03.12.E2 · modern-geometry / homotopy

Singular and cellular homology exercise pack (Hatcher Ch. 2 supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

Hatcher's Chapter 2 builds ordinary homology three ways — simplicial, singular, and cellular — and proves they agree on CW complexes. The chapter's engine is the trio of Eilenberg-Steenrod axioms made operational: the long exact sequence of a pair, excision (packaged as Mayer-Vietoris), and the dimension axiom. With these in hand the chapter computes $H_{*} (S^{n})$ by induction, defines and computes the degree of a self-map of $S^{n}$ , and grinds out the cellular chain complexes of projective spaces, lens spaces, and Moore spaces.

This pack collects ten exercises drawn from §2.1, §2.2, and §2.C — three easy, four medium, three hard — each with a hint and a full solution. The exercises are grouped by technique: $Δ$ -complex and degree warm-ups (easy), Mayer-Vietoris and cellular-chain computations of standard spaces (medium), and the structural results that pin down homology up to isomorphism — the universal coefficient theorem in action and the homology of an infinite product of cells (hard).

The conventions are Hatcher's: $H_{n}$ denotes reduced or unreduced singular homology with $Z$ coefficients unless stated, and the cellular boundary map is computed by the degree formula $d_{n} ([e_{α}^{n}]) = \sum_{β} d_{α β} [e_{β}^{n - 1}]$ , where $d_{α β}$ is the degree of the attaching-then-collapse map $S_{α}^{n - 1} \to X^{n - 1} \to S_{β}^{n - 1}$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as the model solution. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Compute $H_(\mathbb{R}P^n; \mathbb{Z})$ from the cellular chain complex.*

Solution. Real projective $n$ -space has a CW structure with exactly one cell $e^{k}$ in each dimension $0 \leq k \leq n$ , where $R P^{k} = R P^{k - 1} \cup e^{k}$ and the attaching map $S^{k - 1} \to R P^{k - 1}$ is the standard double cover. The cellular chain complex is therefore

0 \to Z d_{n} Z d_{n - 1} \dots d_{1} Z \to 0,

one $Z$ in each degree $0$ through $n$ . The cellular boundary $d_{k}$ is multiplication by the degree of the composite $S^{k - 1} 2-fold cover R P^{k - 1} \to R P^{k - 1} / R P^{k - 2} = S^{k - 1}$ . That composite is $1 + (- 1)^{k}$ (the antipodal identification contributes the antipodal-map degree $(- 1)^{k}$ to the identity): so $d_{k} = 0$ for $k$ odd and $d_{k} = \pm 2$ for $k$ even.

The complex thus alternates $\dots 0 Z 2 Z 0 Z 2 Z 0 Z$ . Reading homology degree by degree:

$H_{0} = Z$ (path-connected).
For $0 < k < n$ odd: $H_{k} = ker (d_{k}) / im (d_{k + 1}) = Z /2 Z$ (kernel of $0$ is $Z$ , image of multiplication-by- $2$ is $2 Z$ ).
For $0 < k < n$ even: $H_{k} = 0$ (kernel of multiplication-by- $2$ is $0$ ).
$H_{n} = Z$ if $n$ is odd (the top map into degree $n$ is $0$ , and $d_{n}$ out of degree $n$ does not exist), and $H_{n} = 0$ if $n$ is even.

Summary: $H_{k} (R P^{n}) = Z$ for $k = 0$ ; $Z /2$ for $k$ odd, $0 < k < n$ ; $0$ for $k$ even, $0 < k < n$ ; and $Z$ in degree $n$ when $n$ is odd, $0$ when $n$ is even. $□$

This computation is the template for every cellular calculation downstream: write the cells, compute each boundary by a degree, read off homology. The lens-space and $C P^{n}$ exercises differ only in the cell dimensions and the boundary degrees.

Exercises Intermediate

Exercise 2 (easy, numeric). Degree of a self-map of $S^{n}$ .

Let $f : S^{n} \to S^{n}$ be a map and $f_{*} : H_{n} (S^{n}) \to H_{n} (S^{n})$ the induced map on top homology, identified with multiplication by an integer $de g f$ . Compute $de g f$ for (a) the identity, (b) a constant map, (c) a reflection through a hyperplane, and (d) the antipodal map.

