Let *X,Y* be two topological spaces, and *I*
the closed unit interval [0,1]. Two maps *f,g* from
*X* to *Y* are called *homotopic*
if there exists a map *F* from *X × I*
to *Y* such that *F(x,0) = f(x)* and
*F(x,1) = g(x)* for all *x*. Here *F* is
called a *homotopy* (from *f* to *g*).
Intuitively, the second argument can be viewed as time, and
then the homotopy describes a continuous deformation from
*f* to *g*.

A map is called *nullhomotopic* when it is
homotopic to a constant map.

A space is called *contractible* when the identity
map from the space to itself is nullhomotopic.
For example, each real topological vector space is contractible -
a suitable homotopy is given by *F(x,t) = (1-t)x*.
More generally, each convex subset of a real topological vector space
is contractible.

If *Y* is contractible, then each map from some
topological space *X* to *Y* is nullhomotopic.
Also, if *X* is contractible, then each map from *X*
into some topological space *Y* is nullhomotopic.

**Exercises**

(i) If *Y* is not pathconnected, then not all constant maps
into *Y* are homotopic.

(ii) Let *S(n)* be the *n*-dimensional sphere
(points of norm 1 in *(n+1)*-space).
If *f,g* are two maps from some topological space
*X* into *S(n)*, and for all *x*
the values *f(x)* and *g(x)* are not antipodal,
then *f* and *g* are homotopic.
In particular, a nonsurjective map into *S(n)* is nullhomotopic.

(iii) A map *f* from *S(n)* into some topological space
*Y* is nullhomotopic if and only if *f* has a
continuous extension to the *(n+1)*-ball (points of norm
at most 1 in *(n+1)*-space).

Let *X,Y* be two topological spaces, and *A* a subspace of *X*.
A homotopy *F* from *X × I* to *Y* is called
homotopy *relative to A* if for each *a* in *A*
the map *F(a,t)* is constant (independent of *t*).

Being homotopic is an equivalence relation, so we have
equivalence classes.
Given two spaces *X,Y*, and a map *f* from *X*
to *Y*, let *[f]* denote the homotopy class of *f*,
that is, the set of all maps from *X* to *Y* homotopic
to *f*.
Let *[X,Y]* denote the set of homotopy classes of maps
from *X* to *Y*.

Under suitable assumptions homotopy classes are precisely the
path-components in the space *C(X,Y)* of continuous functions
from *X* to *Y*.
[E.g., give *C(X,Y)* the compact-open topology, and
assume that *X* is a k-space, i.e., has the topology
defined by the injection maps from its compact subspaces.
Then this statement holds.]

Again, we can look at relative homotopy, and similar things hold.

**Exercise**
Let *f* be a path (a map from the unit interval *I* into
some topological space *X*). Let *p* be a continuous map
from *I* to itself fixing 0 and 1 (a parameter transformation).
Show that *[f] = [fp]* (where juxtaposition denotes composition:
*(fp)(t) = f(p(t))*).

We call two paths *f,g* (maps from *I* into some space
*X*) homotopic when they are homotopic relative to *{0,1}*.
Given a path *f*, the *inverse path* is the path *f'*
defined by *f'(t) = f(1-t)*.
Given two paths *f,g*, where the endpoint *f(1)* of the first
is the starting point *g(0)* of the second, their *product*
is the path *h = f#g* defined by *h(t) = f(2t)* for
*2t < 1* and *h(t) = g(2t-1)* otherwise.

**Theorem**
Let *X* be a topological space and fix a point *x* in it.
Let *e(x)* be the constant map that sends all of *I* to
*x*.
Let *P(X;x)* be the set of homotopy classes of paths starting and ending
in *x*. Then *P(X;x)* is a group
with respect to the multiplication defined by *[f][g] = [f#g]*.
It has inverses defined by *[f]' = [f']* and unit element
*[e(x)]*.
When *X* is path-connected, all groups *P(X;x)* are isomorphic.

**Proof** We have to check that the operations are well-defined. They are.

The group found here is called the *Poincaré group* or
*fundamental group* of *X*
(at the point *x*).

**Exercise**
(i) Show that the fundamental group of the circle *S(1)* is *Z*,
the additive group of the integers.
(ii) Show that the fundamental group of the sphere *S(n)* with *n > 1*
is trivial.

**Theorem**
Let *X,Y* be two spaces, and let *f* be a map
from *X* to *Y*. We find a homomorphism *f**
from *P(X;x)* to *P(Y;f(x))* by sending a path *p*
to the path *fp* (defined by *fp(t) = f(p(t))*).

