TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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.

Precomposition and postcomposition by continuous maps preserve homotopies, including their relative form

Statement

Let $f,g:X\to Y$ be continuous and suppose $f\simeq_A g$ for a subspace $A\subseteq X$ .

If $u:W\to X$ is continuous, $B\subseteq W$ , and $u[B]\subseteq A$ , then $f\circ u\simeq_B g\circ u$ .
If $v:Y\to Z$ is continuous, then $v\circ f\simeq_A v\circ g$ .

Facts & Assumptions

Given: A homotopy $H:X\times I\to Y$ from $f$ to $g$ rel $A$ , a continuous $u:W\to X$ with $u[B]\subseteq A$ , and a continuous $v:Y\to Z$ .

[A1]

The endpoint and relative conditions are those of Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints.

[L1]

Product projections are continuous, and a map into a product is continuous exactly when its components are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[L2]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ , condition (b)).

Proof

technique · direct

1.1

Define $U:W\times I\to X\times I$ by $U(w,t)=(u(w),t)$ . Its components are $u$ after the first projection and the second projection, so $U$ is continuous by [L1].

1.2

The composite $v\circ H:X\times I\to Z$ is continuous by the same preimage calculation, and its endpoints are $v\circ f$ and $v\circ g$ ; for $a\in A$ it has the fixed value $v(f(a))=v(g(a))$ . Thus it is a homotopy from $v\circ f$ to $v\circ g$ rel $A$ .

A1L2

2.1

The composite $H\circ U$ is continuous: for every open $V\subseteq Y$ , $(H\circ U)^{-1}[V]=U^{-1}[H^{-1}[V]]$ is open by [L2]. Its endpoints are $(f\circ u)(w)$ and $(g\circ u)(w)$ ; if $b\in B$ , then $u(b)\in A$ , so $(H\circ U)(b,t)=f(u(b))=g(u(b))$ . Thus it is a homotopy from $f\circ u$ to $g\circ u$ rel $B$ .

step 1.1A1L2

3.1

Steps 2.1 and 1.2 prove the two claims.

step 2.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 12 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

A. Hatcher, Algebraic Topology, Section 0 (standard reference, not scraped)
Algebraic Topology lecture notes (UC Riverside) (standard reference, not scraped)