Hint

The identity induces the identity; a constant map factors through a point. A reflection is one coordinate sign-change; the antipodal map is $n + 1$ of them.

Answer

(a) $de g (id) = 1$ , since $id_{*} = id$ on $H_{n} (S^{n}) = Z$ .

(b) A constant map factors as $S^{n} \to pt \to S^{n}$ . Since $H_{n} (pt) = 0$ for $n \geq 1$ , $f_{*} = 0$ , so $de g f = 0$ .

(c) A reflection $r$ swaps the two hemispheres and reverses orientation; on $H_{n} (S^{n})$ it acts by $- 1$ , so $de g r = - 1$ .

(d) The antipodal map $a$ is the composite of $n + 1$ coordinate reflections, so by multiplicativity of degree $de g a = (- 1)^{n + 1}$ . This is Hatcher §2.2.

Exercise 3 (easy, proof). Simplicial equals singular for a $Δ$ -complex.

State the theorem that simplicial homology of a $Δ$ -complex agrees with its singular homology, and use it to compute $H_{*} (T^{2})$ from the standard two-triangle $Δ$ -complex structure on the torus.

Hint

The torus is one vertex, three edges $a, b, c$ , two 2-simplices $U, L$ with $\partial U = \partial L = a + b - c$ .

Answer

Theorem (Hatcher 2.27). For a $Δ$ -complex $X$ , the inclusion of simplicial into singular chains induces an isomorphism $H_{n}^{Δ} (X) ≅ H_{n} (X)$ for all $n$ .

The standard torus $Δ$ -structure has one vertex $v$ , three edges $a, b, c$ (all loops), and two triangles $U, L$ with $\partial U = \partial L = a + b - c$ . Chain groups: $C_{0} = Z$ , $C_{1} = Z^{3}$ , $C_{2} = Z^{2}$ .

$\partial_{1} = 0$ (every edge is a loop, $\partial (edge) = v - v = 0$ ). $\partial_{2} (U) = \partial_{2} (L) = a + b - c$ , so $im \partial_{2} = Z ⟨ a + b - c ⟩$ and $ker \partial_{2} = Z ⟨ U - L ⟩$ .

$H_{0} = Z$ .
$H_{1} = ker \partial_{1} / im \partial_{2} = Z^{3} / ⟨ a + b - c ⟩ ≅ Z^{2}$ .
$H_{2} = ker \partial_{2} = Z$ (generated by $U - L$ , the fundamental class).

This recovers $H_{*} (T^{2}) = (Z, Z^{2}, Z)$ .

Exercise 4 (medium, numeric). Homology of the lens space $L (p; q)$ .

For coprime integers $p \geq 1$ and $q$ , the three-dimensional lens space $L (p; q) = S^{3} / Z_{p}$ has a CW structure with one cell in each dimension $0, 1, 2, 3$ . The cellular boundary maps are $d_{1} = 0$ , $d_{2} = \times p$ , $d_{3} = 0$ . Compute $H_{*} (L (p; q); Z)$ .

Hint

Write the chain complex $Z 0 Z p Z 0 Z$ and read homology degree by degree.

Answer

The cellular chain complex is

0 \to C_{3} Z 0 C_{2} Z p C_{1} Z 0 C_{0} Z \to 0.

$H_{0} = Z$ (connected).
$H_{1} = ker (0) / im (p) = Z / p Z$ .
$H_{2} = ker (p) / im (0) = 0$ (multiplication by $p$ is injective on $Z$ ).
$H_{3} = ker (0) = Z$ (the fundamental class; $L (p; q)$ is a closed orientable 3-manifold).

So $H_{*} (L (p; q)) = (Z, Z / p, 0, Z)$ . Note the homology is independent of $q$ — the integers $q$ and $q^{'}$ give homotopy-distinct but homology-equivalent lens spaces, which is why lens spaces are Hatcher's flagship example that homology does not determine homotopy type. Hatcher §2.2 Example 2.43.

Exercise 5 (medium, proof). Mayer-Vietoris for the wedge.

Use the Mayer-Vietoris sequence to prove that for well-pointed spaces, $\tilde{H}_{n} (X \lor Y) ≅ \tilde{H}_{n} (X) \oplus \tilde{H}_{n} (Y)$ for all $n$ .

