LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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Two homotopies relative to the same subspace concatenate after piecewise-linear reparametrisation

Statement

Let $A \subseteq X$ and let $f, g, h : X \to Y$ be continuous. If $F$ is a homotopy from $f$ to $g$ rel $A$ and $G$ is a homotopy from $g$ to $h$ rel $A$ , then

$K (x, t) : = {\begin{cases} F (x, 2 t), & 0 \leq t \leq \frac{1}{2}, \\ G (x, 2 t - 1), & \frac{1}{2} \leq t \leq 1 \end{cases}$

is a continuous homotopy from $f$ to $h$ rel $A$ .

Facts & Assumptions

Given: Topological spaces $X, Y$ , a subspace $A \subseteq X$ , continuous maps $f, g, h : X \to Y$ , and homotopies $F : f ≃_{A} g$ and $G : g ≃_{A} h$ .

[A1]

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

[L1]

A map is continuous exactly when preimages of closed sets are closed (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 (c)).

[L2]

In a subspace, closed sets are exactly traces of ambient closed sets; restrictions of continuous maps to subspaces are continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L3]

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).

[L4]

A finite union of closed sets is closed, and the complement of an open set is closed (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L5]

In the usual topology of $R$ , open balls are open intervals; $I = [0, 1]$ has the subspace topology (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct

1.1L1L3L4L5

The sets $I_{0} = [0, \frac{1}{2}]$ and $I_{1} = [\frac{1}{2}, 1]$ are closed in $I$ : their complements are respectively $I \cap (\frac{1}{2}, \frac{3}{2})$ and $I \cap (- \frac{1}{2}, \frac{1}{2})$ , traces of open intervals of $R$ . Hence $D_{0} = X \times I_{0}$ and $D_{1} = X \times I_{1}$ are closed in $X \times I$ , because they are the preimages of $I_{0}, I_{1}$ under the continuous time projection.

1.2L5

The maps $a_{0} : I_{0} \to I$ , $a_{0} (t) = 2 t$ , and $a_{1} : I_{1} \to I$ , $a_{1} (t) = 2 t - 1$ , are continuous. Indeed, at any $t_{0}$ and for any ambient open interval of radius $ε$ about $a_{j} (t_{0})$ , the relative interval of radius $ε / 2$ about $t_{0}$ maps into it, since $∣ a_{j} (t) - a_{j} (t_{0}) ∣ = 2 ∣ t - t_{0} ∣$ ; [L5] turns these intervals into the required subspace neighbourhoods.

2.1step 1.2L2L3

Define $r_{j} : D_{j} \to X \times I$ by $r_{j} (x, t) = (x, a_{j} (t))$ . The first component is the restricted product projection and the second is $a_{j}$ after the time projection, so $r_{j}$ is continuous by [L2], step 1.2 and [L3].

3.1step 2.1A1L1

The maps $K_{0} : = F \circ r_{0} : D_{0} \to Y$ and $K_{1} : = G \circ r_{1} : D_{1} \to Y$ are continuous: for every closed $C \subseteq Y$ , $K_{j}^{- 1} [C] = r_{j}^{- 1} [F^{- 1} [C]]$ or $r_{j}^{- 1} [G^{- 1} [C]]$ , which is closed by [L1]. On $D_{0} \cap D_{1} = X \times {\frac{1}{2}}$ they agree, since $K_{0} (x, \frac{1}{2}) = F (x, 1) = g (x) = G (x, 0) = K_{1} (x, \frac{1}{2})$ .

4.1step 1.1step 3.1L1L2L4

Thus the displayed clauses define one function $K : X \times I \to Y$ . If $C \subseteq Y$ is closed, then $K^{- 1} [C]$ is the union of $K_{0}^{- 1} [C]$ , regarded as a closed subset of $X \times I$ through the closed subspace $D_{0}$ , and $K_{1}^{- 1} [C]$ , regarded likewise through $D_{1}$ . Each is closed by [L2] and steps 1.1 and 3.1, so their union is closed by [L4]. Hence $K$ is continuous by [L1].

5.1step 4.1A1∎

At $t = 0$ the first clause gives $K (x, 0) = F (x, 0) = f (x)$ , and at $t = 1$ the second gives $K (x, 1) = G (x, 1) = h (x)$ . If $a \in A$ , both clauses give the common value $f (a) = g (a) = h (a)$ for every $t$ . Therefore $K$ is a homotopy from $f$ to $h$ rel $A$ .

Remarks

The continuity argument uses only a cover by the two closed sets $D_{0}, D_{1}$ and proves the finite pasting step directly. No assertion about an infinite closed cover is used.

Two homotopies relative to the same subspace concatenate after piecewise-linear reparametrisation

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources