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Homework 1, Algebraic Topology, 215A, Fall 2006
(1) Products. Let X and Y be topological spaces. We define their product X × Y
to be the set of pairs (x, y), x ∈ X, y ∈ Y , with the following product topology: A
subset W ⊂ X × Y is defined to be open if for every point (x, y) ∈ W there exist
open neighborhoods U of x and V of y such that U × V ⊂ W . It is not hard to
check that this indeed defines a topology on X × Y . Equivalently, we could define
the product topology on X × Y as the topology whose base is given by all products
U × V of open sets U ⊂ X, V ⊂ Y .
(a) Show that for all m, n ∈ N we have Rm × Rn ∼
= Rm+n .
(b) Show that a map (f1 , f2 ) : Z −→ X × Y is continuous if and only if the maps
f1 : Z −→ X and f2 : Z −→ Y are continuous.
(c) Conclude that this implies the following universal property of the product X ×Y :
Given maps fi , i = 1, 2, as in (b) there exists a unique map g such that the following
diagram commutes:
X ×Y
projX
.
f1
↑g
projY
&
f2
X ←−
Z
−→ Y
(d) Use this to show that there is a bijection of sets
Hom(Z, X × Y ) ∼
= Hom(Z, X) × Hom(Z, Y ),
where Hom(Z, W ) denotes the set of continuous maps from Z to W .
Note: A particularly nice aspect of the bijection of part (d) is that it is natural
in the following sense: For all continuous maps φ : Z 0 −→ Z, ψ1 : X −→ X 0 , and
ψ2 : Y −→ Y 0 we get a commutative diagram
∼
=
Hom(Z, X × Y ) −→ Hom(Z, X) × Hom(Z, Y )
↓
↓
∼
=
Hom(Z 0 , X 0 × Y 0 ) −→ Hom(Z 0 , X 0 ) × Hom(Z 0 , Y 0 ).
Here the vertical arrows are defined by pre- and post-composition with the given
functions φ and ψi (how exactly?). One uses the word natural here to indicate that
the isomorphism in (d) does not reflect specific properties of the spaces X, Y , and
Z, but rather describes a general feature of the product construction.
(2) Manifolds. In this problem we want to define an important class of topological spaces, so-called manifolds, and look at some examples. Let M be topological
space and recall that M is Hausdorff if for any two distinct points x, y ∈ M there
exist disjoint open neighborhoods U of x and V of y, i.e. two points in M can be
“separated” by open sets. M is second countable if there exists a countable base for
the topology on M .
We define a n-dimensional (topological) manifold to be a second countable topological Hausdorff space that is locally homeomorphic to Rn . By this we mean the
following: For each point x ∈ M there exists an open neighborhood U of x and a
∼
=
homeomorphism φ : U −→ Rn . We call such a φ a chart for M .
(a) Show that every open subset of Rn is a manifold. This implies that we could
replace ’Rn ’ in the above definition with ’an open subset of Rn .’
P
2
(b) Let S n := {x ∈ Rn+1 / |x|2 = n+1
i=1 xi = 1} be the n-dimensional sphere. Show
n
that for each x ∈ S we have a homeomorphism S n \{x} ∼
= Rn . Conclude that S n
is an n-dimensional manifold.
(c) Show that the product of two manifolds is a manifold. We see for example the
torus/donut S 1 × S 1 is a manifold.
(d) Give an example of a topological space M which is locally homeomorphic to Rn ,
but which is not Hausdorff.
Note: The second countability axioms is useful when one wants to construct interesting functions and geometric structures on a manifold. The empty set is a
manifold of any dimension (a surprisingly important fact), otherwise the dimension
of a manifold is uniquely determined, see the comment on the invariance of dimension in the note on problem 3.
(3) A Peano curve. The heart of this problem is the construction of a continuous
map from the unit interval I = [0, 1] onto the square I 2 ⊂ R2 . Consider a continuous
function f : R −→ [0, 1] such that
(
0 if 0 ≤ t ≤ 13
f (t) =
1 if 32 ≤ t ≤ 1
and f (t + 2) = f (t) for all t ∈ R. Set
X
X
x(t) :=
f (32n−1 t)2−n and y(t) :=
f (32n t)2−n
n≥1
n≥1
Define
γ : I −→ I 2 , γ(t) := (x(t), y(t))
(a) Show that γ is continuous. Here you can use the fact that a uniform limit of
continuous functions is again continuous.
(b) Prove that γ is surjective.
Hint: A point (x0 , y0 ) ∈ I 2 can be represented as
X
X
x0 (t) :=
a2n−1 2−n , y0 (t) :=
a2n 2−n ,
n≥1
n≥1
P
−k−1
where ak = 0 or 1. Then γ maps t0 := k≥1 2ak 3
to (x0 , y0 ).
(c) Show that γ is not injective.
(d) Conclude that there is a surjective map I −→ I n for all n ∈ N.
Note: These examples raise the question whether there exists a continuous, bijective
map from the interval I to the n-dimensional cube I n . However, it is known that
such a map doesn’t exsist. The background is the so-called theorem on invariance of
dimension: The existence of a homeomorphism φ : U −→ V between open subsets
U ⊂ Rm and V ⊂ Rn implies m = n. (We’ll prove this theorem this semester.)
Now assume there existed a continuous bijective map γ : I −→ I n . Then γ is
automatically a homeomorphism (this follows using the compactness of I and I n ),
and restriction of γ to the subset γ −1 (γ((0, 1))∩(0, 1)n ) would give a homeomorphism
between open subsets of Euclidian spaces of different dimensions which is impossible
according to the invariance of dimension.
Please return in class on Tuesday, Sept. 5.