Download Let X be a path-connected space and suppose that every map f: S^1

Let X be a path-connected space and suppose that
every map f: S^1 --> X is homotopically trivial but not necessarily
by a homotopy leaving the base point x_0 fixed. Show that 1(X, x0)
= 0.
Need a detailed proof outline and a summary of background
information necessary for the proof
Solution:
Some standard results:
1) Suppose X is a space, and x0 is a point of X. We define 1(X, x0) as
homotopy classes of maps f: [0, 1]  X, such that f (0) = f (1) = x0.
In other-words, homotopy means that one views as equivalent two maps
which lie on a 1-parameter family of functions ft [0,1]  X which satisfy the
boundary conditions ft(0) = ft(1) = x0.
So, we call 1(X, x0) as fundamental group of X, if X is path connected.
2) We say a space is “simply connected” if its fundamental group is trivial.
3) Covering space: A map p: X  Y is a covering map if around each point
y in Y there exists a neighborhood Ny of b, so that p-1(Ny) is a disjoint union
of sets Ai each of which is mapped homeomorphically onto Ny by map p.
3) If X is a contractible space then the fundamental group is trivial.
4) A topological space X is called simply connected if it is path-connected
and any continuous map f : S1 -> X (where S1 denotes the unit circle in
Euclidean 2-space) can be contracted to a point in the following sense: there
exists a continuous map F : D2 -> X (where D2 denotes the unit disk in
Euclidean 2-space) such that F restricted to S1 is f.
An equivalent formulation is this: X is simply connected if and only if it is
path connected, and whenever p : [0,1] → X and q : [0,1] → X are two paths
(i.e.: continuous maps) with the same start and endpoint (p(0) = q(0) and
p(1) = q(1)), then p and q are homotopic relative {0,1}. Intuitively, this
means that p can be "continuously deformed" to get q while keeping the
endpoints fixed. Hence the term simply connected: for any two given points
in X, there is one and "essentially" only one path connecting them.
A third way to express the same: X is simply connected if and only if X is
path-connected and the fundamental group of X is trivial, i.e. consists only of
the identity element.
So it is enough to show that X is simply connected (0r) the map is an
identity map such that i(x0) = x0 leaving x0 as fixed point. It is enough to
show that x0 is indeed a fixed point and i is the identity element of 1(X,
x0). To show that X is simply connected, we use the preposition given
below.
Preposition:
A space X is simply connected iff there is a unique homotopy class of paths
connecting any two points in X.
Proof:
We are already given that X is a path-wise connected, so it enough to prove
the uniqueness of connecting paths. Suppose 1(X,x0) = 0[where x0 is the
base point]. If f and g are two paths connecting from x0 to x1, then
f  f  g’  g  g since the loops fg’ and gg’ are homotopic to constant
loops under the assumption that 1(X) = 0.
Conversely if there is only one homotopic class of paths connecting a base
point x0 to itself, then all the loops to x0 are homotopic to a constant loop
and hence 1(X, x0) =0.
Hope the above result helps you here… 
Thanks!