### Quick description

This will eventually become a list of many different techniques for computing homology and cohomology, including passing between different flavours (simplicial, singular, Cech, de Rham, sheaf), the advantages and disadvantages of each, the use of various long exact sequences and spectral sequences, universal coefficient theorems, and so on.

### Prerequisites

A knowledge of various points of view on algebraic topology, and in particular on homology and cohomology, as well as some algebra (especially computing with exact sequences, and related methods of homological algebra). The precise requirements vary from example to example.

### Example 1

If is path-wise connected, then and are both isomorphic to . More generally, if denotes the set of path components of , then (a direct sum of many copies of ), and (a direct product of many copies of ). (More generally, if you are computing (co)homology with coefficients in an abelian group , then replace by .)

The isomorphisms are quite concrete: if is a particular path component of , then the "basis vector" which is in the th component and in all other components is generated by the class of any point lying in , while if , then is represented by the -cochain which takes the value on all the points in the path component .

### Example 2

As a general remark, it is useful when trying to compute (co)homology to have a geometric picture of the (co)chains that represent the various (co)homology classes that you are working with. The explicit descriptions of the isomorphisms in Example 1 are the most elementary examples of this. The discussion of the boundary map in Example 6 below, and the related computation in Example 7, give slightly more substantial examples.

### Example 3

If has dimension (or is the union of pieces of dimension at most ), then and vanish if .

### Example 4

Homology and cohomology are homotopy invariants, i.e. if is a homotopy equivalence, then the maps induced by on all homology and cohomology groups are isomorphisms.

For example, retracts onto the -sphere , and so has the same (co)homology as . (For a computation of the homology of , see Example 7 below.)

### Example 5

As a special case of the preceding example, a contractibe space has vanishing homology and cohomology in all degrees above , and in degree (as an instance of Example 1), or more generally, in degree , if we compute (co)homology with coefficients in the abelian group .

One way to show that a space is contractible is to embed it homeomorphically as a convex subspace of a Euclidean space.

### Example 6

If the space you are studying can be broken up in a reasonable way into a union of two pieces, say then the Mayer-Vietoris sequence can be used. This long exact sequence has the form

(Here I am working with homology, for definiteness, and to make the subsequent discussion somewhat more concrete. The Mayer-Vietoris sequence for cohomology works similarly, with all arrows reversed.) As with any long exact sequence arising in topology that one wants to work with, it is important to understand the maps.

The map is just the direct sum of the maps and induced by the inclusions and .

The map is the difference of the two maps and induces by the inclusions and .

The boundary map is the most interesting map, since it carries the geometric information describing exactly how has been glued together from and .

Imagine that and meet along an "interface" . If a class is represented by an -chain (say a simplicial or singular chain), then we can cut into two along the interface , to write it as where lies entirely in , and entirely in . Now was a cycle, i.e. since it represents a homology class. But neither nor will typically be a cycle, i.e. typically, their boundary is non-zero. (Imagine taking something without boundary, e.g. a closed loop, and slicing it in half. The two pieces you obtain will each have boundary.) Since we find that . (This reflects that geometrically and share a common boundary, along which we can reglue them to obtain the cycle .) The boundary map is then defined via (Note that makes sense, since , so is a cycle. On the other hand, although is a boundary in , it typically won't be a boundary in , and so typically is non-zero in .)

### Example 7

We can apply the Mayer-Vietoris sequence to compute the homology of an -sphere. If we slice an -sphere down the middle, it gets decomposed into two -disks glued along there common boundary, which is an -sphere.

Our computation will thus be inductive, beginning with . Since is just the subset , i.e. a pair of points, we have from Example 1 that while Example 3 shows that for .

Let us now consider the case . If we cut the circle into two, it becomes the union of two intervals (or if you prefer, -disks), say and , glued along their common boundary (which is a pair of points, i.e. ). Intervals are contractible (being convex; see Example 5), and so have vanishing homology above degree , and homology equal to in degree .

The Mayer-Vietoris sequence thus becomes

(All homology vanishes in degrees , as we see either by contemplating the higher degree parts of the Mayer-Vietoris sequence, or by appealing to Example 3.) Thus everything hinges on understanding the third map. If this were an isomorphism, we would conclude that vanishes for all . Of course, this is impossible, because Example 1 tells us that . We conclude that the third arrow must have an image of rank , and hence, by the rank-nullity theorem, it must also have a rank kernel. Consequently, .

We could also directly compute the structure of the third arrow. Namely, if and are the two points on the boundary of , its source is spanned by and (the homology classes of these two points). On the other hand, each of and is path connected, and so all points in or represent the same homology class. Thus and map to the same element in , and also map to the same element in . Thus the image does inded have rank . And the kernel has an explicit description: it is spanned by .

This also has a geometric meaning: if we consider the description of the boundary map in the Mayer-Vietoris sequence given in Example 6, we see that must be generated by the class of some -cycle such that, when we cut into two and thus decompose it as we have (or at least, the two should be homologous). But there is one obvious such -cycle, namely, the circle itself! Thus is generated by .

Proceeding with the induction, one finds that vanishes unless or . Of course, is generated by the class of any point, while is generated by .

### Example 8

If is any connected, compact, orientable -dimensional manifold, then is isomorphic to , and is generated by , the class of itself. (To be precise, we have to an orientation of in order for this class to be well-defined; reversing the orientation changes its sign.)

## Comments

## Inline comments

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## (minor) correction?

Tue, 08/12/2009 - 10:44 — Anonymous (not verified)If I recall correctly, the Mayer-Vietoris sequence tells you something about a space decomposed as two sets _whose interiors cover the space_. So I think U and V must intersect in two intervals (which are homotopically trivial, so the rest of the proof runs the same).

## Inline comments

The following comments were made inline in the article. You can click on 'view commented text' to see precisely where they were made.

## Helpful tip

Tue, 08/12/2009 - 10:48 — Anonymous (not verified)Maybe it'd be nice to mention that if your space is being decomposed into U and V and these have non-empty intersection, then the end terms (H_0 and such) can be taken as reduced homology instead of just normal homology, simplifying some calculations.