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Lemma.
Let {Ui} be an open covering of the space X having the following
properties:
(a) There exists a point x0 such that x0 Ui for all i.
(b) Each Ui is simply connected.
(c) If i≠j, then Ui Uj is arcwise connected.
Then X is simply connected.
Prove the lemma using the following approach:
To prove that any loop f: IX based at x0 is trivial, first consider the open covering
{f-1(Ui)} of the compact metric space I and make use of the Lebesgue number of this
covering.
We say  is a Lebesgue number of a covering of a metric space X if the following
condition holds: any subset of X of diameter <  is contained in some set of the covering.
Restate the lemma for the following special cases:
(1) A covering by two open sets
(2) The sets {Ui} are linearly ordered by inclusion
Using the restated lemma for special case (1), prove that the unit n-sphere Sn, n ≥ 2, is
simply connected.
Solution with each definition clearly
Connected space: A connected set is a set that cannot be partitioned into two
nonempty subsets which are open in the relative topology induced on the set.
Equivalently, it is a set which cannot be partitioned into two nonempty subsets such that
each subset has no points in common with the set closure of the other.
Let be a topological space. A connected set in is a set
which cannot be
partitioned into two nonempty subsets which are open in the relative topology induced on
the set . Equivalently, it is a set which cannot be partitioned into two nonempty subsets
such that each subset has no points in common with the set closure of the other. The
space is a connected topological space if it is a connected subset of itself.
The real numbers are a connected set, as are any open or closed interval of real numbers.
The (real or complex) plane is connected, as is any open or closed disc or any annulus in
the plane. The topologist's sine curve is a connected subset of the plane. An example of a
subset of the plane that is not connected is given by
Geometrically, the set is the union of two open disks of radius one whose boundaries are
tangent at the number 1.
Simply connected space:
Here it is given that every Ui simply connected.
A path-connected domain is said to be simply connected (also called 1-connected) if any
simple closed curve can be shrunk to a point continuously in the set. If the domain is
connected but not simply, it is said to be multiply connected. In particular, a bounded subset
of
is said to be simply connected if both
are connected.
and
, where
denotes a set difference,
A space is simply connected if it is path-connected and if every map from the 1-sphere to
extends continuously to a map from the 2-disk. In other words, every loop in the space is
contractible.
An arc connecting two points and of a topological space is not simply connected (like a
path) a continuous function
such that
and
, but must also have a
continuous inverse function, i.e., that it is a homeomorphism between
and the image of
.
The difference between the two notions can be clarified by a simple example. The set
with the trivial topology is pathwise-connected, but not arcwise-connected since the
function
defined by
for all
, and
, is a path from to , but
there exists no homeomorphism from
to , since even injectivity is impossible.
So
To prove that any loop f: IX based at x0 is trivial,
Then if we consider the open covering
{f-1(Ui)} of the compact metric space I and make use of the Lebesgue number of this
covering.
We say  is a Lebesgue number of a covering of a metric space X if the following
condition holds: any subset of X of diameter <  is contained in some set of the covering.
Since Every open covering of a sequentially compact set has a Lebesgue number s>0.
If M, N are metric spaces with M compact (since covering compact and sequentially
compact are equivalent)
then any continuous f : M->N is uniformly continuous. Indeed, e>0, the (e/2)neighborhoods in N constitute an open cover of M,
which has a Lebesgue number s>0 from the Lebesgue number lemma.
Thus, each f(N_s(x)) with x in M is contained in some N_(e/2)(y), so
d(x,x')<s which implies that d(f(x),f(x'))<e, as claimed.
Hence it is Simply connected.
Restating of the lemma:
Let U1,U2 be an open covering of the space X having the following properties:
(d) There exists a point x0 such that x0 Ui for all i. i = 1,2
(e) Each Ui is simply connected. I = 1,2
(f) If i≠j, then Ui Uj is arcwise connected. i,j =1,2.
Then X is simply connected.
T o prove that the unit n-sphere Sn, n ≥ 2, is simply connected.
The unit n – Sphere is Sn .
If we take a mapping from f : Ui
Sn
Here there exits a homeomorphism f from Ui to Sn also f-1 is a continuous function
so the connection is very clear. Such that Ui Will form the covering with inclusion order
and each Ui is connected so it covers Sn .so is Sn.
Here of importance is for continuity and Homeomorphism. Once these two hold good
yhen Sn is automatically connected.
If you have any doubts fell free to ask them..
Thank u.