Questions: Euler Characteristic via Homology

Question 1

The 2-sphere S² has H_0 ≅ Z, H_1 = 0, H_2 ≅ Z. What is its Euler characteristic?

• A. 0
• B. 1
• C. 2
• D. 3

Answer: (C)

chi(S²) = b_0 - b_1 + b_2 = 1 - 0 + 1 = 2. This matches the classical formula V - E + F for any triangulation of the sphere: the boundary of a tetrahedron has V = 4, E = 6, F = 4, giving 4 - 6 + 4 = 2. The homological definition explains WHY V - E + F is always 2 regardless of triangulation: it equals the alternating sum of Betti numbers, which depends only on the topology.

Question 2

A compact orientable surface of genus g has Euler characteristic 2 - 2g.

• T. True
• F. False

Answer: True

A genus-g surface has H_0 ≅ Z (connected), H_1 ≅ Z^{2g} (2g independent loops), and H_2 ≅ Z (closed orientable surface). So chi = 1 - 2g + 1 = 2 - 2g. The sphere (g=0) has chi = 2, the torus (g=1) has chi = 0, the genus-2 surface has chi = -2, and so on. Each handle added to the sphere decreases the Euler characteristic by 2, because it adds two new independent 1-cycles.

Question 3

The Euler characteristic of a disjoint union X ⊔ Y equals chi(X) + chi(Y). Why?

• A. Because V - E + F is additive over disjoint pieces
• B. Because H_n(X ⊔ Y) ≅ H_n(X) ⊕ H_n(Y) for all n, so Betti numbers add
• C. Because the Euler characteristic is always multiplicative
• D. This is only true for connected spaces

Answer: (B)

Homology respects disjoint unions: H_n(X ⊔ Y) ≅ H_n(X) ⊕ H_n(Y). Therefore b_n(X ⊔ Y) = b_n(X) + b_n(Y), and chi(X ⊔ Y) = sum(-1)^n(b_n(X) + b_n(Y)) = chi(X) + chi(Y). This is the homological explanation of the additivity. The Euler characteristic is additive for disjoint unions but NOT generally additive for unions with overlap (one needs an inclusion-exclusion correction from the Mayer-Vietoris sequence).

Question 4

The torus T² has chi = 0. Does this mean the torus has no topological features detectable by homology?

• T. True
• F. False

Answer: False

The Euler characteristic is a single number that combines all Betti numbers with alternating signs. The torus has b_0 = 1, b_1 = 2, b_2 = 1, and these happen to give chi = 1 - 2 + 1 = 0. But the individual Betti numbers carry much more information: two independent 1-cycles (the meridian and longitude) and one 2-cycle (the fundamental class). A point also has chi = 1 - 0 + 0 = 1, not 0. Euler characteristic zero means the Betti numbers cancel in the alternating sum, not that the space is topologically trivial.

Question 5

Explain why the combinatorial formula V - E + F and the homological formula b_0 - b_1 + b_2 always agree for triangulated surfaces.

Model answer: The rank-nullity theorem applied to the boundary maps gives the connection. If we let c_n be the number of n-simplices (so V = c_0, E = c_1, F = c_2), then c_n = rank(d_n) + nullity(d_n) = rank(d_n) + dim(ker(d_n)). Since b_n = dim(ker(d_n)) - rank(d_{n+1}), the alternating sum of Betti numbers telescopes to the alternating sum of simplex counts: sum(-1)^n b_n = sum(-1)^n c_n.

This is not a coincidence but a theorem in linear algebra. The chain complex links the combinatorial data (numbers of simplices) to the algebraic data (homology groups) through the boundary operators. The alternating sum is preserved because ranks of adjacent boundary operators cancel when you expand b_n = dim(Z_n) - dim(B_n). This is why the Euler characteristic is simultaneously computable from a triangulation AND independent of the choice of triangulation.