PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Zero-th singular homology is free on path components

Statement

For every topological space $X$ , $H_{0}^{s i n g} (X; Z) ≅ ⨁_{C \in π_{0}^{p a t h} (X)} Z e_{C},$ the free abelian group on the path components of $X$ .

Facts & Assumptions

Given: A topological space $X$ .

[L1]

Singular homology in degree $0$ is $Z_{0} (X; Z) / B_{0} (X; Z)$ , and the reduced complex uses the augmentation kernel in degree $0$ (The singular chain complex and singular homology, Augmentation at 0-simplices and reduced singular homology).

[L2]

A path component is the equivalence class of the path relation (Paths, path-connected spaces and path components).

Proof

technique · direct

1.1L1L2given

Let $Φ : C_{0} (X; Z) \to ⨁_{C \in π_{0}^{p a t h} (X)} Z e_{C}$ send each singular $0$ -simplex $σ$ to the basis vector indexed by the path component of its unique image point. If $τ : Δ^{1} \to X$ is a singular $1$ -simplex, then the map $s \mapsto τ (1 - s, s)$ is a path in $X$ from the image point of $τ δ_{1}$ to the image point of $τ δ_{0}$ . Hence $Φ (\partial_{1} τ) = 0$ , so $Φ$ kills $B_{0} (X; Z)$ and descends to a homomorphism $\overline{Φ} : H_{0}^{s i n g} (X; Z) \to ⨁_{C \in π_{0}^{p a t h} (X)} Z e_{C} .$

2.1step 1.1L2

The map $\overline{Φ}$ is surjective: if $X = \emptyset$ there is nothing to prove, and otherwise any finite sum $\sum_{k = 1}^{m} a_{k} e_{C_{k}}$ is the image of the $0$ -cycle $\sum_{k = 1}^{m} a_{k} σ_{k}$ , where $σ_{k}$ is any singular $0$ -simplex landing at a chosen point of $C_{k}$ .

2.2L1L2step 1.1algebra

Let $z = \sum_{k = 1}^{m} a_{k} σ_{k}$ be a $0$ -cycle with $\overline{Φ} ([z]) = 0$ . For each path component that meets the finite support of $z$ , choose one base point simplex $σ_{C}$ . Since $\overline{Φ} ([z]) = 0$ , the total coefficient in each such component is zero. If $σ_{k}$ and $σ_{C}$ lie in the same path component, [L2] gives a path $γ_{k} : I \to X$ from the image point of $σ_{k}$ to the image point of $σ_{C}$ . Define the singular $1$ -simplex ${\overline{γ}}_{k} : Δ^{1} \to X$ by ${\overline{γ}}_{k} (t_{0}, t_{1}) = γ_{k} (t_{1})$ . Then $\partial_{1} {\overline{γ}}_{k} = σ_{C} - σ_{k}$ . Therefore $[z] = \sum_{C} (\sum_{σ_{k} \in C} a_{k}) [σ_{C}] = 0$ in $H_{0}^{s i n g} (X; Z)$ . Thus $\overline{Φ}$ is injective.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 show that $\overline{Φ}$ is an isomorphism, so $H_{0}^{s i n g} (X; Z)$ is free on the path components of $X$ .

Depends on

Used by

Dependency tree · two levels

18 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

Allen Hatcher, Algebraic Topology (standard reference, not scraped)