MATH 131 SOLUTION SET, WEEK
8
ARPON RAKSIT AND ALEKSANDAR MAKELOV
1. Normalisers
We rst claim H NG (H). Let h H. Since H is a subgroup, for all k H we have
hkh1 H and h1 kh H. Since h(h1 kh)h1 = k, it follows that hHh1 = H,
proving the claim. That H is in
MATH 131 SOLUTION SET, WEEK
4
ARPON RAKSIT
1. 1 as an unbased functor
Let X a space and x, x X such that there is a path : [0, 1] X from x to x .
Recall from lecture that there is an isomorphism : 1 (X, x) 1 (X, x ) sending
[] [ 1 ].
1.1. Remark. Some peo
MATH 131 SOLUTION SET, WEEK 2
ARPON RAKSIT
1. Application of Baire category
1.1. Remark. I thought this one was pretty dicult! So I asked Alex for help. Then Alex
pointed me to a blog post1 by Terence Tao on this (well, a very similar) problem, where
he w
MATH 131 SOLUTION SET, WEEK 1
1. Two limits at once with uniform convergence
Recall that since the convergence fn g is uniform, g is continuous. Then, since xn x, for all
> 0 there exists N N such that
n > N = dY (g(x), g(xn ) <
On the other hand, again
MATH 131 SOLUTION SET, WEEK
5
ALEX
1. The topologists broom
1.1. A solution with epsilons. We shall show that X is contractible to (0, 1). First, we will
work with the translate of X by 1 down the y axis, which is homeomorphic to X, so that the
common poi
MATH 131 SOLUTION SET, WEEK
6
ARPON RAKSIT
Problem 1. Let G =: (V, E). Let n := |V | and m := |E|. We induct on m. Since g
is undirected we must have n 3 and m 3 if each v V has even degree. Thus
our base case is n = m = 3, i.e., a cycle on 3 vertices. Th
MATH 131 SOLUTION SET, WEEK 7
ALEX
1. Even loops have even degree
Lift f to a path F : [0, 1] R. Observe that if we identify S 1 R/Z, the condition f (x) = f (x)
=
becomes
x [0, 1) : f (x) = f (x + 1/2 (mod 1)
which means
x [0, 1] : nx Z : F (x) = F (x +
MATH 131 SECTION NOTES, SEPTEMBER 8
1. Set theory review
1.1. More fun with the Pigeonhole principle. It can indeed give you surprisingly non-trivial
results. Another such gem, similar to the one we discussed in class (about inverses modulo a prime)
is th
MATH 131 SOLUTION SET, FINAL
Part I
1. Sharp polynomials are very sharp, intermediate value theorem reports.
Let N N. Then, take n large enough so that
fn
1
1
2 4N
< 1/8
and fn
1
2
> 3/8.
By the mean value theorem applied to the polynomial fn (which is co
MATH 131 SECTION, I
ARPON RAKSIT
1. Some more on countability
Professor McMullen didnt get quite as far as he had planned to last Thursday, so lets
talk a little more about countability. (The material for this part of section is all on pp. 811
of Professo
MATH 131 SOLUTION SET, WEEK 3
ALEX
1. Separating the diagonal of a compact metric space
Let C = U c and assume what we want to prove fails. Then, in particular, there is a sequence
(xn , yn ) of pairs of points in X such that d(xn , yn ) 0, but (xn , yn )