BASIC CATEGORY THEORY AND CHAIN COMPLEXES
· What are a chain complex and a homology group?
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A: A category, C, is a class of objects, Ob C, together with a class of morphisms, Mor C, between objects. The morphisms satisfy the following properties:
1) For every two objects X, Y in Ob C, there is a subset, Hom(X, Y), of Mor C. These subsets are all pairwise disjoint. When first referring to an element, f, of Hom(X, Y), we will usually adopt the usual function notation: f: X à Y
2) For every X, Y, Z in Ob C, and morphisms f: X à Y and g: Y à Z, there is a morphism h: X à Z called the composition of g with f. Again, we will adopt the usual functional notation: h = g ○ f.
3) Composition is associative, that is, for f: W à X, g: X à Y, and h: Y à Z, h○(g○f) = (h○g)○f.
4) For every object, Y, there is a morphism 1 in Hom(Y, Y) such that for all f: X à Y and g: Y à Z, 1○f = f and g○1 = g. This morphism is called the identity.
There are many familiar examples of categories. The category of all sets and relations is one such example. Another is the category of groups and homomorphisms, or the category or vector spaces and linear transformations. There is also a category of tangles, whose objects are nonnegative integers and whose morphisms are tangles. This will be explained more in a later section.
Now suppose we have morphisms f: X à Y, f’: X à Y’, g: Y à Z, and g’: Y’à Z such that g’○f’=g○f (that is, the image of X under both transformations is identical). Then we say the following diagram is commutative:

Any diagram is called commutative if any two chains of arrows having the same origin and end yield the same image. If we restrict this to a single object and two morphisms, we see that it corresponds to our usual definition of commutivity:

A: A functor, F, is a map between two categories C and D. It maps objects in C to objects in D. Also, for all X, Y in Ob C, F maps Hom(X, Y) to Hom(F(X), F(Y)) in such a way that compositions are preserved. This automatically means that functors preserve identities. An easy example of a functor is the so-called ‘forgetful functor.’ This functor maps the category of groups with homomorphisms to the category of sets with functions by mapping every group to its underlying set, thus “forgetting” some of the structure of the objects. There are other forgetful functors as well, such as the one from the category of topological spaces with homeomorphisms to that of sets and functions; again it forgets the structure on the objects, mapping them to their underlying sets.
It is possible to define morphisms of functors, too. Let F and G be two functors from C to D. For every X in Ob C, choose a morphism fX in HomD(F(X), G(X)) such that for all X, Y in Ob C, g in Hom(X, Y), the following diagram commutes.

The morphism between the functors F and G is the set of fX. With this definition, we can now think about the category of functors Funct(C, D) from a category C to a category D.
A: A chain complex are indexed objects, Cn, of a category along with a boundary operator, dn: Cn à Cn-1 such that dn○dn+1 = 0, that is, the image of one boundary operator is inside the kernel of the next. The homology group is just the quotient of these two subspaces. Hn = ker dn / im dn+1. A chain complex, C*, is usually represented by the following diagram:
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It is possible to think of “finite” chain complexes, as well. These are chain complexes in which all but finitely many objects are 0. Also, we can think of chain complexes that “terminate.” These are complexes that may have infinitely many non-zero objects, but eventually (for small enough n), every object is 0.
There are some really easy geometric examples of chain complexes. For instance, take a triangle, with its edges oriented in the standard counter-clockwise fashion. From this triangle, we can define the following chain complex:

The triangle
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The corresponding chain complex
Here, the boundary operators give the oriented geometric boundary of the path: for instance, if f represents the face, the image of f under the boundary is a+b+c. Likewise, the boundary of the edge, a, is C – B. It still remains to show that the composite of the boundary is 0. There is only one non-trivial case to examine: the boundary of the face of the triangle is the complete oriented boundary, an oriented closed loop, which clearly has no boundary. Hence this is a chain complex. In fact, this example can be extended to that of an n-dimensional simplex, when you adopt the standard orientation for the faces.
What are the homologies of this complex?
· The image of the first boundary is 0, and the second boundary is injective, hence possessing trivial kernel. Thus the first homology is 0.
· The image of the second boundary is k*(a+b+c) where a, b, and c are the positively oriented edges of the triangle. Under the third boundary, an edge gets set to its final point – its initial point. Hence, the kernel is comprised of complete cycles. Again, the homology is 0.
· Because A – C is a combination of B – A and C – B, the image of the third boundary is the span of B – A and C – B. Since the entire span of A, B, and C is the kernel of the fourth boundary, the homology is 1-dimensional, a copy of K, where K is whatever ring or field over which the linear combinations were formed.
Next, we can define a morphism between two chain complexes, f: C* à D*, by choosing morphisms fn: Cnà D n such that the following diagram commutes for all n:

It is easy to verify that that due
to this commutativity, f sends kernels into kernels and images into
images. Hence, f induces a map between
the homologies of C* and D*, f*: HC à HD. If f* is an isomorphism on every homology
group, then we call f a quasi-isomorphism between the complexes.
One can also talk about cochain complexes. A cochain complex is exactly the same thing as a chain complex with one minor change: everything is renumbered into ascending rather than descending order. This leads immediately to the definition of cohomology groups, defined in the analogous way to that of homology. Since there is no real difference between chains and cochains, we may use the terms interchangeably.