A *graded Abelian group* *C = {C_i}* is a collection
of Abelian groups, indexed by the integers.
A *homomorphism of degree* *e* from *{C_i}*
to *{D_i}* is a collection *f = {f_i}* of
homomorphisms of Abelian groups, where *f_i : C_i -> D_(i+e)*.
A *morphism* of graded Abelian groups is a homomorphism
of degree 0.

A *differential* on a graded Abelian group
is a homomorphism of some degree with square 0.
A *chain complex* *(C,d)* is a graded Abelian
group *C* with fixed differential *d* of degree -1.
A *morphism* of chain complexes (a *chain map*)
is a morphism of graded Abelian groups that commutes with
the differential.

The *homology group* *H(C)* of a chain complex
*(C,d)* is the graded Abelian group *H(C) = {H_i(C)}*
with *H_i(C) = ker d_i / im d_(i+1)*.
(We put *Z_i(C) = ker d_i*, the *cycles*, and
*B_i(C) = im d_(i+1)*, the *boundaries*.)
Given a morphism *f : C -> D* of chain complexes
we find an induced morphism *f* : H(C) -> H(D)*.
Thus, *H* is a functor (with *H(f) = f**).

**Exercise** Check that *f** is well-defined and that
*H* commutes with composition.

One has arbitrary sums, products, direct and inverse limits for chain complexes. Taking homology commutes with sums, products and direct limits.

**Exercise** Show the homology of an inverse limit may
differ from the inverse limit of the homology of a sequence of
chain complexes.

Given a set *V*, let the *exterior algebra*
*E(V)* be the Abelian group generated by associative
expressions *u_1 ^ ... ^ u_m* (where the empty
product is allowed and denoted by 1), where the *u_i*
are in *V*, subject to the relations *u ^ u = 0*
and *u ^ v + v ^ u = 0*. Let *E'(V)* be the
subgroup of the elements with an expression not involving 1.
The group *E(V)* has a natural grading
(say, with *u_0 ^ ... ^ u_m* having degree *m*)
and a natural differential *d* of degree -1 defined by
*d_m(u_0 ^ ... ^ u_m) =
Sum_i (-1)^i u_0 ^ ... ^ u_(i-1) ^ u_(i+1) ... ^ u_m*.
Also *E'(V)* has a natural differential, defined by the
above expression for *m > 0* and zero on degree 0 and smaller.
Thus, we find homology in both cases.

**Exercise** Show that *E(V)* has zero homology.
Show that *C = E'(V)* has all homology groups zero, except
for *H_0(C)*, which is isomorphic to the integers when
*V* is nonempty.

[Maybe it feels as if the grading is wrong, shifted by one.
I agree. But this is tradition. Similarly, one usually
defines homology *H* using *E'(V)*, notices that
there are formal complications in degree 0, and then looks at the
*augmented* chain complex *E(V)* with *reduced*
homology *H*-tilde, which equals *H* everywhere,
except in degree 0, in case *V* is nonempty, where
*H_0* has an additional summand *Z*.
The theory is simplified by never mentioning *E'(V)*
and only using reduced homology.]

A *simplicial complex* is a set *V*
together with a collection *K* of nonempty finite subsets
called *simplices* such that any nonempty subset of a simplex
(called *face* of the simplex) is again in *K*.
Each simplicial complex (with underlying set *V*)
gives rise to the subgroup *E'(K)* of *E'(V)* generated
by the products *u_1 ^ ... ^ u_m* for each simplex
*{ u_1, ..., u_m }* in *K*, and *E'(K)*
is invariant for the differential *d*.
The corresponding homology is called simplicial homology.

Let *X* be a topological space. A *singular simplex*
in *X* is a map from a Euclidean (solid) simplex (with ordered
set of vertices) into *X*, where two singular simplices are
considered equal when they differ by the unique affine transformation
of the domain simplices that preserves vertex order.

(One does not need a Euclidean space to start with; given an ordered
*m*-set *M*, one constructs a solid *(m-1)*-simplex
by taking all functions from *M* to the unit interval with values
summing to 1. The coordinates obtained in this way are called
*barycentric coordinates*. Now two singular simplices
are considered identical when they only differ by an order preserving
bijection of the domains of their domains. Thus, there is no loss in
generality to take the set of integers *0..m* as domain for all
*m*-simplices.)

We obtain a simplicial complex (called *S(X)*) by letting
the faces of a singular simplex be the restrictions of the map
to the faces of the domain simplex. The homology obtained in this way
is called the *singular homology* of the topological space.

