Introduction to Topology and Geometry 2nd (Instructor's Solution Manual)

166 Pages·2013·1.887 MB·English

by  Saul Stahl, Catherine Stenson

A SOLUTION MANUAL
to
INTRODUCTION TO
TOPOLOGY AND GEOMETRY
Second Edition
by
SAUL STAHL
Department of Mathematics
University of Kansas
Lawrence, KS 66045
and
CATHERINE STENSON
Department of Mathematics
Juniata College
Huntingdon, PA 16652
Copyright 2004 and 2013
EXERCISES 1.1 1
CHAPTER 1
EXERCISES 1.1
1. Letting a,b,c,d,...,z refer to the 26 elements of Figure 1.18, we have
c ≈ i ≈ j ≈ j ≈ l ≈ m ≈ n ≈ s ≈ u ≈ v ≈ w ≈ z,
d ≈ o,e ≈ f ≈ g ≈ t ≈ y,q ≈ r,x ≈ k.
2. a ≈ b ≈ c ≈ e,d ≈ f,h ≈ k,i ≈ j.
3. a ≈ b.
4. a,b, and d.
5. (n+1)−2n+(n+1) = 2.
6. 2n−3n+(n+2) = 2.
7. Itisreasonabletointerpretsuchanexcursionasonewhichbeginsand
endsinsomeone’shome. Wemightaswellassumethatthestrollbegins(and
ends) on the larger of two islands. Note that once one bridge was used to
leave the island, two bridges are left at that island. This makes it impossible
for the excursion to end at that island. Hence no such excursion is possible.
8. Thedeletionofanypointfromthestraightlineleavestwocomponents,
but the removal of any point from the plane leaves it connected.
9. Thedeletionofanypointfromthestraightlineleavestwocomponents,
but the removal of any point from the circle leaves it connected.
10. The removal of any two distinct points from the circle leaves it
disconnected, but the removal of any two distinct points leaves the plane
connected.
EXERCISES 2.1 1
CHAPTER 2
1 EXERCISES 2.1
1. (a) (4,4,4,4,4). (b) n (n (cid:0) 1)0s. (c) (4,4,4,3,3,3,3). (d) Assuming
m (cid:21) n, n m0s followed by m n0s. (e) (5,4,4,3,3,3). (f) Assuming l (cid:21)
m (cid:21) n, n (l + m)0s, followed by m (l + n)0s followed by l(m + n)0s. (g)
(18,18,18,18,17,17,17,17,17,16,16,16,16,16,16,15,15,15,15,15,15,15). (h) Let
m = k n . Assuming n (cid:21) n (cid:21) ::: (cid:21) n , the degree sequence consists
i i=1 i 1 2 k
0 0 0
of nk mPks followed by nk(cid:0)1 mk(cid:0)1s followed by nk(cid:0)2 mk(cid:0)2s and so on.
2. (a) Yes. (b) No. (c) Yes. (d) Yes. (e) No. (f) Yes. (g) No. (h) Yes.
(i) No. (j) Yes. (k) Yes. (l) Yes.
3. (a) Yes. (b) No. (c) No. (d) No. (e) No. (f) No. (g) No. (h) Yes. (i)
No. (j) Yes. (k) No. (l) No.
4. a) Proposition 2.1.1 entails 4q = n(n+1) and hence the sequence is
not graphical if n (cid:17) 1;2(mod4). Moreover, if G realizes (n;n(cid:0)1;:::;1),
n
then a graph G realizing (n+4;n+3;:::;1) is obtained by adding two
n+4
Hamiltonian cycles to the arcs of G as well as a completely disjoint copy
n
of the easily constructed G . Since G and G are easily constructed, G
4 3 4 n
exists for all n(cid:17) 3;4(mod4).
b) Proposition 2.1.1 entails 4q = n(n + 3) and hence the sequence is
not graphical if n (cid:17) 2;3(mod4). If n (cid:17) 0 then add to Gn(cid:0)1 (see part a
above) two disjoint copies ofthe bouqueton n=2 loops. If n(cid:17)1 thenaddto
Gn(cid:0)1 a disjoint copy of the connected graph that consists of two bouquets
on (n(cid:0)1)=2 loops joined by a single arc
c) Proposition 2.2.1 entails 4q = n(n+1) and hence the sequence is not
graphical if n(cid:17) 1;2(mod 4). For the other values of n add an isolated node
to the G of part a.
n
d) Proposition 2.2.1 entails 4q = n(n+3) and hence the sequence is not
graphical if n(cid:17) 2;3(mod 4). For the other values of n add an isolated node
to the graph of part b.
e) Use two copies of the graph of part d.
f) Suppose the given sequence has length p. It follows from Proposition
2.1.1 the sequence is realizable only if kp is even. If k is even then the
sequence is realized by p disjoint bouquets, each of which has k=2 loops.
