1Definitions
II Algebraic Topology
1.1 Some recollections and conventions
We will start with some preliminary definitions and conventions.
Definition (Map). In this course, the word map will always refer to continuous
maps. We are doing topology, and never care about non-continuous functions.
We are going to build a lot of continuous maps in algebraic topology. To do
so, we will often need to glue maps together. The gluing lemma tells us that
this works.
Lemma (Gluing lemma). If
f
:
X → Y
is a function of topological spaces,
X
=
C ∪ K
,
C
and
K
are both closed, then
f
is continuous if and only if the
restrictions f|
C
and f|
K
are continuous.
Proof. Suppose f is continuous. Then for any closed A ⊆ Y , we have
f|
−1
C
(A) = f
−1
(A) ∩ C,
which is closed. So f|
C
is continuous. Similarly, f |
K
is continuous.
If f|
C
and f|
K
are continuous, then for any closed A ⊆ Y , we have
f
−1
(A) = f|
−1
C
(A) ∪ f|
−1
K
(A),
which is closed. So f is continuous.
This lemma is also true with “closed” replaced with “open”. The proof is
the same proof with “closed” replaced with “open”.
We will also need the following technical lemma about metric spaces.
Lemma. Let (
X, d
) be a compact metric space. Let
U
=
{U
α
}
α∈A
be an open
cover of
X
. Then there is some
δ
such that for each
x ∈ X
, there is some
α ∈ A
such that B
δ
(x) ⊆ U
α
. We call δ a Lebesgue number of this cover.
Proof.
Suppose not. Then for each
n ∈ N
, there is some
x
n
∈ X
such that
B
1/n
(
x
n
) is not contained in any
U
α
. Since
X
is compact, the sequence (
x
n
)
has a convergent subsequence. Suppose this subsequence converges to y.
Since
U
is an open cover, there is some
α ∈ A
such that
y ∈ U
α
. Since
U
α
is open, there is some
r >
0 such that
B
r
(
y
)
⊆ U
α
. But then we can find a
sufficiently large n such that
1
n
<
r
2
and d(x
n
, y) <
r
2
. But then
B
1/n
(x
n
) ⊆ B
r
(y) ⊆ U
α
.
Contradiction.