LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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Direct and inverse image satisfy Beck–Chevalley for pullback squares of sets

Statement

Let

$\begin{array}{ccc} P & \overset{q}{\to} & Y \\ p ↓ & ↓ g \\ X & \overset{f}{\to} & Z \end{array}$

be a pullback square of sets. For every $A \subseteq X$ ,

$q [p^{- 1} [A]] = g^{- 1} [f [A]] .$

Equivalently, direct image and inverse image satisfy $q_{!} p^{*} = g^{*} f_{!}$ on power sets. The formula remains valid for empty fibres and identity pullbacks.

Facts & Assumptions

Given: The displayed pullback square and a subset $A \subseteq X$ .

[L1]

A pullback of $X \overset{f}{\to} Z \overset{g}{\leftarrow} Y$ has projections satisfying $f p = g q$ and the universal property for every compatible pair (Pullbacks and pushouts as limits and colimits of cospans and spans).

[L2]

Membership in a direct image or inverse image is witnessed by the corresponding relation equation (The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

Proof

technique · direct

1.1

If $y \in q [p^{- 1} [A]]$ , then some $w \in P$ satisfies $q (w) = y$ and $p (w) \in A$ . The pullback equation gives $g (y) = g q (w) = f p (w)$ , so $g (y) \in f [A]$ and $y \in g^{- 1} [f [A]]$ .

L1L2

1.2

Conversely, if $y \in g^{- 1} [f [A]]$ , choose $x \in A$ with $f (x) = g (y)$ . The pullback universal property supplies the unique $w \in P$ with $p (w) = x$ and $q (w) = y$ , so $y \in q [p^{- 1} [A]]$ . If the fibre is empty, both existential conditions fail.

L1L2construct

2.1

Steps 1.1 and 1.2 prove equality. Identity squares give the identity direct and inverse images; if one map is a section of the other, the same equality specializes to the usual section–retraction image formulas.

step 1.1step 1.2∎

Depends on

Used by

The contravariant power-set functor is monadic Theorem

Dependency tree · two levels

10 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

E. Riehl, Category Theory in Context, 2nd ed., Lemma 5.5.10 (standard reference, not scraped)
D. Mehrle, Category Theory Part III, Exercise 5.22 (standard reference, not scraped)