If k is odd and p is even, use p=2 disjoint copies of the graph obtained by
joining two nodes with k "parallel" arcs.
5. a) For no n because p = n and hence every node must have degree
less than n.
EXERCISES 2.1 2
b)Onlyforn= 1. Otherwise,thetwonodesofdegreenmustbeadjacent
to every other node. This means that no node can have degree 1.
c) For no value of n, for a similar reason as in a).
d) Only for n= 1 for the same reason as in b).
e) Let H consist of the isolated nodes u ;v . Assume that H has been
0 0 0 n
de(cid:12)ned so that it has nodes u ;v ;:::;u ;v so that deg u = deg v = i.
0 0 n n i i
De(cid:12)ne Hn+1 by adding to Hn the arcs uivn(cid:0)i;i = 0;1;2;:::;n as well as two
new isolated nodes u(cid:0)1;v(cid:0)1. Of course, it is now necessary to add 1 to each
index. Note that each new generation of arcs is centered in a new location
and so this is indeed a simple graph.
f) It is clearly necessary that kp should be even and p > k. Conversely,
suppose (cid:12)rst that k is even and p > k. Join distinct u and u with an arc
i j
provided the cyclical distance of i and j mod p is at most k=2 . If k is odd
and p is even, p > k, join distinct u and u provided the cyclic distance of
i j
i and j mod p is at most (k(cid:0)1)=2 and also join each u to u (addition
i i+p=2
mod p).
6. By Proposition 2.2.1, the sum of all the degrees of a graph is even. It
follows that the number of odd degrees must be even.
n n
7. Let n = k n . Then q = (cid:0) k i . The degree se-
i=1 i 2 i=1 2
! !
quencehask moPnotonesubsequencessuchthatPthei0thsubsequenceconsists
of n repetitions of n(cid:0)n .
i i
8. If the sequence is graphical then the evenness of the sum follows from
Proposition 2.1.1. Conversely, suppose the sum is even. It follows that the
number of odd a ’s is even. Each of the even a ’s is realized by a bouquet
i i
of loops at one vertex. Two odd terms, say a and a can be realized be
i j
joining a bouquet of (a (cid:0)1)=2 loops to a bouquet of (a (cid:0)1)=2 loops by a
i j
single arc.
9. See S. Hakimi, "On the realizability of a set of integers as the degrees
of the vertices of a graph", J. SIAM appl. Math 10(1962), 496 - 506.
10. See Frank Harary’s book.
11. For each k = 1;2;3;::: let A = (k;k2;k3) 2 <3. It will su(cid:14)ce to
k
prove that for distinct k;l;m;n the straight lines A A and A A do not
k l m n
intersect. This, in turn,follows from the fact that thevectors A (cid:0)A ;A (cid:0)
l k m
A ;A (cid:0)A are linearly independent. This linear indepenence follows from
k n k
the fact that the 3 (cid:2) 3 matrix formed by these three vectors is nonsingular.
2 EXERCISES 2.2 3
2 EXERCISES 2.2
1. a) Yes b) Yes c) No d) No
2. a) No b) No
3. a) Yes b) No c) Yes d) Yes e) No f) No g) No h) No i) Yes j) No k)
Yes l) No.
4. By Theorem 2.2.2, if and only if n is odd.
5. By Theorem 2.2.2, if an only if n and n are both even.
1 2
6. By Theorem 2.2.2, if andonly if n +n , n +n , n +n are all even,
1 2 1 3 2 3
which happens if and only if n ;n ;n all have the same parity.
1 2 3
7. By Theorem 2.2.2, if and only if the terms n(cid:0)n , where n = n +
i 1
n +:::+n are all even, which implies that the n all have the same parity.
2 m i
0
If m is even, this means all the n s are even. If m is odd either all of them
i
are odd or all of them are even.
8. Let G be the (4, 7) graph whose nodes represent the two banks and
the two islands and whose arcs represent the seven bridges. Since a stroll
through the portions of the city corresponds to a trail of G, it follows from
Theorem 2.2.2 that the required stroll is impossible.