If *X* and *Y* are topological spaces,
and *f* is a map from *X* to *Y*, then
composition with *f* defines a map from *S(X)*
into *S(Y)* that commutes with taking faces, and hence
defines a homomorphism *f** (of degree 0) from
*H(X)* to *H(Y)*.

[

- unfortunately ISO-8859-1 does not have a hacek.] An entirely different way of finding a simplicial complex in a topological space is by taking an open cover and letting a simplex be a subset of the cover with nonempty intersection. This simplicial complex is called the

Let *f,g : C -> D* be morphisms of chain complexes.
A *chain homotopy* from *f* to *g*
is a homomorphism *e : C -> D* of degree 1 such
that *f - g = de + ed*.

If *f,g* are homotopic, i.e., if there is a chain
homotopy from *f* to *g*, then *f* = g**,
i.e., *f* and *g* induce the same map
*f* : H(C) -> H(D)*.

[Indeed, if *z* is a cycle in *C*, then
*ed(z) = 0* and hence *f(z) - g(z) = de(z)*
is a boundary.]

**Exercise** Being homotopic is an equivalence relation.
If *f,g : C -> D* are homotopic, and
*f',g' : D -> E* are homotopic, then *f' f* and
*g' g* are homotopic.

A chain complex *C* is called *contractible* when
the identity on *C* is homotopic to 0.
It follows that *H(C) = 0*, that is, *C* is *acyclic*.

For example, a simplex, or, more generally, a simplicial complex
all of whose simplices contain some fixed vertex *v*,
is acyclic: the map *e* defined by
*e(v_0 ^ ... ^ v_m) = v ^ v_0 ^ ... ^ v_m* is a
chain homotopy from the identity to 0.

Let us look at .Cech homology again.
Given an open cover *U* and a refinement *V*,
there are in general many ways of assigning to each member of *V*
a member of *U* containing it. Thus, we find many morphisms
of the corresponding chain complexes. However, these are all
homotopic, and hence induce the same mapping on the homology groups.
[Indeed, if the two maps assign *u* and *u'* (respectively)
to some *v* in *V*, then we can define a chain homotopy
*e* by *e(v_0 ^ ... ^ v_m) =
Sum_i (-1)^i u_0 ^ ... ^ u_i ^ u'_i ^ ... ^ u'_m*, so that
*(de+ed)(v_0 ^ ... ^ v_m) = u'_0 ^ ... ^ u'_m - u_0 ^ ... ^ u_m*.]

**Exercise**
Show that given topological spaces *X,Y* and a map *f:X->Y*
we find a map *f(V)*: H_U(X) -> H_V(Y)* on the .Cech homology
groups corresponding to open covers *U* of *X* and *V*
of *Y*, where *U* is the collection of inverse images of *V*.
Show that this induces a map *f*: H(X) -> H(Y)*, on the .Cech homology
groups of *X* and *Y* (the inverse limit of the groups
*H_U(X)* and *H_V(Y)*).

Let *f : C -> C'* be a morphism of chain complexes.
The *mapping cone* of *f* is the chain complex *C''*
with *C''_i = C_(i-1) + C'_i* (direct sum) and
*d'' (c,c') = (-dc, fc+d'c')*.
Then *f* is a chain equivalence (i.e., there exists a
morphism of chain complexes *f' : C' -> C* such that
both *ff'* and *f'f* are homotopic to 1),
if and only if *C''* is contractible.

**Exercise** Verify all details. More precisely:
(i) *C''* is a chain complex, and
(ii) If *C''* is contractible, so that *d''e''+e''d'' = 1*
for some *e''*, then *e''(c,0) = (ec,*)* and
*e''(0,c') = (f'c', -e'c')* define a morphism of chain
complexes *f'* and chain homotopies *e* and *e'*
from *f'f* and *ff'* to 1.
(iii) If *f' : C' -> C* is a morphism of chain complexes
and *e* and *e'* are chain homotopies from *f'f*
and *ff'* to 1, then *e''(c,c') = ((e+f'e'f-f'fe)c+f'c',
(e'fe-e'e'f)c-e'c')* defines a contraction of *C''*.

Let us look at singular homology again. Instead of oriented simplices we can use ordered simplices (where the generating elements are sequences of points, possibly with repetitions, that occur in a common simplex).

**Exercise** The map that assigns to an ordered simplex
*(u_0,...,u_m)* the oriented simplex *u_0 ^ ... ^ u_m*
is a chain equivalence.
Answer

**Exercise** A simplicial complex *K* and
its barycentric subdivision (that has as vertices the
simplices of *K*, and as *m*-simplices the chains
consisting of *i*-simplices of *K* for
*i=0,...,m*, totally ordered by inclusion)
are chain equivalent.

