HW 1 SOLUTIONS, MA525
1. Problem 1
Let C be a closed subset of Y . Then f 1 (C ) = fA 1 (C ) fB 1 (C ). Since fA and fB are
1
1
continuous, the sets fA (C ) and fB (C ) are closed in A and B respectively, and hence in X .
This implies f 1 (C ) is closed i
Math 525: Takehome Midterm 2 Solutions
Here is a detailed solutions for the problem that caused people the most diculty.
Problem 3: Hatcher 2.1 #22
Let X be a nite-dimensional CW-complex. Throughout, I freely us that excision gives
Z if n = k
k
k1
k
k1
k
Math 525: Takehome Midterm 1
Due date: In class on Wednesday, September 23.
Disclaimer, Terms, and Conditions: You may not discuss the exam with anyone except myself.
You may only consult the following:
The beloved(?) text, Hatchers Algebraic Topology.
Yo
HW 10 SOLUTIONS, MA525
Hatcher 2.2 Problem 29
The Mayer-Vietoris sequence gives
Hn (Mg ) Hn (R) Hn (R) Hn (X ) Hn1 (Mg )
The compact space R deformation retracts on a wedge of g circles. So H0 (R) = Z, H1 (R) =
g
g=1 Z, Hn (R) = 0 for n > 1. Also we kno
HW 9 SOLUTIONS, MA525
Problem 1
(a): First, the exact sequence of pairs:
Hn (S n \ pt) Hn (S n ) Hn (S n , S n \ pt) Hn1 (S n \ pt)
Since S n \ pt is contractible, the above sequence gives the isomorphism Hn (S n ) Hn (S n , S n \ pt).
Excision of the s
HW 8 SOLUTIONS, MA525
Hatcher 2.1 Problem 11
Let r : X A be the retraction and let i : A X be the inclusion. Then r i is the identity
map on A. If r and i are maps induced at the level of homology, then r i has to be the identity
map on H (A). This implie
HW 7 SOLUTIONS, MA525
1. Hatcher 2.1, Problem 1
Cut along the edge marked with double arrows, and then ip vertically the triangle on the right,
move it to the other side and glue. This shows it to be a mobius band.
2. Hatcher 2.1, Problem 12
Letting F = 0
HW 6 SOLUTIONS, MA525
Hatcher 1.3 Problem 20
The fundamental group G of the Klein bottle X has the presentation < a, b | abab1 >. Since
= a2 (abab1 )ba2 b1 = a2 ba2 b1 = a(abab1 )bab1 = abab1 = 1, the map a a3 , b b
induces a homomorphism : G G. Consider
HW 4 SOLUTIONS, MA525
Problem 5
Suppose X is a tree. Fixing some base-vertex x, for a positive integer k , call all vertices
distance k from x as level k vertices. Because of the tree structure, all edges in the tree have
endpoints in adjacent levels. We
HW 4 SOLUTIONS, MA525
1.2 Problem 7
Let Z denote the quotient space of S 2 obtained by identifying the north pole to the
south pole. The space Z is the space marked as X/A in the gure next to Example 0.8 of
Hatcher. The cell complex structure is as follow
HW 3 SOLUTIONS, MA525
1. Problem 2
For t [0, 1], let ft : S 1 I S1 I be dened by ft (, s) = ( + 2st, s). Then f0 = id
and f1 = f . Moreover, ft restricted to S 1 cfw_0 is the identity map.
Glue S 1 cfw_0 to S 1 cfw_1 by (, 0) (, 1) to get a torus T 2 . No
HW 2 SOLUTIONS, MA525
1. Problem 2
(a): If p is not onto then the image of p and its complement disconnect X . So we can just
focus on the image without changing the problem. Hence assume that p is onto.
Let x be a point in X . We claim that the set of pr
Math 525: Takehome Midterm 2
Due date: In class on Wednesday, November 11.
Disclaimer, Terms, and Conditions: You may not discuss the exam with anyone except myself.
You may only consult the following:
The beloved(?) text, Hatchers Algebraic Topology.
You