Commutative Algebra/Diagram chasing within Abelian categories

Exact sequences of Abelian groups
Note that if one of the objects is the trivial group, we denote it by $$0$$ and simply leave out the caption of the arrows going to it and emitting from it, since the trivial group is the zero object in the category of Abelian groups.

There are also infinite exact sequences, indicated by a notation of the form
 * $$A_1 \overset{\varphi_1}{\longrightarrow} A_2 \overset{\varphi_2}{\longrightarrow} \cdots \overset{\varphi_{n-2}}{\longrightarrow} A_{n-1} \overset{\varphi_{n-1}}{\longrightarrow} A_n \overset{\varphi_n}{\longrightarrow} \cdots$$;

it just goes on and on and on. The exact sequence to be infinite means, that we have a sequence (in the classical sense) of objects and another classical sequence of morphisms between these objects (here, the two have same cardinality: Countably infinite).

There is a fundamental example to this notion.

The exactness of this sequence means, considering the form of the image and kernel of the zero morphism:
 * 1) $$f$$ injective
 * 2) $$\ker g = \operatorname{im} f$$
 * 3) $$g$$ surjective.

Example 4.4:

Set $$A := \mathbb Z / 3 \mathbb Z$$, $$B := \mathbb Z / 15 \mathbb Z$$, $$C := \mathbb Z / 5 \mathbb Z$$, where we only consider the additive group structure, and define the group homomorphisms
 * $$f: A \to B, f(n + 3 \mathbb Z) := 5n + 15 \mathbb Z$$ and $$g: B \to C, g(n + 15 \mathbb Z) := n + 5 \mathbb Z$$.

This gives a short exact sequence
 * $$0 \longrightarrow A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} C \longrightarrow 0$$,

as can be easily checked.

A similar construction can be done for any factorisation of natural numbers $$k = m \cdot j$$ (in our example, $$k = 15$$, $$m = 3$$, $$j = 5$$).

Diagram chase: The short five lemma
We now should like to briefly exemplify a supremely important method of proof called diagram chase in the case of Abelian groups. We shall later like to generalize this method, and we will see that the classical diagram lemmas hold in huge generality (that includes our example below), namely in the generality of Abelian categories (to be introduced below).

Proof:

We first prove that $$f$$ is injective. Let $$f(b) = 0$$ for a $$b \in B$$. Since the given diagram is commutative, we have $$0 = t(f(b)) = h(q(b))$$ and since $$h$$ is an isomorphism, $$q(b) = 0$$. Since the top row is exact, it follows that $$b \in \operatorname{im} p$$, that is, $$b = p(a)$$ for a suitable $$a \in A$$. Hence, the commutativity of the given diagram implies $$0 = f(b) = f(q(a)) = s(g(a))$$, and hence $$a = 0$$ since $$s \circ g$$ is injective as the composition of two injective maps. Therefore, $$b = q(a) = q(0) = 0$$.

Next, we prove that $$f$$ is surjective. Let thus $$b' \in B'$$ be given. Set $$c' := t(b')$$. Since $$h \circ q$$ is surjective as the composition of two surjective maps, there exists $$b \in B$$ such that $$h(q(b)) = c'$$. The commutativity of the given diagram yields $$t(f(b)) = c'$$. Thus, $$t(f(b) - b') = 0$$ by linearity, whence $$f(b) - b' \in \ker t = \operatorname{im} s$$, and since $$g$$ is an isomorphism, we find $$a \in A$$ such that $$s(g(a)) = f(b) - b'$$. The commutativity of the diagram yields $$f(b) - b' = s(g(a)) = f(p(a))$$, and hence $$f(b - p(a)) = b'$$.

Additive categories
Although additive categories are important in their own right, we shall only treat them as in-between step to the definition of Abelian categories.

Abelian categories
We now embark to obtain a canonical factorisation of arrows within Abelian categories.

Lemma 4.8:

Let $$\mathcal C$$ be a category with a zero object and kernels and cokernels for all arrows. Then every arrow $$f$$ of $$\mathcal C$$ admits a factorisation
 * $$f = kq$$,

where $$k = \ker(\operatorname{coker} f))$$.

Proof:

The factorisation comes from the following commutative diagram, where we call $$u := \operatorname{coker} f$$ and $$k := \ker(\operatorname{coker} f)$$:
 * Factorisation.svg

Indeed, by the property of $$k$$ as a kernel and since $$u \circ f = 0$$, $$f$$ factors uniquely through $$k$$.

In Abelian categories, $$q$$ is even a monomorphism:

Lemma 4.9:

Let $$\mathcal C$$ be an Abelian category. If $$k = \ker(\operatorname{coker} f)$$ and we have any factorisation $$f = kq$$, then $$q$$ is an epimorphism.

Proof:

Exact sequences in Abelian categories
We begin by defining the image of a morphism in a general context.

Construction 4.13:

We shall now construct an equivalence relation on the set $$P_c$$ of all morphisms whose codomain is a certain $$c \in \mathcal C$$, where $$\mathcal C$$ is a category. We set
 * $$f \le g :\Leftrightarrow f = gf'$$ for a suitable $$f'$$ (that is, $$f$$ factors through $$g$$).

This relation is transitive and reflexive. Hence, if we define
 * $$f \sim g :\Leftrightarrow f \le g \wedge g \le f$$,

we have an equivalence relation (in fact, in this way we can always construct an equivalence relation from a transitive and reflexive binary relation, that is, a preorder).

With the image at hand, we may proceed to the definition of sequences, exact sequences and short exact sequences in a general context.

Diagram chase within Abelian categories
Now comes the clincher we have been working towards. In the ordinary diagram chase, we used elements of sets. We will now replace those elements by arrows in a simple way: Instead of looking at "elements" "$$x \in a$$" of some object $$a$$ of an abelian category $$\mathcal C$$, we look at arrows towards that element; that is, arrows $$x: d \to a$$ for arbitrary objects $$d$$ of $$\mathcal C$$. For "the codomain of an arrow $$x$$ is $$a$$", we write
 * $$x \in_m a$$,

where the subscript $$m$$ stands for "member".

We have now replaced the notion of elements of a set by the notion of members in category theory. We also need to replace the notion of equality of two elements. We don't want equality of two arrows, since then we would not obtain the usual rules for chasing diagrams. Instead, we define yet another equivalence relation on arrows with codomain $$a$$ (that is, on members of $$a$$). The following lemma will help to that end.

Lemma 4.18 (square completion):

Construction 4.19 (second equivalence relation):

Now we are finally able to prove the proposition that will enable us doing diagram chases using the techniques we apply also to diagram chases for Abelian groups (or modules, or any other Abelian category).

We have thus constructed a relatively elaborate machinery in order to elevate our proof technique of diagram chase (which is quite abundant) to the very abstract level of Abelian categories.