This is a functor from pointed topological spaces with maps to groups
with homomorphisms. In particular: *(gf)* = g* f**
and the identity map on *X* is sent to the identity
homomorphism on *P(X;x)*.

It follows that if *f* is a homotopy equivalence from
*X* to *Y* (that is, a map such that there is
a map *g* from *Y* to *X* such that
both *fg* and *gf* are nullhomotopic),
then *f** is an isomorphism.

A topological space *X* is called *simply connected*
when it is path-connected and its fundamental group *P(X)* is trivial.

**Exercise**
A path-connected space *X* is simply connected if and only if
each map from the circle *S(1)* into *X* can be extended to
a map from the 2-ball (circle plus interior) into *X*.

**Theorem**
The fundamental group of a product is the product of the fundamental groups
of the factors.

**Proof**
If *X* is a product of factors *Xi*, and *pi* is
the projection onto *Xi*, then *pi** maps *P(X;x)*
into *P(Xi;xi)*,
and the homomorphism *p* with *i*-th coordinate *pi*
maps *P(X;x)* into the product of the *P(Xi;xi)*.
Now *p* is trivially surjective and injective, so is an isomorphism.

**Exercise**
Show that the product of simply connected spaces is simply connected again.

Let *X* be a topological space. A subspace *A* of *X*
is called a *retract* of *X* if there is a map
*r* from *X* onto *A* that extends the identity on *A*.
The subspace *A* is called a *deformation retract* of *X*
if there is a map *r* from *X* onto *A* that is homotopic
to the identity on *X* relative to *A*.

**Exercise** Clearly, every deformation retract is a retract.
Show that the converse does not hold.

Given a topological space *X* and a subspace *A*, let *i*
be the inclusion map. Let *a* be a point of *A*.
Then *i** is a homomorphism from *P(A;a)* into *P(X;a)*.

**Exercise** Show that *i** need not be injective.

**Theorem**
If *A* is a retract of *X*, then *i** is injective.

**Theorem**
If *A* is a deformation retract of *X*, then *i**
is an isomorphism.

**Exercise** Show that a retract of a simply connected space is again
simply connected.

Let *X* be a topological space that is the union of two
path-connected subspaces *A* and *B*, where the
intersection of *A* and *B* is nonempty and path-connected.
Then the fundamental group of *X* is generated by (the images of)
the fundamental groups of *A* and *B*.

More generally we have:
**Theorem** (Van Kampen)
Under these same hypotheses, let *f* and *g* be homomorphisms
from *P(A;c)* (resp. *P(B;c)*) into some group *G*
that agree on the intersection *C* of *A* and *B*.
(Here *c* is a point of *C*.)
Then there is a unique homomorphism *h* from *P(X;c)* into
*G* that agrees with *f* and *g* on *A* and
*B*, respectively.

Perhaps more hypotheses are required?

If *X* is a topological group then we have a multiplication of paths
given by *(f.g)(t) = f(t).g(t)* where *.* denotes the group
operation.

**Theorem**
The group operation on *X* induces a group operation on *P(X;x)*
that coincides with the old group operation, and *P(X;x)* is
commutative.

**Proof**
First of all *[f].[g] = [f.g]* is well-defined.
Next, if *e = e(x)*, then
*[f].[g] = [f#e].[e#g] = [f#g] = [f][g]*
and *[f].[g] = [e#f].[g#e] = [g#f] = [g][f]*.

The fundamental group of *SO(2,R)*, the group of orthogonal
transformations of determinant 1 of the real plane, is isomorphic
to *Z*, the additive group of the integers.
(Indeed, this group is isomorphic to *S(1)*.)
The fundamental group of *SO(n,R)* with *n > 2* is isomorphic to
*Z2*, the additive group of the integers mod 2.

A *Hopf space* is a space in which the proof given above
for the statement that the fundamental group of a topological group
is Abelian still works. Thus, by definition, the fundamental group
of a Hopf space is Abelian.
What do we need? A multiplication *.* on the space that preserves
homotopy so that it induces a multiplication on homotopy classes,
and a unit element *e* for this multiplication such that
*[e#f] = [f] = [f#e]*.

Instead of using maps from *I* into *X*, with the condition
that both endpoints are mapped to a given point, we consider maps from
*I^n* into *X*, with the condition that the entire boundary
(the set of all points that have at least one coordinate 0 or 1) is mapped
to a given fixed point *x*. Now again we have a composition #
(acting on the first coordinate only) that induces a group operation
on the homotopy classes. The group obtained is called *pi_n(X;x)*.
(Thus, *pi_1(X;x) = P(X;x)*.)

