Middle East Technical University Spring 1998 Math 422 Elementary Topology and Geometry

MIDTERM EXAM

MARCH 26, TIME: 75 MIN

Problem 1. (30 pts) Assume that Γ is a graph on the projective plane ℝP^{2} such that all the regions of its complement in ℝP^{2} are triangles (i.e., cells bounded by 3 edges) and all the vertices of Γ are m-valent.

(1) Determine the number of vertices for m = 5.

(2) For which values of m such a graph Γ may exist ? Motivate your answer.

(3) Sketch such a graph Γ in ℝP^{2} for m = 4.

Solution: Denote by v, e, r the number of vertices, edges and regions of a graph. In our case, 2e = mv since all vertices are m-valent, 2e = 3r since all regions are triangles, and by Euler’s formula v − e + r = χ(ℝP^{2}) = 1, since all the regions are cells. Elliminating e and r

we obtain v − \frac{mv}{2} + \frac{mv}{3} = v(1 − \frac{m}{6}) = 1 or

v(6 − m) = 6

Thus 0 < m < 6 and 6 − m divides 6, which allows only m = 3, m = 4 and m = 5. For m = 5 we obtain v = \frac{6}{6−5} = 6. For m = 4 we have v = 3, the graph Γ is sketched below.

Problem 2. (15 pts) Sketch a Seifert surface for the knot shown below. Determine its Euler characteristic. Is this surface orientable ?

Solution: The Seifert surface, F, is sketched below

χ(F) = r − c = −9, where r = 4 is the number of the shaded regions and c = 13 the number of crossings. F is orientable because the loops around 10 white regions make even number (namely 2) of twists each.

Problem 3. (15 pts) Assume that B is a closed subset of A and C is a closed subset of B. Prove that C is a closed subset of A.

Solution: The proof mimics the proof of a similar statement about open subsets, which was proved in the class. Namely, B closed in A means that B = A ∩ F where F is closed in ℝ^{n}; C closed in B means that C = B ∩ F′ where F′ is closed in ℝ^{n}. Thus C = (A ∩ F) ∩ F′ = A ∩ (F ∩ F′), which implies that C is closed in A since F ∩ F′ is closed as an intersection of finitely many closed sets.

Problem 4. (20 pts) Consider a subset B ⊂ A ⊂ ℝ^{n} with the inclusion map f: B → A. Prove that f is an open map if and only if B is relatively open in A.

Solution: f is open means that f(C) = C is open in A for any C ⊂ B open in B. In particular, if f is open, then f(B) = B is open in A, since B is open in itself.

Assume now that B open in A. We proved in the class that any subset C ⊂ B which is open in B is open also in A, that is f(C) = C is open in A. This means that the inclusion map f is open.

Problem 5. (20 pts) Prove that ℝ^{2} is homeomorphic to (1) a square (0, 1) × (0, 1)

(2) an open disc D^{2} = {(x, y) ∈ ℝ^{2} | x^{2} + y^{2} < 1}

Solution: (1) We have constructed on the lectures a homeomorphism f: (0, 1) → ℝ. Consider a direct product map h = f × f: [0, 1] × [0, 1] → ℝ × ℝ defined as h(x, y) = (f(x), f(y)). It is continuous, since the both coordinates are continuous. It is bejective with the inverse map h^{−1} = f^{−1} × f^{−1}, which is also continuous for the same reasons, since f^{−1} is continuous.

Remark. The above arguments show that more generally, a direct product of homeomorphisms is a homeomorphism.

(2) Using identification of ℝ^{2} with ℂ and letting z = x + iy, define h: D^{2} → ℝ^{2} as h(z) = \frac{z}{\sqrt{1−|z|^{2}}} . This map is well-defined and continuous since 1 − |z|^{2} > 0 for z ∈ D^{2}. It

is easy to check that h^{−1}(z) = \frac{z}{\sqrt{1+|z|^{2}}} is the inverse map, which is obviously well-defined and

continuous in ℂ ≅ ℝ^{2}. Therefore h is a homeomorphism.

Remark. Certainly, there exist many functions like the above h, take for instance any homeomorphism f: [0, 1) → [0, ∞) and define h in the polar coordinates (r, θ) as h(r, θ) = (f(r), θ).