Homotopy is a finer invariant than homology. We give two results in this direction.

**Theorem**
Let *X,Y* be two topological spaces, and
*f,g : X -> Y* two homotopic maps. Then
we find for singular homology that the maps
*f*,g* : H(X) -> H(Y)* coincide.

Let us set up a little bit of machinery, so that the proof
will be obvious. Given a solid simplex *A*, let
*i,j : A -> A × I* be the two injections sending
*a* to *(a,0)* and *(a,1)*, respectively.
Let us define a canonical subdivision (triangulation) *T(A)*
of *A × I* as follows.
Let *c* be the center of *A*. Then *T(A)*
consists of the simplices *<c,iA>* and
*<c,jA>* (and *<c,iF>* and
*<c,jF>* for all faces *F* of *A*)
and all simplices *<c,S>*
with *S* in *T(F)* for some face *F* of *A*.
(Given a point *x* distinct from *c* in *A × I*,
the line *cx* hits the boundary of *A × I*
either in *iA* or in *jA* or in *F × I*
for some face *F* of *A*, so we do indeed cover
*A × I*.)

Given a simplicial complex *K*, let *T(K)* be the
simplicial complex that is the union of *T(A)* for all
simplices *A* of *K*. The maps *i* and *j*
extend linearly to chain maps (denoted by the same symbols) from
the chain complex defined by *K* to that defined by *T(K)*.

**Lemma** The chain maps *i,j : K -> T(K)* are
chain homotopic.

**Proof**
Consider the chain map *k : K -> T(K)* that assigns to a
simplex *A* the element
*<c,iA> - <c,jA> - <c,kdA>*
(defined by induction of dim *A*; *<c,*>*
extended linearly).
We see by induction that *dk+kd = i-j*. Indeed,
*(dk+kd)(A) = iA - jA - <c,idA> + <c,jdA> +
<c,dkdA> = iA - jA - <c,kddA> = iA - jA*. QED

**Proof of the theorem**
Let *i,j : X -> X × I* be the two injections sending
*x* to *(x,0)* and *(x,1)*, respectively.
Since *f* and *g* are homotopic,
there is a map *F : X × I -> Y*
such that *f = Fi* and *g = Fj*.
Thus, it suffices to show that *i* = j**.
We do this by exhibiting a chain homotopy *h*
from *i* to *j* considered as maps from *S(X)* to
*S(X × I)*. Thus, *h*
will be a map from *S(X)* to *S(X × I)* of degree 1,
such that *dh+hd = i-j*.
Elements of *S(X)* are (linear combinations of) maps *s*
from a (solid) simplex *A* into *X*.
Define *h(s) = (s × 1) k(A)* where *k* is the
map defined in the proof of the lemma, viewed as a linear combination
of maps into *A × I* and 1 is the identity on *I*.
We find *(dh+hd)(s) = (s × 1)(dk+kd)(A) = (s × 1)(i-j)(A) =
(i-j)(s)*. QED

The second result here is that H_1 equals pi_1 made Abelian, that is, the first homology group of a space (for singular homology) equals the quotient of the fundamental group by its commutator subgroup.

**Theorem**
*H_1(X)* is the quotient of *P(X,x)*
by its commutator subgroup.

**Proof**
Let *P(X;x)* be the fundamental group of *X*
and let *H_1(X)* be the first homology group. We find a
homomorphism *h : P(X;x) -> H_1(X)* as follows.
A map *f : I -> X* with *f(0) = f(1) = x* is a singular
simplex of dimension 1, so we can let *h* map the homotopy
class of *f* to the homology class of *f*.
This is well-defined: we can view *f* as a map from the circle
*S(1)* into *X*, and then find a map
*f* : H(S(1)) -> H(X)*, and as we just saw, *f** only
depends on the homotopy class of *f*. Let *e : I -> S(1)*
be the map sending *t* to *exp(2 pi i t)*. It is a
singular simplex and hence represents an element of *H(S(1))*.
The image *f*(e)* is an element of *H(X)*, that has
representative *f*. Thus *h* is well-defined.

Moreover, *h* is a homomorphism. Indeed, we have to check that
*h(f#g) = h(f)+h(g)*, that is, that *f + g - f#g = 0*
in *H_1(X)*, that is, that *f + g - f#g* is a boundary,
and clearly it is.

