751 Problem Set I JWR. Due Sep 28, 2004

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1 751 Problem Set I JWR Due Sep 28, 2004 Exercise 1. For any space X define an equivalence relation by x y iff here is a path γ : I X with γ(0) = x and γ(1) = y. The equivalence classes are called the path components of X. Show that the following are equivalent. (i) The space X is locally path connected, i.e. for every x 0 X and every neighborhood V of x 0 there is an open set U with x 0 U V and such that any two points of U can be joined by a path in U. (ii) For every x X and every neighborhood V of x in X there is a a neighborhood U of x with U V and such that any two points of U can be joined by a path in V. (iii) The path components of every open set are open. Solution. That (i) = (ii) is immediate: a path in U is a fortiori a path in V. Assume (ii). Choose an open set V and a path component C of V. (We will show C is open.) Choose x 0 C. By (ii) there is an open set U containing x 0 such that any two points of U lie in the same path component of V. In particular any point of U lies in the same path component of C as x 0, i.e. U C. Thus C is open. This proves (iii). Finally assume (iii). Choose x 0 X and a neighborhood V of x 0. Let U be the path component of V containing x 0. By (iii), U is open. This proves (i). Remark. The whole space X is an open set so it follows from (iii) that if X is locally path connected, then the path components are open. Exercise 2. Construct explicit homotopy equivalences between the wedge of circles ) ) X = (( 1, 0) + S 1 ((1, 0) + S 1, the spectacles Y = ) ) ( ) (( 2, 0) + S 1 ((2, 0) + S 1 [ 1, 1] 0, and the space ( ) Z = S 1 0 [ 1, 1] obtained from a circle by adjoining a diameter. 1

2 Solution. Let p = ( 1, 0), q = (1, 0), and C be the convex hull of X. It suffices to show that each of the spaces X, Y, Z is homeomorphic to a deformation retract of C \ {p, q}. 1. The space X is a deformation retract of C \ {p, q}. To see this radially project the insides of the two circles from their centers, radially project the portion of E outside the circles and above the x-axis from (0, 2), and radially project the portion of E outside the circles and below the x-axis from (0, 2). 2. The space Y := {(x, y) R 2 : (x ± 1) 2 + y 2 = 1/4} ( [ 1/2, 1/2] 0 ) is a deformation retract of C \ {p, q}. To see this radially project the insides of the two circles from their centers, radially project the points of C outside the two circles and outside the square [ 1, 1] 2 from the corresponding center, and project the remaining points vertically. 3. The space Z := C ( 0 [ 1, 1] ) is a deformation retract of C \ {p, q}. To see this radially project the left half from p and the right half from q. Each fiber of each of these retractions is a closed or half open interval so it is easy to see that each of these retractions is a deformation retraction. Remark. Note the wording of the last sentence. I didn t write Each fiber of each of these retractions is a closed or half open interval so each of these retractions is a deformation retraction. The latter wording seems to suggest that a retraction whose fibers are (homeomorphic to) intervals is a deformation retraction. I doubt that this is true in complete generality. However a theorem of Smale [5] combined with Whitehead s Theorem (Theorem 4.5 page 346 of [2]) implies that map between finite CW complexes with contractible fibers is a homotopy equivalence. (Then we can apply and Corollary 0.20 page 16 of [2].) Exercise 3. Let P be a polygon with an even number of sides. Suppose that the sides are identified in pairs in any way whatsoever. Prove that the quotient space is a compact surface. A surface is a space which is locally homeomorphic to R 2. That the sides are identified in pairs means the following. There is an enumeration α 1, β 1,..., α n, β n of the edges of P (not necessarily in cyclic order but without repetitions) and for each k = 1, 2..., n a homeomorphism φ k : α k β k so that desired identification space S is obtained from P by identifying x α k P with φ k (x) β k P. Solution. Choose p S. We must construct a neighborhood U of p in S and a homeomorphism h : U R 2. There are three cases. Case 1. Assume p lies in the interior of P. Then there is an open disk centered at p. Any open disk is homeomorphic to R 2. Case 2. Assume that p = {a, b} where a α := α k, b β := β k, and φ(a) = b where φ := φ k for some k = 1, 2..., n. Abbreviate R := ( 1, 1) ( 1, 1), R 1 := ( 1, 1) [0, 1), R 2 := ( 1, 1) ( 1, 1], I := R 1 R 2. Let f and g be a homeomorphisms from R 1 and R 2 onto neighborhoods U 1 and U 2 of a and b in P respectively. Thus a f(i) α and b g(i) β. Shrinking and modifying as necessary we may assume φ f(i) = g(i). Define ψ : I I by ψ(x, 0) = g 1 (φ(f(x, 0)) and Ψ : R R by Ψ(x, y) = ψ(x, 0) + (0, y). Replacing g by g Ψ we may assume that φ f = g. Now define h : R S by h(x, y) = f(x, y) if y 0 and h(x, y) = g(x, y) if y 0. 2

