LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Pushforward sends ultrafilters to ultrafilters and is functorial

Statement

For a function $f : X \to Y$ and an ultrafilter $U$ on $X$ , define

$f_{*} U = {B \subseteq Y : f^{- 1} [B] \in U} .$

Then $f_{*} U$ is an ultrafilter on $Y$ . Moreover, $(1_{X})_{*} = 1_{β X}$ and $(g \circ f)_{*} = g_{*} \circ f_{*}$ , where $β X$ denotes the set of ultrafilters on $X$ .

Facts & Assumptions

Given: A function $f : X \to Y$ and an ultrafilter $U$ on $X$ .

[L1]

An ultrafilter is a proper filter maximal among proper filters (Ultrafilter).

[L2]

A proper filter is an ultrafilter exactly when, for every subset $A$ , it contains either $A$ or its complement (Characterisation of ultrafilters: every set or its complement).

Proof

technique · direct

1.1

Inverse image preserves the whole set, inclusions, and finite intersections and sends the empty set to the empty set. Consequently the displayed family contains $Y$ , excludes $\emptyset$ , is upward closed, and is closed under finite intersections, so it is a proper filter.

2.1

For $B \subseteq Y$ , [L2] applied to $f^{- 1} [B]$ says that either $f^{- 1} [B] \in U$ or $X ∖ f^{- 1} [B] = f^{- 1} [Y ∖ B] \in U$ . Thus $f_{*} U$ decides every subset of $Y$ and is an ultrafilter by [L2].

L2step 1.1

3.1

The equalities $(1_{X})^{- 1} [A] = A$ and $(g \circ f)^{- 1} [C] = f^{- 1} [g^{- 1} [C]]$ show membership-by-membership that $(1_{X})_{*} U = U$ and $(g \circ f)_{*} U = g_{*} (f_{*} U)$ .

step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Riehl, Category Theory in Context, 2nd ed., Exercise 5.1.ii (standard reference, not scraped)