LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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Homotopy relative to a subspace is reflexive and symmetric

Statement

Let $A\subseteq X$ . Every continuous map $f:X\to Y$ is homotopic to itself rel $A$ . If $f\simeq_A g$ , then $g\simeq_A f$ .

Facts & Assumptions

Given: Topological spaces $X,Y$ , a subspace $A\subseteq X$ , continuous maps $f,g:X\to Y$ , and, for symmetry, a homotopy $H:X\times I\to Y$ from $f$ to $g$ rel $A$ .

[A1]

A homotopy rel $A$ is a continuous $K:X\times I\to Y$ with $K(x,0)$ and $K(x,1)$ the prescribed endpoint maps and $K(a,t)$ equal to their common value for every $a\in A$ and $t\in I$ (Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

[L1]

The projections from a product are continuous, and a map into a product is continuous exactly when all of 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)).

[L3]

In the usual topology of $\mathbb R$ , the open balls are the intervals $(u-r,u+r)$ ; $I=[0,1]$ has the subspace topology (The absolute value makes $\mathbb{R}$ 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

1.1

The projection $p_X:X\times I\to X$ is continuous by [L1]. For every open $V\subseteq Y$ , $(f\circ p_X)^{-1}[V]=p_X^{-1}[f^{-1}[V]]$ is open, so $K_f(x,t):=f(x)$ is continuous by [L2].

L1L2

1.2

The map $r:I\to I$ , $r(t)=1-t$ , is continuous: for $t_0\in I$ and an open neighbourhood $V=O\cap I$ of $r(t_0)$ , with $O$ open in $\mathbb R$ , [L3] gives $\varepsilon>0$ with $(r(t_0)-\varepsilon,r(t_0)+\varepsilon)\subseteq O$ ; then $U=(t_0-\varepsilon,t_0+\varepsilon)\cap I$ is an open neighbourhood of $t_0$ and $r[U]\subseteq V$ , since $|r(t)-r(t_0)|=|t-t_0|$ .

2.1

The homotopy $K_f$ has $K_f(x,0)=K_f(x,1)=f(x)$ and $K_f(a,t)=f(a)$ for $a\in A$ , so it is a homotopy from $f$ to itself rel $A$ .

step 1.1A1

2.2

The map $R:X\times I\to X\times I$ , $R(x,t)=(x,r(t))$ , is continuous because its components are continuous by step 1.2 and [L1]. For every open $V\subseteq Y$ , $(H\circ R)^{-1}[V]=R^{-1}[H^{-1}[V]]$ is open, so $\overline H:=H\circ R$ is continuous by [L2].

step 1.2L1L2

3.1

One has $\overline H(x,0)=H(x,1)=g(x)$ and $\overline H(x,1)=H(x,0)=f(x)$ ; for $a\in A$ , $\overline H(a,t)=H(a,1-t)=f(a)=g(a)$ . Hence $\overline H$ is a homotopy from $g$ to $f$ rel $A$ .

step 2.2A1∎

Depends on

Used by

Homotopy relative to a fixed subspace, and path homotopy relative to endpoints, are equivalence relations Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 13 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)
Homotopy lecture notes (University of Padua) (standard reference, not scraped)