Hint

Fatten the wedge point: take $A$ a neighborhood of $X$ deformation-retracting to $X$ , $B$ a neighborhood of $Y$ , with $A \cap B$ contractible.

Answer

Choose open sets $A \supset X$ and $B \supset Y$ in $X \lor Y$ such that $A$ deformation-retracts to $X$ , $B$ deformation-retracts to $Y$ , and $A \cap B$ is contractible (a neighborhood of the basepoint). Well-pointedness guarantees such neighborhoods exist. Then $A \cup B = X \lor Y$ and the reduced Mayer-Vietoris sequence reads

\tilde{H}_{n} (A \cap B) \to \tilde{H}_{n} (A) \oplus \tilde{H}_{n} (B) \to \tilde{H}_{n} (X \lor Y) \to \tilde{H}_{n - 1} (A \cap B) .

Since $A \cap B$ is contractible, $\tilde{H}_{*} (A \cap B) = 0$ , so the outer terms vanish and the middle map is an isomorphism:

\tilde{H}_{n} (X) \oplus \tilde{H}_{n} (Y) = \tilde{H}_{n} (A) \oplus \tilde{H}_{n} (B) ≅ \tilde{H}_{n} (X \lor Y) .

The deformation retractions identify $\tilde{H}_{*} (A) = \tilde{H}_{*} (X)$ and $\tilde{H}_{*} (B) = \tilde{H}_{*} (Y)$ . This is the homology analogue of the van Kampen splitting and underlies the computation of homology of any wedge of spheres. Hatcher §2.2.

Exercise 6 (medium, numeric). Homology of a Moore space.

The Moore space $M (Z / m, n)$ is built from $S^{n}$ by attaching an $(n + 1)$ -cell via a degree- $m$ map $S^{n} \to S^{n}$ . Compute its reduced homology $\tilde{H}_{*} (M (Z / m, n); Z)$ .

Hint

The cellular chain complex has $C_{n + 1} = Z$ , $C_{n} = Z$ , with boundary the degree- $m$ map, i.e. multiplication by $m$ .

Answer

The CW structure has one $0$ -cell, one $n$ -cell, and one $(n + 1)$ -cell (for $n \geq 1$ ). The reduced cellular chain complex has $\tilde{C}_{n + 1} = Z$ , $\tilde{C}_{n} = Z$ , and the boundary $d_{n + 1} : C_{n + 1} \to C_{n}$ is multiplication by the degree $m$ . All other chain groups vanish.

$\tilde{H}_{n + 1} = ker (\times m) = 0$ .
$\tilde{H}_{n} = Z / m Z$ (cokernel of multiplication by $m$ ).
$\tilde{H}_{k} = 0$ for all other $k$ .

So $M (Z / m, n)$ has a single nonzero reduced homology group $Z / m$ concentrated in degree $n$ — exactly the defining property of a Moore space. This is the homology dual of the Eilenberg-MacLane space and the building block for arbitrary homology via wedges and telescopes. Hatcher §2.2.

Exercise 7 (medium, proof). Euler characteristic from homology.

Prove that for a finite CW complex $X$ , the alternating sum of cell counts equals the alternating sum of Betti numbers: $\sum_{n} (- 1)^{n} c_{n} = \sum_{n} (- 1)^{n} rank H_{n} (X)$ .

Hint

Apply rank-nullity to each cellular boundary map. The cellular chain group $C_{n}$ has rank $c_{n}$ .

Answer

The cellular chain complex $(C_{*}, d_{*})$ has $rank C_{n} = c_{n}$ (the number of $n$ -cells). Let $Z_{n} = ker d_{n}$ and $B_{n} = im d_{n + 1}$ , with ranks $z_{n}$ and $b_{n}$ . Rank-nullity for $d_{n} : C_{n} \to C_{n - 1}$ gives

c_{n} = z_{n} + b_{n - 1} .

The rank of $H_{n} = Z_{n} / B_{n}$ is $z_{n} - b_{n}$ (rank is additive on the short exact sequence $0 \to B_{n} \to Z_{n} \to H_{n} \to 0$ , modulo torsion). Now compute the alternating sum:

n \sum (- 1)^{n} c_{n} = n \sum (- 1)^{n} (z_{n} + b_{n - 1}) = n \sum (- 1)^{n} z_{n} + n \sum (- 1)^{n} b_{n - 1} .