Remains to show that the commutator subgroup of *P(X,x)* is
the full kernel of *h*. To this end we construct a left inverse
*h'* to *h* from *H_1(X)* into *AP(X,x)*,
the quotient of *P(X,x)* by its commutator subgroup.
The elements of *H_1(X)* are equivalence classes of
1-cycles in *S(X)*, and these equivalence classes of 1-cycles
have representatives that have boundary contained in *{x}*.
Given such a 1-cycle *z*, let *h'* map it to *z*,
regarded as a path in *X*. We have to show that this is well-defined,
that is, that *z* is null-homotopic (0 in *AP(X,x)*)
when it is a boundary. But that is clear. QED

Let *A -> B -> C* be a chain of Abelian groups, connected
by mappings *f : A -> B* and *g : B -> C*.
The chain is called *exact* at *B* when
the image of *f* equals the kernel of *g*.
A longer chain is called exact when it is exact at the middle
of each 3-term subsequence. An exact sequence of chain complexes
is a sequence of morphisms of chain complexes giving an exact
sequence of Abelian groups for each index.

**Theorem**
Given a short exact sequence *0 -> C -> D -> E -> 0*
of chain complexes, we find a long exact sequence
* ... -> H_i(C) -> H_i(D) -> H_i(E) -> H_(i-1)(C) -> ...*.
The correspondence is functorial: a morphism between two
short exact sequences is mapped to a morphism between two
long exact sequences, where a morphism of exact sequences
is a collection of mappings making all squares commute.

**Proof**
We already saw that a morphism *C -> D* gives a
morphism *H(C) -> H(D)*. Our first task is to define
the map *H_i(E) -> H_(i-1)(C)*.

Given *z* in *Z_i(E)*, we find *y* in
*C_i(D)* of which it is the image. Since *dz = 0*
the element *dy* has zero image in *E_(i-1)*
and hence is the image of a unique element *x* in
*C_(i-1)*. Moreover, *dx* has image
*ddy = 0*, so *dx = 0*, i.e., *x*
is a cycle. Map the homology class of *z*
to that of *x*. This is well-defined:
if we replace *y* by *y'* with the same image *z*,
then *y'-y* is the image of a *x'* in *C_i*
and *x* becomes *x+dx'*, which lies in the same
homology class. And if we change *z* by a boundary:
*z' = z + dz''* with *z''* in *E_(i+1)*, then
*y* is changed by an element *dy''*, where
*z''* is the image of *y''*, so that *dy* is
not changed at all.

Thus we find a function from *H_i(E)* to *H_(i-1)(C)*,
and one quickly verifies that it is a homomorphism.

Remains to verify exactness at each point, and functoriality. Let us leave that as an exercise.

**Exercise** Prove that the above long sequence is exact
at each point. Prove that the construction behaves well under
morphisms.

As a special example of the situation of the previous section
we have the case of two simplicial complexes *B,C*
of which the union again is a simplicial complex *D*.
Let their intersection be *A*. Then we have a short
exact sequence *0 -> A -> B+C -> D -> 0*
(where *+* denotes direct sum) if we define the arrows
by mapping *a* to *(a,-a)* and *(b,c)*
to *b+c*.

Consequently, we find a long exact sequence, called the
*Maier-Vietoris sequence*,
*... -> H_i(A) -> H_i(B)+H_i(C) -> H_i(D) -> H_(i-1)(A) -> ...*.

This enables one to compute the homology *H_i(D)*
of a union of two complexes, given the homology of these
complexes and that of their intersection.

For example, look at the reduced homology of the n-sphere *S = S(n)*.
Let *x,y* be two points of *S*. Then *S\{x}*
and *S\{y}* are homeomorphic to Euclidean n-space *E(n)*
(by `stereographic projection') and hence are contractible
and have zero homology. And *S\{x,y}* is homeomorphic to
*E(n)* minus a point, which is homotopic to *S(n-1)*.
Thus, *H_i(S(n)) = H_(i-1)(S(n-1))*.
At the bottom we have the empty set *S(-1)* with reduced
homology vanishing everywhere except that *H_(-1)* is isomorphic
to the integers. It follows that *H_i(S(n)) = 0* for all *i*
except that *H_n(S(n))* is isomorphic to the integers.

**Exercise** Prove that *E(m)* is not homeomorphic to
*E(n)* when *m < n*.

**Exercise** A map *f* from *S(n)* to itself
has a (mapping) *degree*, namely the number *d* such that
*f** is multiplication by *d* on *H_n(S(n))*.
Viewing the complex plane together with a point infinity
as *S(2)*, show that the mapping degree of a polynomial of
degree *d* equals *d*. Conclude that a nonconstant
polynomial has a root.