Let *L(X;x)* be the space of loops in *X* with base point
*x* (i.e., maps from *I* to *X* sending 0 and 1
to *x*), provided with the compact-open topology.
For suitably nice (say, locally compact) *X* we have
*pi_2(X;x) = P(L(X;x);e)* when *e* is the constant map
that sends *I* to *x*, and more generally
*pi_n(X;x) = P(L_(n-1)(X;x);e)* where
*L_n(X;x) = L(L_(n-1)(X;x);e)* for a suitable constant
*e = e_n*.

For *n > 1* the group *pi_n(X;x)* is Abelian.
Indeed, *L(X;x)* is a Hopf space for the operation #.

Let *E* and *B* be topological spaces
(the *total* and *base* space, respectively),
and let *p : E -> B* be a map (the *projection*).
We say that *p* has the CHP (Covering Homotopy Property)
for *X* if for every map *F : X×I -> B*
and every map *f' : X -> E* with *F(x,0) = pf'(x)*
(for all *x*) there is a map *F' : X×I -> E*
with *F = pF'* and *F'(x,0) = f'(x)* (for all *x*).
In this case *(E,B,p)* is also called a *fiber space*
(for *X*).

We say that *p* has the BP (Bundle Property)
if there is a space *D* such that *B* has an open
cover such that for each member *U* of this cover
there is a homeomorphism *f_U : U × D -> p^(-1) (U)*
with *p f_U (u,d) = u* (for all *u,d*).
In this case *(B,E,p)* is also called *locally trivial*.

**Theorem** If *p* has the Bundle Property,
then it has the CHP for every paracompact Hausdorff space *X*.
If moreover *B* is paracompact, then *p* has the CHP
for every topological space.

For a proof, see e.g. Dugundji, *Topology*, Allyn & Bacon,
1966, Chapter XX.

If *(S(n),B,p)* is a fiber space, and *B* has more
than one point, then the map *p* is not nullhomotopic.
Indeed, otherwise the homotopy could be lifted to one from the
identity to a map with image contained in a single fiber,
so that the identity on *S(n)* would be nullhomotopic.
But it isn't, for example because *S(n)* has nontrivial homology.

Hopf gave as examples locally trivial fiber structures
*S(3) -> S(2)* and *S(7) -> S(4)* and *S(15)->S(8)*
constructed from the norm 1 elements in the complex, quaternion or
octonion plane, mapped to the corresponding point of the projective line.

These maps are algebraically trivial, that is, they induce 0 on the homology and cohomology groups, but homotopically nontrivial. Thus, homotopy is strictly finer than homology.

A table with some results for spheres, taken from
Sze-Tsen Hu, *Homotopy Theory*, Academic Press, 1959.

Let *S(n)* be the *n*-sphere, and consider
*P(m,n) = pi_m(S(n))* for *n > 0*.
If *m < n* then this group is trivial.
If *m = n* then it is isomorphic to *Z*,
the additive group of the integers.
If *n = 1* and *m > n*, then it is trivial.
We have *P(m,2) = P(m,3)* for every *m > 2*.
If *n* is odd and *m > n*, then *P(m,n)*
is finite.

If *n > 2* and *m = n+1* or *m = n+2*
then it is cyclic of order 2.

*P(6,3) = Z(12)*, the cyclic group of order 12.
*P(7,4) = Z + Z(12)*.
If *n > 4* and *m = n+3* then *P(m,n) = Z(24)*.

*P(7,3) = Z(2)*, *P(8,4) = Z(2)+Z(2)*,
*P(9,5) = Z(2)*.
If *n > 5* and *m = n+4* then *P(m,n) = 0*.

*P(8,3) = Z(2)*, *P(9,4) = Z(2)+Z(2)*,
*P(10,5) = Z(2)*, *P(11,6) = Z*.
If *n > 6* and *m = n+5* then *P(m,n) = 0*.

*P(9,3) = Z(3)*, *P(10,4) = Z(24)+Z(2)*.
If *n > 4* and *m = n+6* then *P(m,n) = Z(2)*.

*P(10,3) = P(11,4) = Z(15)*, *P(12,5) = Z(30)*,
*P(13,6) = Z(60)*, *P(14,7) = Z(120)*,
*P(15,8) = Z+Z(120)*.
If *n > 8* and *m = n+7* then *P(m,n) = Z(240)*.

*P(11,3) = P(12,4) = P(13,5) = Z(2)*,
*P(14,6) = Z(24) + Z(2)*,
*P(15,7) = Z(2)+Z(2)+Z(2)*,
*P(16,8) = Z(2)+Z(2)+Z(2)+Z(2)*,
*P(17,9) = Z(2)+Z(2)+Z(2)*.
If *n > 9* and *m = n+8* then *P(m,n) = Z(2)+Z(2)*.

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