9. Let G be traversable. If G is Eulerian then it has no odd nodes. If G
is not Eulerian, connect the endpoints of its traversing walk by a new arc
0 0
to obtain a graph G . Since G is Eulerian it follows from Theorem 2.2.2
that G has exactly two odd nodes, namely, the said endpoints. Conversely,
if G has no odd nodes, we are done by Theorem 2.2.2. If G has exactly two
odd nodes, add to G a new arc joining these two odd nodes. The resulting
0
graph G is, by Theorem 2.2.2 Eulerian. By deleting the new arc from the
0
Eulerian walk of G we obtain a traversing walk of G.
10. a) Yes b) Yes c) No d) Yes.
11. This follows immediately from Proposition 2.2.3.
12. a) Yes b) Yes c) No d) Yes e) No f) No g) No h) Yes i) No
13. Using the notation of Exercise 2.1.5, note that any Hamiltonian
cycle of K must alternate between A and A . It follows that K is
n1;n2 1 2 n1;n2
Hamiltonian if and only if n = n .
1 2
14. Let G = K be Hamiltonian. Assume that n (cid:21) n (cid:21) n .
n1;n2;n3 1 2 3
0
Since no two consecutive nodes of G s Hamiltonian cycle can come from A
1
itfollows thatn (cid:20) n +n . Conversely, supposetheinequality n (cid:20)n +n
1 2 3 1 2 3
holds. Then for any two nonadjacent vertices the inequality of Proposition
2.2.3 isone of the following three: 2(n +n )(cid:21) n +n +n , or2(n +n )(cid:21)
1 2 1 2 3 1 3
n +n +n , or 2(n +n ) (cid:21) n +n +n , all of which are now true.
1 2 3 2 3 1 2 3
15. If and only if n (cid:21) n +n +:::+n (assuming that n (cid:21)n (cid:21)n (cid:21)
1 2 3 m 1 2 3
:::). The proof is similar to that of Exercise 13.
3 EXERCISES 2.3 4
16. Labelthe(cid:12)veoutsidenodesOandthe(cid:12)veinsidenodesI.AHamilton
cycle would then consist of a cyclic string S of 10 O’s and I’s. It is easy
to see that S cannot contain any of the strings OOOOO, IIIII, IOI, OIO,
OIIIIO, IOOOOI as substrings. It follows that it may be assumed without
loss of generality that S = OOIIOOOIII. This, possibility, however, can be
eliminated by trial and error.
17. Seehttp://mathworld.wolfram.com/GeneralizedPetersenGraph.html
or Alspach, B.R. "The Classi(cid:12)cation of Hamiltonian Generalized Petersen
Graphs." J. Combin. Th. Ser. B 34, 293-312, 1983 or Holton, D.A. and
Sheehan, J. "Generalized Petersen and Permutation Graphs." Section 9.13
in The Petersen Graph. Cambridge, England: Cambridge University Press,
pp.45 and 315-317, 1993.
18. See http://mathworld.wolfram.com/TuttesGraph.html.
19. If a is a loop then G (cid:0)a is connected. Otherwise, let u and v be
the distinct endpoints of a. Let G denote the subgraph of G spanned by
u
the set of nodes that can be connected to u by a path that does not contain
v. Let G denote the subgraph of G spanned by the set of nodes that can
v
be connected to u by a path that does not contain v. Both G and G are
u v
connected. If they are disjoint, then they are the two components of G(cid:0)a,
otherwise, G(cid:0)a is connected.
20. a) The old nodes of G retain their degree in S(G) whereas the new
ones all have degree 2. It follows that S(G) is Eulerian if and only if G is
Eulerian. b) Since a Hamiltonian path of S(G) must contain all the new
vertices, it corresponds to an Eulerian trail of G. Since this Eulerian trail
can visit each of the nodesexactly once, it follows that each ofthe old nodes
has degree 2 in G. Thus, G is a cycle.
21. As Figure 2.1 indicates, this graph is really the Petersen graph in
disguise and hence, by Exercise 14, this graph is not Hamiltonian.
3 EXERCISES 2.3
1. a) k (cid:21) 7 b) k (cid:21) 2 c) k (cid:21)2 d) k (cid:21) 3 e) k (cid:21) m f) k (cid:21) 3 g) k (cid:21)3 h) k (cid:21) 4
2. The endpoints of a path of maximum length in the graph must have
degree 1 in the graph. Since every subgraph of an acyclic graph is also
acyclic, it follows that every subgraph of an acyclic graph is 1-degenrate.
By either of Proposition 2.3.2 or Theorem 2.3.1, it is 2-colorable.
3. Note that every diagonal divides such a graph into two subgraphs
of the same type that intersect in that diagonal. Every such division must
leave a vertex with degree 2.