Reindex the second sum: $\sum_{n} (- 1)^{n} b_{n - 1} = - \sum_{n} (- 1)^{n} b_{n}$ . Hence

n \sum (- 1)^{n} c_{n} = n \sum (- 1)^{n} (z_{n} - b_{n}) = n \sum (- 1)^{n} rank H_{n} (X) .

The middle sum telescopes the boundary ranks away, leaving the homological Euler characteristic. This is the homological invariance of $χ$ — the alternating cell count is a homotopy invariant. Hatcher §2.2 Theorem 2.44.

Exercise 8 (hard, proof). Universal coefficients in action.

State the universal coefficient theorem for homology and use it to compute $H_{*} (R P^{3}; Z /2)$ from the integral homology $H_{*} (R P^{3}; Z) = (Z, Z /2, 0, Z)$ .

Hint

The UCT gives $H_{n} (X; G) ≅ (H_{n} (X) \otimes G) \oplus Tor (H_{n - 1} (X), G)$ . Compute each tensor and each Tor against $G = Z /2$ .

Answer

Theorem (UCT for homology). For any space $X$ and abelian group $G$ , there is a natural split short exact sequence

0 \to H_{n} (X) \otimes G \to H_{n} (X; G) \to Tor (H_{n - 1} (X), G) \to 0,

so $H_{n} (X; G) ≅ (H_{n} (X) \otimes G) \oplus Tor (H_{n - 1} (X), G)$ .

Take $G = Z /2$ and $H_{*} (R P^{3}) = (Z, Z /2, 0, Z)$ . Recall $Z \otimes Z /2 = Z /2$ , $(Z /2) \otimes Z /2 = Z /2$ , $Tor (Z, Z /2) = 0$ , $Tor (Z /2, Z /2) = Z /2$ .

$H_{0} = (Z \otimes Z /2) \oplus Tor (0, -) = Z /2$ .
$H_{1} = (Z /2 \otimes Z /2) \oplus Tor (Z, Z /2) = Z /2 \oplus 0 = Z /2$ .
$H_{2} = (0 \otimes Z /2) \oplus Tor (Z /2, Z /2) = 0 \oplus Z /2 = Z /2$ .
$H_{3} = (Z \otimes Z /2) \oplus Tor (0, -) = Z /2$ .

So $H_{*} (R P^{3}; Z /2) = (Z /2, Z /2, Z /2, Z /2)$ — one $Z /2$ in each degree, matching $R P^{3}$ 's mod-2 homology as a closed 3-manifold. The Tor term in degree 2 is the homological shadow of the torsion in $H_{1}$ . Hatcher §3.A.

Exercise 9 (hard, proof). Hurewicz isomorphism for a wedge of spheres.

Let $X = ⋁_{i = 1}^{k} S^{n}$ with $n \geq 2$ . Use the Hurewicz theorem to compute $π_{n} (X)$ , and explain why the theorem does not directly compute $π_{n + 1} (X)$ .

Hint

$X$ is $(n - 1)$ -connected. The Hurewicz map $π_{n} \to H_{n}$ is an isomorphism in the first nonvanishing degree.

Answer

The wedge $X = ⋁_{i = 1}^{k} S^{n}$ is $(n - 1)$ -connected: it is simply connected (for $n \geq 2$ ) and has $H_{i} (X) = 0$ for $1 \leq i \leq n - 1$ , with $H_{n} (X) = Z^{k}$ (by the wedge formula, one generator per sphere).

Hurewicz theorem. If $X$ is $(n - 1)$ -connected with $n \geq 2$ , then $π_{i} (X) = 0$ for $i < n$ and the Hurewicz map $h : π_{n} (X) \to H_{n} (X)$ is an isomorphism.

Applying it in degree $n$ : $π_{n} (X) ≅ H_{n} (X) = Z^{k}$ , with generators the inclusions of the $k$ spheres.

The theorem does not compute $π_{n + 1} (X)$ : Hurewicz only identifies the first nonvanishing homotopy group with the corresponding homology group. In degree $n + 1$ the map $h : π_{n + 1} (X) \to H_{n + 1} (X) = 0$ is generally neither injective nor surjective onto something interesting — $π_{n + 1} (X)$ can be nonzero even though $H_{n + 1} (X) = 0$ . For example $π_{n + 1} (S^{n}) = Z /2$ for $n \geq 3$ (the suspended Hopf map), invisible to homology. Computing $π_{n + 1}$ requires the Whitehead product and the relative/stable refinements beyond plain Hurewicz. Hatcher §4.2.

Exercise 10 (hard, proof). Homology of $S^{n}$ by the long exact sequence of a pair.

Compute $H_{*} (S^{n})$ for all $n \geq 1$ using the long exact sequence of the pair $(D^{n}, S^{n - 1})$ together with excision, by induction on $n$ .

Hint

$\tilde{H}_{k} (S^{n}) ≅ H_{k} (D^{n}, S^{n - 1})$ via the quotient $D^{n} / S^{n - 1} = S^{n}$ , and the LES of $(D^{n}, S^{n - 1})$ relates this to $\tilde{H}_{k - 1} (S^{n - 1})$ .

Answer

The pair $(D^{n}, S^{n - 1})$ is a good pair, so $H_{k} (D^{n}, S^{n - 1}) ≅ \tilde{H}_{k} (D^{n} / S^{n - 1}) = \tilde{H}_{k} (S^{n})$ (the quotient collapses the boundary of the disk to give a sphere).

The long exact sequence of the pair reads

\dots \to \tilde{H}_{k} (D^{n}) \to H_{k} (D^{n}, S^{n - 1}) \partial \tilde{H}_{k - 1} (S^{n - 1}) \to \tilde{H}_{k - 1} (D^{n}) \to \dots

Since $D^{n}$ is contractible, $\tilde{H}_{*} (D^{n}) = 0$ , so the connecting map is an isomorphism:

\tilde{H}_{k} (S^{n}) ≅ H_{k} (D^{n}, S^{n - 1}) \partial ≅ \tilde{H}_{k - 1} (S^{n - 1}) .

This is the suspension isomorphism in homology: $\tilde{H}_{k} (S^{n}) ≅ \tilde{H}_{k - 1} (S^{n - 1})$ .

Induction. Base case $S^{0}$ (two points): $\tilde{H}_{0} (S^{0}) = Z$ , all higher groups zero. Inductively, $\tilde{H}_{k} (S^{n}) = \tilde{H}_{k - 1} (S^{n - 1}) = \dots = \tilde{H}_{k - n} (S^{0})$ , which is $Z$ exactly when $k = n$ and $0$ otherwise.

Result: $\tilde{H}_{k} (S^{n}) = Z$ for $k = n$ and $0$ otherwise; unreduced, $H_{0} (S^{n}) = H_{n} (S^{n}) = Z$ (with $H_{0} = H_{n} = Z^{2}$ collapsing appropriately only at $n = 0$ ). This is the canonical inductive computation, Hatcher §2.1.

Exercise pack EP2. Hatcher Chapter 2 supplement: singular, simplicial, and cellular homology, degree theory, Mayer-Vietoris, and the universal coefficient theorem across §2.1, §2.2, §2.C.

Prerequisites

03.12.11 pending
03.12.12 pending
03.12.13 pending
03.12.14 pending
03.12.15 pending
03.12.18 pending
03.12.19 pending
03.12.10 pending
03.12.23 pending

Tier anchors

beginner
intermediate: Hatcher, Algebraic Topology, Ch. 2 exercises (§2.1 singular/simplicial, §2.2 Mayer-Vietoris and degree, §2.2 cellular homology, §2.C)
master

References

Hatcher — Algebraic Topology · Ch. 2 exercises: §2.1 (simplicial vs singular homology, $\Delta$-complex structures), §2.2 (Mayer-Vietoris, degree of maps $S^n\to S^n$, cellular chain complex of $\mathbb{R}P^n$, $\mathbb{C}P^n$, lens spaces, Moore spaces), §2.C (homology with coefficients, Euler characteristic)
Vick — Homology Theory · Ch. 1-3 — singular homology and the Eilenberg-Steenrod axioms
May — A Concise Course in Algebraic Topology · Ch. 13-14 — cellular and ordinary homology

Estimated time

intermediate: 5h