3 Case 3. Assume that p = {v 1, v 2,..., v k } is a set of vertices of P. Renaming and replacing some of the φ j by φ 1 j we may assume that v j is the common vertex of β j 1 and α j where β 0 := β k. Let R be the open unit disk in C = R 2 and write R = R 1 R k where R j is the sector { 2(j 1)π R j := r exp(iθ) : 0 r < 1, θ 2jπ }. k k It is easy to construct homeomorphisms f j : R j U j where U j is a neighborhood of v j in P, for example we make take f j (r exp(iθ) = v j +c j r exp(i(a j θ+b j )) where c j > 0 is small and a j and b j are judiciously chosen. Reflecting if necessary we can even achieve f j (I j 1 ) α j and f j (I j ) β j where I j := { r exp(iθ) : 0 r < 1, θ = 2jπ k To get a homeomorphism from R to a neighborhood of p in S we must find such homeomorphisms f j satisfying the additional condition that }. φ j f j+1 I j = f j I j. ( ) For this we try f j (r exp(iθ)) = v j + ρ j (θ, r) exp(i(a j θ + b j )). As in case 2 we modify f j+1 I j to achieve ( ). This determines ρ j (2(j 1)π/k, ) and ρ j (2π/k, ). These last two maps are homeomorphisms (i.e. strictly increasing functions) from from the the unit interval to two other intervals and we define ρ j (θ, ) for intermediate values of θ by linear interpolation. Remark. The following representations are called standard: 1. The sphere S 2 is homeomorphic to P/ where P is a square with boundary P = α 1 β 1 α 2 β 2, the sides are are enumerated and oriented in the clockwise direction, and the identifications reverse orientation. 2. The surface M g is homeomorphic to P/ where P is a 4g-gon with boundary P = α 1 α 2 β 1 β 2 α 2g 1 α 2g β 2g 1 β 2g, the sides are are enumerated and oriented in the clockwise direction, and the identifications reverse orientation. 3. The surface N k is homeomorphic to P/ where P is a 2k-gon with boundary P = α 1 β 1 α k β k. the sides are are enumerated and oriented in the clockwise direction, and the identifications preserve orientation. 3

4 The Classification Theorem for Surfaces (See [1] Page 236, [4] page 9, or [3] page 204) implies that any compact surface is homeomorphic to one of these. One can transform any space S = P/ as in Exercise 3 to one of the standard representations with a sequence of elementary moves. The are two kinds of elementary moves: type I which removes adjacent edges α and β that are identified reversing orientation and type II which cuts a polygon in two along a diagonal producing a new pair (α, β) and then reassembling along an old pair (α, β). This process produces a proof of the Classification Theorem from the assertion that any compact surface is homeomorphic to some P/ (not necessarily a standard one). The process has nothing to do with proving that P/sim is a compact surface. Exercise 4. Given positive integers v, e, and f satisfying v e+f = 2 construct a CW complex homeomorphic to S 2 having v 0-cells, e 1-cells, and f 2-cells. Solution. If v e + f = 2 then (v, e, f) = (1, 0, 1) + m(1, 1, 0) + n(0, 1, 1) where m = v 1 and n = f 1. The sphere is a CW complex with v = f = 1 and e = 0 in the only way possible: the attaching map is constant. To increase m by one without changing n add a new vertex and edge and modify the corresponding characteristic map by composing with complex square root. To increase n without changing m insert a loop at a vertex. The result follows by induction. Exercise 5. By a triangulation of a space X we mean homeomorphism from geometrical realization K of a simplicial complex K to X. Show that for any triangulation of a compact surface, show that 3f = 2e, e = 3(v χ), and v 1 2 ( χ ) In the case of the sphere, projective pla.ne and torus, what are the minimum values of the numbers v, e and f? (Here v,e and f denote the number of vertices, edges and triangles respectively; χ := v e + f denotes the Euler characteristic.) Solution. Let P be the set of pairs (E, F ) such that F is a face and E is an edge in the boundary of F, P E = {F : (E, F ) P }, and P F = {E : (E, F ) P }. Then #(P E ) = 2 for each E so #(P ) = 2e. Also #(P F ) = 3 for each F so #(P ) = 3f. Hence 2e = 3f. From v e + f = χ we get e = 3v 3χ. But clearly e ( ) v 2 = v(v 1) 2 so 0 v 2 7v + 6χ so v χ 2. For the sphere S 2 we have χ = 2 and hence v 4 with equality for the tetrahedron. For the projective plane we have χ = 1 so v 6 with equality for the triangulation shown on page 15 of [4]. (This example is obtained by identifying opposite edges of a hexagon and adding a triangle in the interior.) For the torus T 2 we have χ = 0 do v 7 with equality in the following triangulation. Let A 1 = (0, 0), A 2 = (0, 1), A 3 = (1, 1), A 4 = (1, 0), B 1 = (0, 1/3), B 2 = (1, 1/3), C 1 = (0, 2/3), C 2 = (1, 2/3), D 1 = (1/3, 1), D 2 = (1/3, 0), E 1 = (2/3, 1), E 2 = (2/3, 0), F = (1/2, 1/3), G = (1/2, 2/3), and faces A 1 B 1 D 1, B 1 D 2 F, D 2 F E 2, F E 2 B 2, E 2 B 2 A 4, B 1 C 1 F, C 1 F G, F GB 2, GB 2 C 2, A 2 C 1 D 1, C 1 D 1 G, 4

