Congratulations on your study of algebraic topology. It is truly a
beautiful subject, intertwining elegantly the two already beautiful
subjects of group theory and topology.
It took me three tries to understand the subject: first using a really
bad undergraduate textbook in a course I took when I was 14, believe
it or not; then an abysmal episode where I tried to understand
Spanier, who was much too austere for me; and finally I learned the
basics by doing all the exercises in Rotman's textbook on the subject.
I must say that the Hatcher textbook, from what I've seen, looks
really excellent. (I am impressed by how quickly he illustrates with
interesting applications, like Brouwer and the fundamental theorem of
algebra).
I understand that algebraic topology has applications in some aspects
of physics, but I never was able to learn physics to that level. I
looked around for applications in computer science for a while, but
there were not so many.
Anyway, I think the reason that this exercise is a little bit trickier
than it looks is that it is a highly "technical" exercise that depends
on the exact notions of where basepoints are located, details of
parameterizations, and so on. Although fundamental, usually people
sort of read through these types of proofs once, then just use
higher-level theorems. So it's clear that you are reading this book
quite closely to notice that seemingly "obvious" facts are not always
so obvious.
There certainly is an underlying and fairly simple geometric intuition
going on, but I am not sure how to convey that here, so I will just
give the proof.
Let's begin by rephrasing the question:
Part 1: Let X be a topological space, let S^1 be the unit circle, and
let D^2 be the closed unit disk. All maps will be continuous.
Then the following conditions are equivalent:
(a) For each
f:S^1 --> X
there is an x_0 in X such that f is homotopic to the constant map at
x_0 .
(b) For each f:S^1 --> X there is a g:D^2-->X such that g restricted
to S^1 equals f.
(c) For each x_0 in X, the fundamental group pi_1(X,x_0) is the
trivial group.
Part 2. A space X is simply connected if and only if any two maps
f:S_1 --> X and h:S_1 --> X are homotopic.
------------
Preliminary comments
There are two key subtleties I would like to get out of the way.
The first is that the word "homotopy" has two DIFFERENT definitions.
There is homotopy of maps, defined on page 3 of the Hatcher text, and
then there is homotopy of paths defined on page 25.
The main difference between them is that a path homotopy must keep the
endpoints fixed throughout the homotopy. This all can get a little
fuzzy when we want to treat paths as maps, and consider map homotopy
between them.
Therefore, since all out path homotopies will be of loops, in the
following I will use the term "homotopy" for standard homotopy of
maps, and the term "loop homotopy" for the path homotopy that keeps
endpoints fixed.
The second subtlety has to do with parameterizations of S^1 and D^2 .
I will use polar coordinates here, identifying a point on S^1 with its
angle theta, and identifying a point in D^2 with its polar coordinates
(r,theta) .
Technically, I should probably make all these parametrizations
explicit, but the author also routinely identifies S^1 and "theta". If
these implicit parametrizations make the ideas unclear, just let me
know and I can write them out more explicitly.
We will first prove part 1, then prove part 2.
----
PROOF OF PART 1
-----
We show that 1(a) implies 1(b); 1(b) implies 1(c); and 1(c) implies
1(a) .
-----
1(a) implies 1(b)
-----
We suppose that for each
f:S^1 --> X
there is some
x_0 in X
such that f is homotopic to the constant map at x_0.
Let f:S^1 --> X be some map. By (a), there is some x_0 in X such that
f is homotopic to the constant map at x_0. We will construct a g:D^2
-> X such that g restricted to S^1 equals f.
Let h:S^1 -->X be the constant map at x_0, so that h(theta)=x_0 for
all theta in [0,2*pi]
Let I be the closed unit interval. By the definition of homotopy,
there is a map
H: S^1 x I -> X
such that
H(theta,0)=f(theta), and
H(theta,1)=h(theta)=x_0
for all theta.
Define g:D^2 --> X by
H(theta,1-r) if r>0
g(r,theta)=
x_0 if r=0
I claim that
(i) g restricted to S^1 equals f .
(ii) g is continuous.
Once I prove these two things, (a)=>(b) will follow.
(i) is true because
g(1,theta) = H(theta,0) = f(theta)
To show (ii) is true is kind of a textbook "technical" lemma. I will
do a little handwaving, but I can be more explicit if you like.
First, because H is continuous, it is clear that g is continous at all
(r,theta) for r>0 .
Now we want to show that H is continuous at r=0 . We have to be a
little careful here because X might not be Hausdorff, and it's
certainly not a metric space.
So let U be any open set containing x_0 in X .
To show that g is continuous at r=0, it suffices to show that
g^{-1}(U) contains a neighborhood of the origin in D^2 .
Now, for each theta in [0,2*pi], H(theta,1)=x_0.
Hence, for each such theta, there is an epsilon_theta, and some open
subset U_theta of theta in S^1 , such that
H(U_theta, (1-epsilon_theta,1]) is a subset of U .
The
(1-epsilon_theta,1]
term above refers to all t in I such that 1-epsilon_theta < t < 1 .
So, this is true for each theta in S^1 .
Thus, the U_theta cover S^1 .
Since S^1 is compact, a finite number of the U_theta cover S^1 :
U_theta_1, U_theta_2,...,U_theta_n .
Let epsilon be
min(epsilon_theta_1, epsilon_theta_2, ... , epsilon_theta_n )
Now it can be seen that
H(theta,(1-epsilon,1]) is a subset of U.
Indeed, each theta is in some U_theta_i, and since
H(U_theta_i,(1-epsilon_theta_i,1]) is a subset of U ,
we must have H(theta,(1-epsilon,1]) is a subset of U.