A *topological pair* is a pair *(X,A)* where
*X* is a topological space, and *A* a subspace.
A morphism *f : (X,A) -> (Y,B)* is a map *f : X -> Y*
such that *f(A)* is contained in *B*.
The singular chain complex *S(X,A)* is defined as *S(X)/S(A)*
(singular simplices in *X* modulo those entirely
contained within *A*). The homology is called *H(X,A)*.
All concepts generalize in an obvious way. Regard *X* as a pair
by identifying it with *(X,empty)*.

**Exercise** Prove that if *A* is a retract of *X*
then *H(X) = H(A) + H(X,A)*.

**Exercise** Prove that *S(n-1)* is not a retract of
*E(n)*. Conclude that
the two theorems
from the Introduction hold.

Let *A* be an *R*-module. An element *a* of *A*
is called a *torsion element* when *ra = 0* for some
nonzero *r* in *R*.

**Exercise** Give an example to show that the set of torsion elements
need not be a submodule. Show that it is a submodule when *R* is
an integral domain (i.e., is commutative, without zero-divisors,
with nonzero 1).

Assume that *R* is an integral domain.
Let *Tor(A)* be the torsion submodule of *A*, that is,
the submodule consisting of all torsion elements. The module *A*
is called *torsion-free* when *Tor(A) = 0*.
The module *A' = A / Tor(A)* is torsion free.
If *A* is finitely generated, then *A'* is free,
with a finite basis. The cardinality of such a basis is called
the *rank* of *A*.
Let *f : A -> A* be a morphism.
The *trace* *Tr(f)* if the trace of *f*
induced on *A'* (e.g., the sum of the diagonal elements
when written on a basis).

The above applies (with *R* the ring of integers)
to Abelian groups. The *Betti numbers* of a simplicial complex
are the ranks of the homology groups *H_i*.

Let *C* be a finitely generated chain complex, and let
*f : C -> C* be a homomorphism of degree 0.
The *Lefschetz number* *L(f)* is defined as
*L(f) = sum (-1)^i Tr(f*|*C_i)*.

The *Euler characteristic* *chi(C)* is
the Lefschetz number of the identity map.

**Theorem** (Hopf trace formula)
Let *C* be a finitely generated chain complex, and let
*f : C -> C* be a homomorphism of degree 0.
Then *L(f) = L(f*)*.
(In particular, the Euler characteristic is the alternating sum
of the Betti numbers.)

**Exercise** Prove this. (Hint: do the case of the
Euler characteristic first.)

**Theorem** (Lefschetz fixed point theorem)
Let *X* be a nice compact topological space
and let *f : X -> X* be a map.
If *L(f)* is nonzero then *f* has a fixed point.

(Proof: "nice" here may mean carrier of a finite simplicial complex,
or connected orientable *n*-dimensional manifold, or so.
Use a suitable simplicial approximation. If *f* has no
fixed points then all simplices will be moved far away by *f*,
so that *f* will have trace 0.)

Thee are many homology theories (we have seen singular homology
and .Cech homology), and it is possible to develop the theory
axiomatically. See S. Eilenberg & N.E. Steenrod,
*Foundations of Algebraic Topology*, Princeton, 1952.

A *homology theory* is a pair *(H,d)*, where
*H* is a covariant functor from the category of topological pairs
to that of graded Abelian groups (i.e., *H* commutes with
composition of maps and sends the identity to the identity),
and *d* assigns to each pair *(X,A)* a homomorphism
of degree -1 from *H(X,A)* to *H(A)*, where for
*f : (X,A) -> (Y,B)* we have
*d(Y,B) H(f) = H(f*|*A) d(X,A)*,
satisfying the following four axioms:

**Homotopy Axiom**
If *f,g : (X,A) -> (Y,B)* are homotopic, then *H(f) = H(g)*.

**Exactness Axiom**
Consider the pair *(X,A)* and the two inclusion maps
*A -> X -> (X,A)*. We find an exact sequence
*... -> H_i(A) -> H_i(X) -> H_i(X,A) -> H_(i-1)(A) -> ...*
(with arrows given by *...,H,H,d,...*).

**Excision Axiom**
If *U* is an open subset of *X* with closure
contained in the interior of *A*, then the excision map
*i : (X\U,A\U) -> (X,A)* induces an isomorphism *H(i)*
between *H(X\U,A\U)* and *H(X,A)*.

**Dimension Axiom**
If *X* is a single point, then *H_i(X) = 0*
for nonzero *i* and *H_0 (X)* is isomorphic to the
additive group of integers.

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