Questions: Singular Homology Groups

Question 1

Two spaces X and Y are homotopy equivalent (there exist maps f: X -> Y and g: Y -> X with g∘f ≃ id_X and f∘g ≃ id_Y). What can you conclude about their singular homology groups?

• A. H_n(X) and H_n(Y) are isomorphic for all n
• B. H_n(X) and H_n(Y) have the same rank but may differ in torsion
• C. H_0(X) ≅ H_0(Y) but higher homology groups may differ
• D. Nothing — homotopy equivalence does not constrain homology

Answer: (A)

Homotopy invariance is a fundamental property of singular homology: homotopic maps induce identical homomorphisms on homology. If f and g are homotopy inverses, then f_* and g_* are inverse isomorphisms on H_n for every n. This is stronger than topological invariance (homeomorphism invariance) and has powerful consequences: for example, any contractible space has the homology of a point (H_0 ≅ Z, H_n = 0 for n > 0), because contractible means homotopy equivalent to a point.

Question 2

The singular homology of a point is H_0 ≅ Z and H_n = 0 for all n > 0.

• T. True
• F. False

Answer: True

A point has exactly one singular 0-simplex (the unique map from Delta^0 to the point), and for each n > 0, it has exactly one singular n-simplex (the unique constant map from Delta^n to the point). The chain complex is Z -> Z -> Z -> ... with alternating boundary maps that are either zero or the identity. Working out the kernels and images: ker(d_0) = Z, im(d_1) = 0, so H_0 = Z. For n > 0, the cycles and boundaries cancel to give H_n = 0. Since any contractible space is homotopy equivalent to a point, this also gives the homology of R^n, D^n, and any convex subset of Euclidean space.

Question 3

A continuous map f: X -> Y always induces a surjective homomorphism f_*: H_n(X) -> H_n(Y).

• T. True
• F. False

Answer: False

Functoriality guarantees that f induces a homomorphism f_* on homology, but this homomorphism need not be surjective (or injective). For example, the inclusion of a point into S^1 induces the zero map on H_1, which is not surjective. The constant map from any space to a point induces the zero map on all H_n with n > 0. Only special maps (like homotopy equivalences, or maps with degree +/- 1 on spheres) induce surjections. The power of functoriality is that f_* exists and respects composition, not that it preserves all algebraic structure.

Question 4

Explain intuitively why homotopy equivalent spaces must have isomorphic homology groups, using the fact that homology counts 'holes.'

Model answer: Homotopy equivalence means two spaces can be continuously deformed into each other (in a looser sense than homeomorphism — dimensions can collapse as long as they can be re-inflated). If X deforms to Y, every hole in X deforms to a corresponding hole in Y, and vice versa. Since homology counts holes (by dimension: H_0 counts components, H_1 counts tunnels, H_2 counts cavities), the counts must match. Formally, the induced maps f_* and g_* on homology are mutual inverses, giving isomorphisms H_n(X) ≅ H_n(Y) for all n.

The formal proof uses the chain homotopy concept: homotopic maps f ≃ g induce chain-homotopic chain maps, which induce identical maps on homology. The key technical lemma (the prism operator) constructs an explicit chain homotopy from a homotopy between maps. This is one of the deepest results in algebraic topology, because it shows that homology depends only on the 'shape' of a space up to continuous deformation, not on metric, dimension, or other geometric details.

Question 5

For a path-connected space X, H_0(X) ≅ Z. What is the geometric meaning of the isomorphism?

Model answer: The isomorphism sends the homology class of any singular 0-simplex (point) to 1 ∈ Z. This is called the augmentation map. Since X is path-connected, any two points p and q are connected by a path gamma, and d_1(gamma) = q - p, so p and q differ by a boundary and represent the same class in H_0. Thus H_0(X) has a single generator [p] for any point p, and is isomorphic to Z. The generator corresponds to the 'connected component' that is the entire space.

For a space with k path components, H_0 ≅ Z^k, with one generator per component. Two 0-cycles are homologous if and only if they assign the same total 'count' to each component. This is the simplest instance of homology detecting topological structure, and it illustrates the general pattern: homology measures the failure of cycles to be boundaries.