CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31

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A homotopy equivalence induces a bijection between path components

Statement

For a space $X$ , write $π_{0} (X)$ for its set of path components. If $f : X \to Y$ is a homotopy equivalence, then

$f_{*} : π_{0} (X) ⟶ π_{0} (Y), f_{*} (P_{X} (x)) : = P_{Y} (f (x)),$

is a well-defined bijection. A homotopy inverse $g : Y \to X$ induces its inverse $g_{*}$ .

Facts & Assumptions

Given: A homotopy equivalence $f : X \to Y$ with homotopy inverse $g : Y \to X$ .

[A1]

Path components are the equivalence classes for the relation “joined by a path” (Paths, path-connected spaces and path components).

[A2]

One has $g \circ f ≃ {id}_{X}$ and $f \circ g ≃ {id}_{Y}$ (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

[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 $f (\overline{A}) \subseteq \overline{f (A)}$ ).

[L3]

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

Proof

technique · direct

1.1

If a path $γ : I \to X$ joins $x$ to $x^{'}$ , then $f \circ γ$ is continuous because $(f \circ γ)^{- 1} [V] = γ^{- 1} [f^{- 1} [V]]$ for every open $V \subseteq Y$ ; it joins $f (x)$ to $f (x^{'})$ . Hence points in one path component of $X$ have images in one path component of $Y$ , so $f_{*}$ is well defined. The same argument defines $g_{*}$ .

A1L2

1.2

If continuous maps $u, v : X \to Y$ are homotopic, then $u (x)$ and $v (x)$ lie in the same path component for every $x \in X$ : precompose the homotopy by the continuous map from a one-point space selecting $x$ , using [L3]; the resulting homotopy of two maps from a point is exactly a path from $u (x)$ to $v (x)$ .

L1L3A1

2.1

Apply step 1.2 to $g \circ f ≃ {id}_{X}$ . For every $x \in X$ , $g (f (x))$ and $x$ lie in the same path component, so $(g_{*} \circ f_{*}) (P_{X} (x)) = P_{X} (x)$ . Thus $g_{*} \circ f_{*}$ is the identity on $π_{0} (X)$ .

step 1.2A2

2.2

Applying step 1.2 to $f \circ g ≃ {id}_{Y}$ similarly gives $f_{*} \circ g_{*} = {id}_{π_{0} (Y)}$ .

step 1.2A2

3.1

Therefore $f_{*}$ and $g_{*}$ are mutually inverse functions, so $f_{*}$ is a bijection.

step 2.1step 2.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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)