TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Direct and inverse image of subobjects form a Galois connection

Statement

Let $f : A \to A^{'}$ be a morphism in an abelian category. Then the direct-image map

$f_{*} : Sub (A) \to Sub (A^{'})$

and the inverse-image map

$f^{*} : Sub (A^{'}) \to Sub (A)$

form a Galois connection:

$f_{*} B \leq C ⟺ B \leq f^{*} C .$

Facts & Assumptions

Given: A morphism $f : A \to A^{'}$ and subobjects $B \leq A$ , $C \leq A^{'}$ .

[L1]

Direct image and inverse image are defined by image factorization and pullback respectively (Direct and inverse image of a subobject).

[L2]

A Galois connection between preorders is exactly a pair of monotone maps satisfying the displayed biconditional (Galois connection between preorders).

[L3]

The image of a morphism is the least subobject through which that morphism factors (The image is the least subobject through which a morphism factors).

Proof

technique · direct

1.1

Assume $f_{*} B \leq C$ . By [L1], the composite $B ↣ A \overset{f}{\to} A^{'}$ factors through the subobject $C ↣ A^{'}$ . The pullback defining $f^{*} C$ therefore gives a factorization of $B$ through $f^{*} C$ , so $B \leq f^{*} C$ .

L1construct

1.2

Assume $B \leq f^{*} C$ . Composing with the pullback leg $f^{*} C \to C$ shows that the composite $B \to A^{'}$ factors through $C$ . By [L3], the image $f_{*} B$ is the least subobject of $A^{'}$ with that property, so $f_{*} B \leq C$ .

L1L3

2.1

Steps 1.1 and 1.2 prove the displayed biconditional, which is exactly the Galois-connection condition of [L2].

L2step 1.1step 1.2∎

Depends on

Used by

Cited to discharge well-definedness by Direct and inverse image of a subobject.

Dependency tree · two levels

11 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 V.7 (standard reference, not scraped)