5 D 1 GE 1, GE 1 C 2, E 2 C 2 A 3. Note that in all cases the lower bound is achieved when the 1-skeleton is the complete graph on v vertices. It is a consequence of the proof that this complete graph can be embedded in a surface of Euler characteristic χ only when the lower bound is an integer. Exercise 6. Show that a map f : X Y between CW complexes is cellular (i.e. that f(x k ) Y k for all k) if and only if the image of each cell of X is a subset of a finite union of cells of Y of the same or lower dimension. Give an example to show that a cellular map f need not satisfy the stronger condition that the image of each cell of X is a subset of a cell of Y of the same or lower dimension. Solution. If is immediate and only if follows from the image of a closed set is compact and therefore a subset of a finite subcomplex of Y. If X = Y = I, X 0 = {0, 1}, Y 0 = {0, 1/2, 1}, and f is the identity map, then f(x 0 ) Y 0 but f(x) is not contained in a cell of Y. Exercise 7. The CW complexes and the cellular maps between them form a category. This means that the identity map of a CW complex is cellular, and the composition of cellular maps is cellular. A cellular isomorphism is an isomorphism in this category, i.e. a cellular map f : X Y such that there is a cellular map g : Y X satisfying g f = id X and f g = id Y. Show that a map f : X Y between CW complexes is a cellular isomorphism if and only if f is a homeomorphism and maps each cell of X onto a cell of Y. Solution. If f is a CW isomorphism, g = f 1 so f is (in particular) a homeomorphism. For each k map f induces a homeomorphism X k /X k 1 Y k /Y k 1 and the map g induces the inverse homeomorphism. But X k /X k 1 is a wedge of spheres and becomes disconnected if the wedge point is removed. Since the wedge point of X k /X k 1 is mapped to the wedge point of Y k /Y k 1 the components of the complement are preserved, i.e. f maps each open k-cell of X homeomorphicly ont a k-cell of Y and determines a bijection between the k- cells of X and the k-cells of Y. Problem 8. If {x 0 } is a deformation retract of X for some x 0 X, then certainly X is contractible. A still stronger condition is that {x 0 } is a deformation retract of X for every x 0 X. Show that in general the reverse implications do not hold, but that they do hold for CW complexes. Solution. Assume that X is a contractible CW complex and x 0 X. After subdividing we may assume that x 0 is a vertex, i.e. that (X, {x 0 }) is a CW pair. Then by Corollary 0.20 page 16 of [2], x 0 is a deformation retract of X. Exercises 6 and 7 on page 18 of [2] show that these implications don t hold in general. 5

6 References [1] William Fulton: Algebraic Topology, A First Course Springer GTM 153, [2] Allen Hatcher: Algebraic Topology, Cambridge U. Press, [3] Morris W. Hirsch: Differential Topology, Springer GTM 33, [4] William S. Massey: A Basic Course in Algebraic Topology, Springer GTM 127, [5] Stephen Smale: A Vietoris mapping theorem for homotopy. Proc. Amer. Math. Soc. 8 (1957),