Category Theory – Additive Category and Zero Map Explained

abelian-categoriesabelian-groupscategory-theoryhomological-algebra

$A$ has a zero object,
$A$ has finite products and coproducts, and
Every Hom-set is an Abelian group such that composition of morphisms is bilinear.

Question: Is the zero morphism always the additive unit?

Since $A$ has a zero object, say $0$, for any two objects $c$ and $d$, there uniquely exists a morphism $c \to d$ called the zero morphism which can be decomposed as $c \to 0 \to d$.

For typical examples (e.g. (Ab), ($R$-Mod), (Chain complexes in $R$-Mod)), the zero map gives the additive unit of each Hom-set. But the axioms of additive categories do not seem to say that the zero maps give the additive units of Hom-sets.

Is it possible to show that the zero morphism is the additive unit of each Hom-set? Or are there any additive categories such that the zero morphism is not the additive unit of some Hom-set? How about abelian categories?

Best Answer

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.

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.

[Math] Are adjoint functors between additive categories additive

Yes this is true. Specifically:

Proposition. Let $L:\mathcal B\to\mathcal A$ be left adjoint to $R:\mathcal A\to\mathcal B$ where $\mathcal A$ and $\mathcal B$ are additive categories. Then $L$ and $R$ are additive.

Proof. Recall that $F:\mathcal A\to\mathcal B$ is additive if and only if $F$ preserves binary products (prove this!). But $R$ is right adjoint so $R$ preserves limits. In particular, $R$ preserves binary products.

Proving $L$ is additive is similar once combined with the observation that binary products coincide with binary coproducts in an additive category. $\Box$

Of course, if you don't like using continuity properties of adjoint functors, you could directly prove the proposition. This is a nice exercise too.

Side-note. This is a useful result in geometry as a morphism $f:X\to Y$ of spaces defines functors $\DeclareMathOperator{Sh}{Sh}$ \begin{align*} f_* &:\Sh(X)\to\Sh(Y) \\ f^{-1} &: \Sh(Y)\to\Sh(X) \end{align*} One shows that $f^{-1}$ is left adjoint to $f_*$ and the above ensures the additivity of both.

In fact, one may further prove that right adjoint functors between abelian categories are left exact and left adjoint functors between abelian categories are right exact.

Best Answer

Related Solutions

Finite Products vs Finite Coproducts in Additive Categories

[Math] Are adjoint functors between additive categories additive

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