MATH4530–Topology. HW5 solutions

Document technical information

Format pdf
Size 62.3 kB
First found May 22, 2018

Document content analysis

Category Also themed
not defined
no text concepts found





MATH 4530 – Topology. HW 5 solutions
Please declare any collaborations with classmates; if you find solutions in books or online,
acknowledge your sources in either case, write your answers in your own words.
Please attempt all questions and justify your answers.
Write the proofs in complete sentences.
(1) Show that any infinite set with the finite complement topology is connected.
Solution: If X = C1 t C2 where C1 , C2 are non-empty closed sets, since C1 and C2 must be
finite, so X is finite. This is a contradiction.
(2) Let T and T 0 be topologies on a set X. If T is finer than T 0 , then the connectedness of which
topology will implies the one of the other? Justify your answer.
Solution: We have a condituous map idX : (X, T ) → (X, T 0 ). Since the image of a connected space is connected, the connectedness of T implies T 0 .
(3) Let p : X → Y be a quotient map. Show that, if p−1 (y) is connected for each y ∈ Y and Y is
connected, then X is connected.
Solution: Suppose that X is not connected. Say, X = U1 t U2 for some non-empty open
sets U1 and U2 . By Theorem 5.5, every p−1 (y) is contained in either U1 or U2 . Let V1 =
{y ∈ Y | p−1 (y) ⊂ U1 } and V2 = {y ∈ Y | p−1 (y) ⊂ U2 }. Since U1 and U2 are non-empty,
disjoint and p is surjective, we have Y = V1 t V2 and V1 and V2 are non-empty. Also,
π−1 (V1 ) = U1 , p−1 (V2 ) = U2 by surjectivity of p, so by the definition of quotient maps, V1
and V2 are open sets in Y. It follows that Y is not connected. Contradiction.
(4) Let f : X → Y be a continuous map. Show that if X is path-connected, then Im f is pathconnected.
Solution: Let x, y ∈ Im f . Let x1 ∈ f −1 (x) and y1 ∈ f −1 (y). Since X is path connected,
there is a path p : [0, 1] → X connecting x1 and y1 . Then f ◦ p is a path connecting x and y.
(5) Show that there is no homeomorphism between (0, 1) and (0, 1] by using the connectedness. Hint:
if we remove a point from each of the spaces, what happens?
Solution: Suppose that there is a homeomorphism f : (0, 1] → (0, 1). If f : X → Y is a
homeomorphism, then the restriction f |X−{x} : X − {x} → Y − { f (x)} is a homeomorphism
too. If we apply this to (0, 1] − {1} = (0, 1), then we have a homeomorphism (0, 1) (0, 1) − { f (1)}. LHS is connected but the right hand side is (0, f (1)) t ( f (1), 1) which is
disconnected. So we have a contradiction.
(6) Show that if X is connected, then any continuous map f : X → Y where Y is a topological space
with discrete topology is a constant map, i.e. f (X) = {y} for some y ∈ Y.
Solution: If f (X) has more than one element, then since Y has the discrete topology, f (X)
is disconnected. Since the image of a connected space must be connected, we have a contradiction. Thus f (X) has only one element, i.e. f is a constant map.
(7) Consider the quotient space of R2 by the identification (x, y) ∼ (x + n, y + n) for all (n, m) ∈ Z2 .
Show that it is connected and compact.
Solution: Since R2 is conencted, the quotient space must be connencted since the quotient
space is the image of a quotient map from R2 . Consider E := [0, 1] × [0, 1] ⊂ R2 , then the
restriction of the quotient map p : R2 → R2 / ∼ to E is surjective. Since [0, 1] × [0, 1] is
compact, the image R2 / ∼ must be compact.
(8) Recall that a square matrix M is orthogonal if MM t = I. This condition is equivalent to “the set
~ i = h~v, w
~ i. In particular, if M is
of row vectors form an orthonormal basis” and also to hM~v, M w
orthogonal, then det M = ±1. Let O(n, R) be the set of orthogonal matrices of size n. Show that it
is not connected.
Solution: The determinant of an orthogonal matrix is ±1. Since M1 := {M : det M = 1}
and M−1 := {M : det M = −1} are closed subsets of the set of matrice, we have
O(n, R) = (M1 ∩ O(n, R)) t (M−1 ∩ O(n, R))
is a disjoint union of closed sets. Each of them is non-empty, since we have the identity
matrix In ∈ M1 ∩ O(n, R) and the identity matrix with −1 multiplied to exactly one of the
diagonal entry is in M−1 ∩ O(n, R).
Munkres, Topology.
Basic Set Theory,∼matsumura/math4530/basic set theory.pdf
Lecture notes, available at∼matsumura/math4530/math4530web.html

Similar documents


Report this document