751 Problem Set I - University of Wisconsin–Madison

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 : I X with (0) = x and (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 x0 X and every neighborhood V of x0 there is an open set U with x0 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 x0 C. By (ii) there is an open set U containing x0 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 x0, i.e. U C. Thus C is open. This proves (iii). Finally assume (iii). Choose x0 X and a neighborhood V of x0. Let U be the path component of V containing x0. 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) + S1 (1, 0) + S1 ,

the spectacles

Y = (-2, 0) + S1 (2, 0) + S1 [-1, 1] ? 0 ,

and the space

Z = S1 0 ? [-1, 1]

obtained from a circle by adjoining a diameter.

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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) R2 : (x ? 1)2 + y2 = 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 R2. That the sides are identified in pairs means the following. There is an enumeration 1, 1, . . . , n, n of the edges of P (not necessarily in cyclic order but without repetitions) and for each k = 1, 2 . . . , n a homeomorphism k : k k so that desired identification space S is obtained from P by identifying x k P with k(x) k P .

Solution. Choose p S. We must construct a neighborhood U of p in S and a homeomorphism h : U R2. 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 R2.

Case 2. Assume that p = {a, b} where a := k, b := k, and (a) = b where := k for some k = 1, 2 . . . , n. Abbreviate R := (-1, 1) ? (-1, 1), R1 := (-1, 1) ? [0, 1), R2 := (-1, 1) ? (-1, 1], I := R1 R2. Let f and g be a homeomorphisms from R1 and R2 onto neighborhoods U1 and U2 of a and b in P respectively. Thus a f (I) and b g(I) . Shrinking and modifying as necessary we may assume f (I) = g(I). Define : I I by (x, 0) = g-1((f (x, 0)) and : R- R- by (x, y) = (x, 0) + (0, y). Replacing g by g we may assume that 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.

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Case 3. Assume that p = {v1, v2, . . . , vk} is a set of vertices of P . Renaming and replacing some of the j by -j 1 we may assume that vj is the common vertex of j-1 and j where 0 := k. Let R be the open unit disk in C = R2

and write R = R1 ? ? ? Rk where Rj is the sector

2(j - 1)

2j

Rj := r exp(i) : 0 r < 1,

.

k

k

It is easy to construct homeomorphisms fj : Rj Uj where Uj is a neighborhood of vj in P , for example we make take fj(r exp(i) = vj +cjr exp(i(aj+bj)) where cj > 0 is small and aj and bj are judiciously chosen. Reflecting if necessary we can even achieve fj(Ij-1) j and fj(Ij) j where

2j Ij := r exp(i) : 0 r < 1, = k .

To get a homeomorphism from R to a neighborhood of p in S we must find such homeomorphisms fj satisfying the additional condition that

j fj+1|Ij = fj |Ij .

()

For this we try

fj(r exp(i)) = vj + j(, r) exp(i(aj + bj)).

As in case 2 we modify fj+1|Ij to achieve (). This determines j(2(j - 1)/k, ?) and j(2/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 j(, ?) for intermediate values of by linear interpolation.

Remark. The following representations are called standard: 1. The sphere S2 is homeomorphic to P/ where P is a square with boundary

P = 1 1 2 2,

the sides are are enumerated and oriented in the clockwise direction, and the identifications reverse orientation.

2. The surface Mg is homeomorphic to P/ where P is a 4g-gon with boundary P = 1 2 1 2 ? ? ? 2g-1 2g 2g-1 2g, the sides are are enumerated and oriented in the clockwise direction, and the identifications reverse orientation.

3. The surface Nk is homeomorphic to P/ where P is a 2k-gon with boundary

P = 1 1 ? ? ? k k.

the sides are are enumerated and oriented in the clockwise direction, and the identifications preserve orientation.

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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 and that are identified reversing orientation and type II which cuts a polygon in two along a diagonal producing a new pair (, ) and then reassembling along an old pair (, ). 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 S2 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

tvrian12g(ul7at+ion 4o9f

a -

compact surface, show 24 ) In the case of the

that 3f = 2e, e sphere, projective

= 3(v pla.ne

- ), and 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; := 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 , PE = {F : (E, F ) P }, and P F = {E : (E, F ) P }. Then #(PE) = 2 for each E so #(P ) = 2e. Also #(P F ) = 3 for each F so

#(P ) = 3f . Hence 2e = 3f . From v - e + f = we get e = 3v - 3. But

clearly e

v 2

=

v(v-1) 2

so

0

v2

- 7v

+ 6

so

v

7+

49-24 2

.

For

the

sphere

S2 we have = 2 and hence v 4 with equality for the tetrahedron. For the

projective plane we have = 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 = 0 do v 7 with equality in the following triangulation. Let

A1 = (0, 0), A2 = (0, 1), A3 = (1, 1), A4 = (1, 0), B1 = (0, 1/3), B2 = (1, 1/3),

C1 = (0, 2/3), C2 = (1, 2/3), D1 = (1/3, 1), D2 = (1/3, 0), E1 = (2/3, 1),

E2 = (2/3, 0), F = (1/2, 1/3), G = (1/2, 2/3), and faces A1B1D1, B1D2F , D2F E2, F E2B2, E2B2A4, B1C1F , C1F G, F GB2, GB2C2, A2C1D1, C1D1G,

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D1GE1, GE1C2, E2C2A3. 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 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 (Xk) Yk 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, X0 = {0, 1}, Y0 = {0, 1/2, 1}, and f is the identity map, then f (X0) Y0 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 = idX and f g = idY . 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 Xk/Xk-1 Yk/Yk-1 and the map g induces the inverse homeomorphism. But Xk/Xk-1 is a wedge of spheres and becomes disconnected if the wedge point is removed. Since the wedge point of Xk/Xk-1 is mapped to the wedge point of Yk/Yk-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 kcells of X and the k-cells of Y .

Problem 8. If {x0} is a deformation retract of X for some x0 X, then certainly X is contractible. A still stronger condition is that {x0} is a deformation retract of X for every x0 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 x0 X. After subdividing we may assume that x0 is a vertex, i.e. that (X, {x0}) is a CW pair. Then by Corollary 0.20 page 16 of [2], x0 is a deformation retract of X. Exercises 6 and 7 on page 18 of [2] show that these implications don't hold in general.

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