4 EXERCISES 2.4 5
Figure 1:
5. Suppose G is not k-colorable, and let h > k be the smallest integer
such that G is h-colorable. Let A ;:::;A be the color classes of some h-
1 h
coloring of G. Then for any 1 (cid:20) i < j (cid:20) h there is an arc that joins some
nodeinA tosomenodeinA , sinceotherwiseG wouldbe(h(cid:0)1)-colorable.
i j
Thus, G contains at least h(h(cid:0)1)=2 (cid:21) k(k+1)=2 arcs.
6. Since w is adajcent to every other node of W , W is k-colorable if
n n
andonlyiftherimC is(k(cid:0)1)-colorable. HenceW isk-colorable fork (cid:21) 3
n n
or 4 according as n is even or odd.
7. See West’s book.
8. The subdivision graph S(G) is clearly homeomorphic to G. It is
2-colorable with all the old vertices colored 1 and the new ones colored 2.
9. Every complete bipartite graph is 2-colorable with A receiving color
i
i for i = 1;2: Hence so are their subgraphs 2-colorable. Conversely, if G is
0
2-colorable, with color sets A and B, let G be the graph obtained from G
by making sure that every vertex of A is joined to every vertex of B. Then
0
G is a complete bipartite graph and G is its subgraph.
4 EXERCISES 2.4
1. a, b, c, f are planar. d is really K and so is nonplanar. e contains K
5 3;3
and so is nonplanar.
2. See Figures 2.2, 2.3.
a is non planar as demonstrated by the cycle 123456789 with "chords"
16, 208, 49.
b is planar.
c is planar.
d is nonplanar because it has p = 8, q = 20 > 3p(cid:0)6 = 18.
EXERCISES 2.4 6
e is non planar because it has p = 6, q = 14 > 3p(cid:0)6 = 12.
f is planar
g is planar
h is nonplanar with cycle 12348765 and chords 14, 27, 36.
i is planar.
3. Let P be Hamiltonian cycle. The regions in the interior of P are
formedbydiagonalarcsandhencetheyare2-colorable. Dittofortheregions
outside P. Thus, the regions are 4-colorable.
4. That G is 3-colorable is Exercise 2.3.3. The map formed by the
regions inside the polygon is clearly 2-colorable. Adding the outside region
merely simply requires one more color.
5. The 3 large regions and the outside of the Tutte graph all share
borders and hence require di(cid:11)erent colors. Hence, the Tutte graph is not
3-map-colorable.
6. Just assign 1 to any region and go on to its neighbors.
7. Figure 2.4 demonstrates that K is planar forevery n. Onthe other
n;2
hand, K contains the nonplanar K whenever m;n(cid:21) 3. It follows that
m;n 3;3
K is non planar if and only if m;n(cid:21) 3.
m;n
8. Figure 2.5 shows that K K and K are planar. If n (cid:21) 3
n;1;1 2;2;1 2;2;2 1
then K contains K . It follows that these three graphs are the only
n1;2;n3 3;3
K , n (cid:21) n (cid:21) n (cid:21) 1 that are planar.
n1;n2;n3 1 2 3
9. Figure2.6demonstratesthatK andK areplanar. Ifn (cid:21)3,
1;1;1;1 2;1;1;1 1
then K (cid:19) K (cid:19) K which is nonplanar. Also both K ,
n1;n2;n3;n4 n1;3 3;3 2;2;2;2
K , K all contain K = K . Hence the 4-partite graphs of
2;2;2;1 2;2;1;1 2+1;2+1 3;3
Figure 2.6 are the only ones that are planar.
10. Every node of the plane map has even degree because the regions
that surround the node receive the two colors alternatingly. Now apply
Theorem 2.2.2.
11. This follows from the fact that the graph obtained by deleting a
single arc from K is planar (see Figure 2.6).
5
12. For each straight line, select one of its half-planes as its preferred
half-plane. Note that for any two abutting regions the numbers of preferred
regions that contain them di(cid:11)er by 1. Hence, coloring each region by the
parity of the number of preferred regions that contain it is a 2-coloring of
the map.
13. Yes. The addition of parallel arcs does not a(cid:11)ect the node colorings.
14. This is a special case of Proposition 2.4.10. Alternatively, the proof
of Exercise 12 works here if the notion of the preferredside of a straight line
is replaced by the interior of circle.
EXERCISES 2.4 7
Figure 2:
EXERCISES 2.4 8
Figure 3:
...
Figure 4:
Figure 5:
Figure 6: