SYLLABUS OF MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
Basic Information
Instructor: Hongbin Sun
E-mail: hongbins at math.berkeley.edu
Class: MWF 8-9am in 2 Evans.
Oce: Evans 751
Oce Hours: Monday 9-10am, Wednesday 9-10am, Friday 10-11am, or by appointment.
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 5
(1) P78, 14
Proof. Since the inverse function i : G G is a homeomorphism and i(H) = H, we
know that i(H) is a closed set in G containing H. Since H is the smallest closed
set containing H, we have H i(H).
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 3
(1) P50, 8
Since X has a countable basis, let = cfw_Uk be a basis with countably many
k=1
open sets. Let F be an open cover of X. Then for each k N, either Uk is not
contained in any open set in F, or Uk
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 4
(1) P63, 38
Proof. It suces to show that for any x S n , there is a path x : [0, 1] S n
with x (0) = (1, 0, , 0) and x (1) = x. Then for any x, y S n , there is a path
: [0, 1] S n dened by
(t) =
x (1 2t
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 7
(1) P111, 33
Proof. (a) The 2-sphere does not have the xedpoint property. f : S 2 S 2 dened
by f (x) = x for any x S 2 does not have a xed point.
(b) The torus does not have the xedpoint property. Paramet
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 2
(1) P31, 3.
a) For cfw_(x, y)|1 < x2 + y 2 2, the interior is cfw_(x, y)|1 < x2 + y 2 < 2, the closure
is cfw_(x, y)|1 x2 + y 2 2, the frontier is cfw_(x, y)|x2 + y 2 = 1 or x2 + y 2 = 2.
b) For E2 with b
MATH 142 FINAL SOLUTION
1. (10 points) Determine whether the following statements are true of false, no
justication is required.
(1) For any continuous map f : X Y , and any subset E X, f (E) f (E)
always holds.
True
(2) If A E3 is a closed bounded subset
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 6
(1) P102, 18
Proof. Let the path be : [0, 1] Y . Then for any t [0, 1], (t) is a point in Y ,
so there is a canonical neighborhood Ut Y containing f (t), and we can suppose
that Ut is open in Y . Let Vt =
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 8
(1) P140, 22
Proof. Since a triangle is pathconnected, its identication space (the dunce hat)
is also pathconnected.
The dunce hat can be obtained by the union of following two spaces A and B
along a circ
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 9
(1) P152, 2
Proof. X is not Hausdor, by considering two dierent points p and (0, 0, 1) in
X. For any neighborhood of p, it is in the form of (U cfw_(0, 0, 1)) cfw_p for some
neighborhood U of (0, 0, 1) in
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
REVIEW OF IMPORTANT CONCEPTS AND THEOREMS
Short Version
Chapter 2: Topological space and continuity.
Topological space: dened by open sets. Neighborhoods, closed sets, interior,
closure, limit point, denseness, bas
MATH 142 MIDTERM 2 SOLUTION
1. (10 points) Determine whether the following statements are true of false, no
justication is required.
(1) A path-connected component of a topological space may not be a closed subset.
True
(2) The identication space of a Hau
NOTES FOR MATH 142
SEPARATION AXIOMS (09/15/2014)
Recall that the proof of the Tietze extension theorem only need the topological space
X to satisfy the following property:
For any two disjoint closed set A, B X, there is a continuous map f : X E1 such
th
MATH 142 MIDTERM 1 SOLUTION
1. (30 points) For a topological space X and two subsets A, B X, show that
AB = AB
and
AB AB
hold, then give an example to show that
AB = AB
may not hold.
Proof. We rst show that A B = A B holds.
The rst step is to show A B A B
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 10
(1) P178, 2
Proof. For any n ai i Z1 (K) (ai Z, i are oriented 1-simplexes in K), we
i=1
need to show that n ai i equals a nite sum of cycles given by simple closed
i=1
oriented polygonal curves.
At rst,
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 11
(1) P188, 21
Proof. We need only to show that for any qsimplex , () = (). Suppose
that the stellar subdivision is respect to A = (v0 , v1 , , vk ).
If A is not a face of , then A is not a face of any fac
MATH 142: ELEMENTARY ALGEBRAIC TOPOLOGY
SOLUTION 1
(1) P22, 6.
Suppose the polyhedron P has e edges. Since each edge is shared by 2 faces, and
each face has p edges, P has 2e faces. Since each vertex is shared by q faces, it is
p
also adjacent to q edges,