Math 2.0

Algebraic topology Definition

Reduced homology

Builds on Singular homologyRelative homology

Definition

Let XX be a nonempty topological space with singular chain complex C∙(X)C_\bullet(X). The augmentation is the homomorphism ε ⁣:C0(X)→ℤ\varepsilon \colon C_0(X) \to \Z given by ε(∑iniσi)=∑ini\varepsilon\left(\sum_i n_i \sigma_i\right) = \sum_i n_i. The augmented chain complex is ⋯→C2(X)→∂2C1(X)→∂1C0(X)→εℤ→0,\cdots \to C_2(X) \xrightarrow{\partial_2} C_1(X) \xrightarrow{\partial_1} C_0(X) \xrightarrow{\varepsilon} \Z \to 0, with ℤ\Z placed in degree −1-1; it is a chain complex because ε∘∂1=0\varepsilon \circ \partial_1 = 0. The reduced homology H~n(X)\tilde H_n(X) is the homology of the augmented chain complex in degree nn.

Remarks

For n≥1n \ge 1, H~n(X)=Hn(X)\tilde H_n(X) = H_n(X), where HnH_n is singular homology. In degree 00, H0(X)≅H~0(X)⊕ℤH_0(X) \cong \tilde H_0(X) \oplus \Z, and H~−1(X)=0\tilde H_{-1}(X) = 0 because ε\varepsilon is onto. For any point x0∈Xx_0 \in X, H~n(X)≅Hn(X,{x0})\tilde H_n(X) \cong H_n(X, \set{x_0}) for all nn, a relative homology group.

Example: for a one-point space XX, Cn(X)≅ℤC_n(X) \cong \Z for every n≥0n \ge 0, and ε\varepsilon is an isomorphism, so H~n(X)=0\tilde H_n(X) = 0 for all nn.

Non-example: the hypothesis that XX is nonempty matters. For X=∅X = \emptyset the map ε ⁣:0→ℤ\varepsilon \colon 0 \to \Z is not onto, so the augmented complex gives H~−1(∅)=ℤ\tilde H_{-1}(\emptyset) = \Z, and the formula H0≅H~0⊕ℤH_0 \cong \tilde H_0 \oplus \Z fails.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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  1. Relative homology concept
  2. Singular homology concept
  3. Reduced homology definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Reduced_homology
About this page
TagSA493EH7DT
URIm2://SA493EH7DT
Kinddefinition
WritingW0 stub
MSC55N10
DefinesReduced homology
UsesSingular homology, Relative homology
Changed2 November 2026

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