Then image of the disk of radius epsilon centered at the origin has
image in U under g .
If we call this little disk of radius epsilon E, then
g(E)
= g({r,theta:r<epsilon})
= H({theta,1-r})
subset of U by the above.
It follows that g is continuous at [0,0], and hence g is continuous
everywhere.
(a) => (b) follows
--------------
Proof of 1(b) ==> 1(c)
We assume (b), so every f: S^1 -> X extends to a g:D^2 -> X .
We want to show that pi_1(X,x_0) = 0 . To do this, it suffices to show
that any loop in X based at x_0 is loop homotopic to the constant
loop.
Let h:I -> X be a loop with h(0)=h(1)=x_0 .
We will show h is loop homotopic to the constant loop.
Let
u:S^1 -> X with
u(theta)=h(theta/(2*pi)) .
It is clear that u is continuous, so u extends to a map g:D^2 -> X
because we assume (b) .
Now we know that since D^2 is convex it certainly has a trivial
fundamental group. In particular, let's let
v:I->D^2
be the map that sends
t |-> (1,2*pi*t).
Then there is a loop homotopy in D^2 from v to the constant (1,0).
H:IxI -> D^2
H(t,0) = v(t)
H(t,1) = (1,0)
H(0,s)= (1,0)
H(1,s)= (1,0)
Now, the map
G: IxI -> X = g o H
is a loop homotopy between h and x_0.
Indeed, G is certainly continuous, and
G(t,0)
= g(H(t,0)) , by definition of g
= g(v(t)) , since H is a homotopy
=g(1,2 pi t) , by the definition of v
=u(2 pi t), since g equals u on S^1
=h(t) by the definition of u
Similarly,
G(t,1)
=g(H(t,1)) , by definition of G
=g(1,0) , since H is a loop homotopy
=u(0) , since g equals u on S^1
=h(0) , by definition of u
=x_0 since h is a loop.
To check that G is a real loop homotopy and keeps the endpoints fixed,
we have:
G(0,s)
=g(H(0,s)) by definition of G
=g(1,0) since H is a loop homotopy
=x_0 by the above .
G(1,s)
=g(H(1,s)) by definition of G
=g(1,0) since H is a loop homotopy
=x_0 by the above
So G is indeed a loop homotopy between h and the constant loop, and so
the fundamental group is trivial at x0, and so (b)==>(c) follows.
------------
1(c) => 1(a)
Suppose the fundamental group of X is trivial at each x_0 and let
f:S^1 -> X have f(0)=x_0 . We want to show that f is homotopic to the
constant map at x_0 .
Let
u: I -> X
send t into f(2 pi t) .
The u is a loop in X at x_0, and so there is a loop homotopy at x_0
between u and x_0, that is, there is some
H:I x I -> X
H(t,0) = u(t)
H(t,1) = x_0
H(0,s) = x_0
H(1,s) = x_0 .
Then the map:
G:S^1 x I -> X
that sends
(theta,s) |--> H(theta/(2 pi), s)
is a homotopy between f and the constant x_0 .
Indeed, G is certainly continuous, and we have
G(theta,0)
=H(theta/(2 pi), 0) by the definition of H
=u(theta/(2 pi)) since H is a homotopy
=f(theta) by the definition of u.
G(theta,1)
=H(theta/(2 pi), 1) by the definition of H
=x_0 since H is a homotopy
So G is a homotopy between f and x_0, and (c)=(a) follows, as does
part 1.
--------------------
PART 2
-------------------
Here, we need to show that X is simply connected if an only if all
maps S^1 -> X are homotopic.
----------
Part 2: "if" direction
--------
Suppose all maps S^1 to X are homotopic. Then X is path-connected: If
there are two points x and y in X, let f:S^1 --> X and g:S^1 -> X be
constant maps with values x_0 and x_1 respectively. There is a
homotopy
H:S^1 x I into X
between f and g such that
H(theta, 0)= f(theta)=x_0
H(theta,1) = g(theta)=x_1
So the map
p:I -> X
defined by
p(t)=H(0,t)
is a path in X from x_0 to x_1 .
Hence, X is path-connected.
Now, to show X is simply connected, we just need to show it has
trivial fundamental group.
But this follows from 1 above: since all maps from S^1 to X, 1(a)
holds, hence 1(c) holds, hence X has trivial fundamental group.
------
Part 2: "only if" direction
We need to show that if X is simply connected, then any two maps from
S^1 to X are homotopic. Let x_1 in X , and let f:S^1 -> X .
We show that f is homotopic to the constant g:S^1->X sending theta to
x_1 .
Now, we know that for some x_0, f is homotopic to the constant map at
x_0 , because of 1(c) implies 1(a) above.
But it is clear that, since X is path connected, the constant map at
x_0 is homotopic to the constant map at x_1 . Indeed,
if p:I -> X
is a path with p(0)=x_0 and p(1) =x_1, then
H:IxI ->X
H(t,s) = p(s)
is a homotopy between these constant functions.
Since each map f:S^1 -> X is homotopic to the a constant path, and any
two constant paths are homotopic, any two such maps are homotopic,
(since homotopy is an equivalence relation).
Therefore, if X is simply connected, any two maps from S^1 to X are
homotopic. The "only if" direction now follows, as does part 2, as
does the entire problem.
----------
References:
Algebraic Topology, By Allen Hatcher. Cambridge University Press, 2001
(ISBN: 0521795400). Online edition at:
http://www.math.cornell.edu/~hatcher/AT/ATpage.html
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