kernel as left adjoint and cokernel as right adjoint

 

Let $\mathcal{A}$$\mathcal A$ be an abelian category. Consider the arrow category

$$\overrightarrow{\mathcal{A}}={\mathcal{A}}^{[1]}.$$

An object of $\overrightarrow{\mathcal{A}}$$\vec{\mathcal A}$ is an arrow

$$f:X\to Y.$$

A morphism

$$(X\stackrel{f}{\to }Y)⟶(X^{\prime }\stackrel{g}{\to }{Y}^{\prime })$$

is a commutative square

$$\begin{array}{ccc}X& \stackrel{u}{\to }& X^{\prime }\\ \downarrow f& & \downarrow g\\ Y& \stackrel{v}{\to }& Y^{\prime }\end{array}\text{with }gu=vf.$$

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1. Short exact sequences in the arrow category

Suppose we have a commutative diagram in $\mathcal{A}$$\mathcal A$:

$$\begin{array}{ccccccccc}0& \to & A& \to & B& \to & C& \to & 0\\ & & \downarrow f_1& & \downarrow f_2& & \downarrow f_3\\ 0& \to & A^{\prime }& \to & B^{\prime }& \to & C^{\prime }& \to & 0.\end{array}$$

Equivalently, we have three objects of the arrow category:

$$f_1:A\to A^{\prime },f_2:B\to B^{\prime },f_3:C\to C^{\prime }.$$

If the two rows are short exact, then this gives a short exact sequence in $\overrightarrow{\mathcal{A}}$$\vec{\mathcal A}$:

$$0\to f_1\to f_2\to f_3\to 0.$$

Applying kernel and cokernel to the vertical arrows gives the diagram

$$\begin{array}{ccccccc}0& \to & \ker f_1& \to & \ker f_2& \to & \ker f_3\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& \to & A& \to & B& \to & C& \to & 0\\ & & \downarrow f_1& & \downarrow f_2& & \downarrow f_3\\ 0& \to & A^{\prime }& \to & B^{\prime }& \to & C^{\prime }& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow \\ & & \mathrm{coker}{f}_1& \to & \mathrm{coker}{f}_2& \to & \mathrm{coker}{f}_3& \to & 0.\end{array}$$

The left part comes from left exactness of the kernel functor:

$$0\to \ker f_1\to \ker f_2\to \ker f_3.$$

The right part comes from right exactness of the cokernel functor:

$$\mathrm{coker}{f}_1\to \mathrm{coker}{f}_2\to \mathrm{coker}{f}_3\to 0.$$

The connecting morphism

$$\ker f_3⟶\mathrm{coker}{f}_1$$

is the extra content of the Snake Lemma.

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2. The kernel functor on the arrow category

Define

$$\ker :\overrightarrow{\mathcal{A}}\to \mathcal{A}$$

by

$$(X\stackrel{g}{\to }Y)⟼\ker g.$$

This functor admits a left adjoint

$$L:\mathcal{A}\to \overrightarrow{\mathcal{A}},T⟼(T\to 0).$$

Indeed, for every object $T\in \mathcal{A}$$T\in\mathcal A$ and every arrow $g:X\to Y$$g:X\to Y$, we have a natural bijection

$${\mathrm{Hom}}_{\overrightarrow{\mathcal{A}}}((T\to 0),(X\stackrel{g}{\to }Y))\cong {\mathrm{Hom}}_{\mathcal{A}}(T,\ker g).$$

A morphism

$$(T\to 0)⟶(X\stackrel{g}{\to }Y)$$

in the arrow category is a commutative square

$$\begin{array}{ccc}T& \stackrel{\alpha }{\to }& X\\ \downarrow & & \downarrow g\\ 0& \to & Y.\end{array}$$

Commutativity says

$$g\alpha =0.$$

Therefore $\alpha :T\to X$$\alpha:T\to X$ factors uniquely through $\ker g$$\ker g$:

$$\begin{array}{ccc}T& \stackrel{\tilde{\alpha }}{\to }& \ker g\\ & {\searrow }_{\alpha }& \downarrow \\ & & X.\end{array}$$

Hence

$${\mathrm{Hom}}_{\overrightarrow{\mathcal{A}}}((T\to 0),(X\stackrel{g}{\to }Y))\cong {\mathrm{Hom}}_{\mathcal{A}}(T,\ker g).$$

So

$$L⊣\ker .$$

Therefore $\ker$$\ker$ is a right adjoint, hence preserves limits. In particular, it is left exact.

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3. The cokernel functor is dual

Similarly, define

$$\mathrm{coker}:\overrightarrow{\mathcal{A}}\to \mathcal{A}$$

by

$$(X\stackrel{g}{\to }Y)⟼\mathrm{coker}g.$$

It has a right adjoint

$$R:\mathcal{A}\to \overrightarrow{\mathcal{A}},T⟼(0\to T).$$

Indeed,

$${\mathrm{Hom}}_{\overrightarrow{\mathcal{A}}}((X\stackrel{g}{\to }Y),(0\to T))\cong {\mathrm{Hom}}_{\mathcal{A}}(\mathrm{coker}g,T).$$

So

$$\mathrm{coker}⊣R.$$

Therefore $\mathrm{coker}$$\operatorname{coker}$ is a left adjoint, hence preserves colimits. In particular, it is right exact.

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4. Relation with the Snake Lemma

If the diagram

$$\begin{array}{ccccccccc}0& \to & A& \to & B& \to & C& \to & 0\\ & & \downarrow f_1& & \downarrow f_2& & \downarrow f_3\\ 0& \to & A^{\prime }& \to & B^{\prime }& \to & C^{\prime }& \to & 0\end{array}$$

has short exact rows, then the vertical arrows form a short exact sequence in the arrow category:

$$0\to f_1\to f_2\to f_3\to 0.$$

Applying $\ker$$\ker$ and $\mathrm{coker}$$\operatorname{coker}$ gives the left and right exact pieces:

$$0\to \ker f_1\to \ker f_2\to \ker f_3,$$

$$\mathrm{coker}{f}_1\to \mathrm{coker}{f}_2\to \mathrm{coker}{f}_3\to 0.$$

The Snake Lemma inserts the connecting morphism

$$\partial :\ker f_3\to \mathrm{coker}{f}_1$$

and gives the exact sequence

$$0\to \ker f_1\to \ker f_2\to \ker f_3\stackrel{\partial }{\to }\mathrm{coker}{f}_1\to \mathrm{coker}{f}_2\to \mathrm{coker}{f}_3\to 0.$$

In the weaker Snake Lemma diagram, the rows need not both be short exact. Then the three vertical arrows need not form a short exact sequence in $\overrightarrow{\mathcal{A}}$$\vec{\mathcal A}$, so kernel left exactness alone does not apply directly. The connecting morphism is obtained by the usual snake-chasing construction.