TheoremStatement: AI-adaptedProof: AI-adaptedprecheck 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 $f ≃_{A} g$ for a subspace $A \subseteq X$ .

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

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 · two levels

19 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

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