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Kernel and image are the inverse and direct images along a morphism

Statement

Let $f : A \to A^{'}$ be a morphism in an abelian category.

The inverse image of the zero subobject of $A^{'}$ along $f$ is $\ker (f)$ .
The direct image of the identity subobject of $A$ along $f$ is $im (f)$ .

Facts & Assumptions

Given: A morphism $f : A \to A^{'}$ .

[L1]

Inverse image is defined by pullback and direct image by ordinary image factorization (Direct and inverse image of a subobject).

[L2]

A subobject is represented by a monomorphism; in particular $0 \to A^{'}$ and $1_{A} : A \to A$ represent the zero and total subobjects (Subobject and quotient object as mutual-factorisation classes of monomorphisms and epimorphisms).

[L3]

The ordinary image of a morphism is defined as the kernel of a cokernel (Image and coimage in a category with kernels and cokernels).

Proof

technique · direct

1.1

Pulling back the zero subobject $0 \to A^{'}$ along $f$ produces exactly the kernel square of $f$ , so by [L1] and [L2] the inverse image $f^{*} (0)$ is $\ker (f)$ .

L1L2

1.2

The direct image of the identity subobject $1_{A}$ is, by [L1], the image of the composite $A \overset{1_{A}}{\to} A \overset{f}{\to} A^{'}$ , which is just the image of $f$ in the sense of [L3].

L1L2L3

2.1

Therefore kernels and images are exactly inverse and direct images along the morphism $f$ .

step 1.1step 1.2∎

Depends on

Used by

Images and preimages of submodules form a concrete Galois connection Example

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Saunders Mac Lane, Categories for the Working Mathematician, Section VIII.3 (standard reference, not scraped)