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 respectiv
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,
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 followin
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) =
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 isomo
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, th
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
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
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 e
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
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 cf
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 on
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