Download Solutions to MMA100 Topology, March 13, 2010. 1. Assume ¯A

Solutions to MMA100 Topology, March 13, 2010.
1. Assume Ā − A = ∅, then A = Ā is closed in X and hence compact, since X is.
Now A is discrete, so sets consisting of a single point in A are open in A. These form
an open cover of A and thus has a finite subcover. This implies that A is finite. A
contradiction.
Hence Ā − A ̸= ∅.
∪
2. Let Oi , i ∈ I be a collection of open sets in X, Then O = i∈I Oi is open in A.
Further, if x ∈ O and |x| = 1 then x ∈ Oi some i ∈ I and 2x ∈ Oi , since Oi is open
in X. Consequently, 2x ∈ O.
∩
Let O1 , O2 , . . . , On be open in X. Then O = ni=1 Oi is open in A. Also, if x ∈ O
and |x| = 1, then x ∈ Oi , i = 1, 2, . . . , n and so 2x is in Oi , 1, 2, . . . , n since each Oi
is open in X. Consequently, 2x ∈ O.
The empty set and X obviously satisfies the criterion of being open in X. Hence X
is a topological space.
The two circles considered individually as subspaces of X have the same topology
as when considered as subspaces of R2 . As S 1 is⨿known to be connected, so is each
of the circles as subspaces of X. Thus if X = U V, where U and V are open in X
the circle |x| = 1 has to be contained in a single one of them, let’s say U . But then
also 2x ∈ U, whenever |x| = 1. This shows that U contains all of X.
Take a point x ∈ X with |x| = 1. Then any open set containing x will also contain
2x. Hence it is not possible to simultaneously separate x and 2x and X is not
Hausdorff. As any metric space is Hausdorff it is not possible to find a metric on X
inducing its topology.
3. As X is Hausdorff and gx ̸= x we can find open sets Vgx containing gx and Vx
containing x such that Vgx ∩ Vx = ∅.
Denote the action map by µ : G × X → X. Then µ−1 (Vgx ) is open and contains
(g, x). By definition of the product topology on G × X, there are then open sets O
in G containing g and Vx′ in X containing x ∈ X such that O × Vx′ ⊂ µ−1 (Vgx ), i.e.
O · Vx′ ⊂ Vgx .
If we put V = Vx′ ∩ Vx we will have an open set containing x such that O · V ⊂ Vgx
and V ⊂ Vx with
O · V ∩ V ⊂ Vgx ∩ Vx = ∅.
⨿
4. Let X = S 3 − (S11 S21 ) so that (z, w) exactly when both z and w are non-zero.
Define a map f : X → S 1 ×√S 1 by√f (z, w) = (z/|z|, w/|w|). Define a map g :
S 1 × S 1 → X by g(u, v) = (u/ 2, v/ 2).
√
√
Then f (g(u, v)) = (u, v), where as g(f (z, w)) = (z/( 2|z|), w/( 2|w|)).
Define the map ρ : X × I → (0, ∞) by
√
√
ρ(z, w, t) = |((1 − t) + t/( 2|z|))z, ((1 − t) + t/( 2|w|))w)| =
(
√
√ )1/2
=
((1 − t)|z| + t/ 2)2 + ((1 − t)|w| + t/ 2)2
Next define F : X × I → X by
F (z, w, t) =
(
)
√
√
1
((1 − t) + t/( 2|z|))z, ((1 − t) + t/( 2|w|))w
ρ(z, w, t)
√
√
And note that F (z, w, 0) = (z, w) as ρ(z, w, 0) = 1 and F (z, w, 1) = (z/( 2|z|), w/( 2|w|)),
as ρ(z, w, 1) = 1.
This shows that f : X → S 1 × S 1 is a homotopy equivalence. Hence the two spaces
have isomorphic fundamental groups, so π1 (X) ∼
= Z×Z as we know the corresponding
result for the 2-tours S 1 × S 1 .
5. We can view Pn × Pn as the orbit space of S n × S n by the action of G = C2 × C2
(C2 is cyclic of order two) specified by
σ(x, y) = (−x, y) and τ (x, y) = (x, −y)
where σ and τ are generators of the two factors of C2 .
The quotient map q : S n × S n → P × P is therefore open.
We have q −1 (X) = Y, where Y ⊂ S n × S n is the subspace consisting of those
pairs (x, y) where x · y ̸= 0. This is an open subset as the dot-product is a map.
Consequently the restriction of q, q| : Y → X is open and thus an identification.
Define a map f˜ : Y → S n as follows:
 x+y


|x + y|
f (x, y) =
x−y


|x − y|
if
x·y >0
if
x·y <0
It’s defined in two different ways on two disjoint open subsets of Y and it is obviously
continuous and well defined on each of this. Thus f is continuous. Note that f (x, y) is
a directional vector for the line bisecting the acute angle between the lines generated
by x and y.
Let q ′ : S n → Pn denote the usual quotient map. Then q ′ f˜ is constant on the fibers
of q and hence induces a map f : X → Pn and f (l1 , l2 ) bisects the acute angle
between the two lines (which are not orthogonal).
This answers the first question.
For the second note that the diagonal map ∆ : Pn → Pn × Pn has image in X and
that f (∆(l)) = f (l, l) = l.
Suppose f −1 (A) is open. Then so is A = ∆−1 (f −1 (A)). Hence f is an identification.
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