1}=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ be the inverse with components some appropriate morphisms. Then one finds $$ \begin{pmatrix}a+fc & b+fd \\ c & d \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} $$ so that $c=0$, hence $a=1$, $d=1$, and $b+f=0$, so $f$ has an inverse under $+$.

Related Solutions

Finite Products vs Finite Coproducts in Additive Categories

Note that a product $A \times B$ is the same as a pull-back diagram

A x B -> B
  |      |
  v      v
  A ---> 0

with maps $p:A \times B \to A$ and $q: A \times B \to B$. In particular there is a map $i: A \to A \times B$ such that $pi = 1_{A}$ and $qi = 0$ as well as a map $j: B \to A \times B$ such that $qj = 1_{B}$ and $pj = 0$. Now note that $p(ip + jq) = pip + 0 = p$ and $q(ip + jq) = q$ so that $ip + jq = 1_{A \times B}$.

Let us check that $i: A \to A \times B$ and $j: B \to A \times B$ define a coproduct. Given $f: A \to D$ and $g: B \to D$ we get a map $d: A \times B \to D$ by setting $d = fp + gq$. Since $di = (fp + gq)i = fpi +gqi = f$ and $dj = g$ it remains to prove uniqueness of $d$. But this is clear as any other such map will satisfy $(d-d')1_{A \times B} = (d-d')(ip+jq) = 0$. Summing up, we have proved that the product of two objects is also a coproduct.

Now if we want to say that finite products and coproducts exist and coincide, we need a zero object, since otherwise the empty product (the terminal object) and the empty coproduct (the initial object) would not coincide.

Category Theory – Additive Category and Zero Map Explained

There is only one map $g : 0 \to d$. Therefore there is only one possibility for $g + g$. Thus,

$$gf + gf = (g+g)f = gf $$

for any map $f$ composable with $g$.

Strictly speaking, composition should be defined as abelian group homomorphisms $\hom(B,C) \otimes \hom(A,B) \to \hom(A,C)$ (with additional properties), so the image of $\hom(0,d) \otimes \hom(c,0) \to \hom(c,d)$ is automatically the zero map.

But as in group theory, some of the requirements of being a homomorphism follow automatically from others, and one often sees this 'simplified' definition instead.

Best Answer

Related Solutions

Finite Products vs Finite Coproducts in Additive Categories

Category Theory – Additive Category and Zero Map Explained

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