How to compute the (co)homology of a space

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How to compute the (co)homology of a space

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[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.

=== See also ===

[[How to use exact sequences]]

[[How to use spectral sequences]]

[[How to compute the cohomology of a sheaf]]

[EXAMPLE connected] If $X$ is [[w:connected space | path-wise connected]], 
then $H_0(X)$ and $H^0(X)$ are both
isomorphic to $\Z$.  More generally, if $\pi_0(X)$ denotes the set
of path components of $X$, then $H_0(X) = \Z^{\oplus \pi_0(X)}$ (a direct sum
of $|\pi_0(X)|$ many copies of $\Z$), and $H^0(X) = \Z^{\pi_0(X)}$ (a direct
product of $|\pi_0(X)|$ many copies of $\Z$).  (More generally, if you are computing (co)homology with coefficients in an abelian group $G$,
then replace $\Z$ by $G$.)

The isomorphisms are quite concrete: if $X_i$ is a particular path component
of $X$, then the "basis vector" $e_i \in \Z^{\oplus \pi_0(X)} = H_0(X)$ 
which is $1$ in the $i$th
component and $0$ in all other components is generated by the class of any
point lying in $X_i$, while
if $f = (f_i) \in \Z^{\pi_0(X)} = H^0(X)$, then $f$ is represented by the $0$-cochain
which takes the value $f_i$ on all the points in the path component $X_i$.

[EXAMPLE concrete]  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 [ref Example #connected]
are the most elementary examples of this.  The discussion of the boundary
map in [ref Example #Mayer_Vietoris] below,
and the related computation in [ref Example #n_sphere], give slightly more substantial
examples.

[EXAMPLE dimension]  If $X$ has dimension $n$ (or is the union of pieces
of dimension at most $n$), then $H_i(X)$ and $H^i(X)$ vanish if $i > n$.

[EXAMPLE homotopy_invariance] Homology and cohomology are homotopy invariants, i.e. if $f: X \rightarrow Y$ is a homotopy equivalence, then the maps induced
by $f$ on all homology and cohomology groups are isomorphisms.

For example, $R^n \setminus \{0\}$ retracts onto the $n-1$-sphere $S^{n-1}$,
and so has the same (co)homology as $S^{n-1}$.  (For a computation of
the homology of $S^{n-1}$, see [ref Example #n_sphere] below.)

[EXAMPLE contractible] As a special case of the preceding example, a [[w:contractible | contractible space]] has vanishing homology and
cohomology in all degrees above $0$, and $\Z$ in degree $0$ (as an instance of
[ref Example #connected]), or more generally, $G$ in degree $0$, if we compute (co)homology
with coefficients in the abelian group $G$.

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

[EXAMPLE Mayer_Vietoris]
If the space $X$ you are studying can be broken up in a reasonable way
into a union of two pieces, say $X = U \cup V,$ then the 
[[w:Mayer-Vietoris sequence | Mayer-Vietoris sequence]] can be used.
This long exact sequence has the form<comment thread="758" />
[maths]
\cdots \rightarrow H_i(U\cap V) \rightarrow H_i(U) \oplus H_i(V)
\rightarrow H_i(X) \rightarrow H_{i-1}(U\cap V) \rightarrow \cdots
[/maths]
(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 $H_i(U\cap V) \rightarrow H_i(U) \oplus H_i(V)$ is just 
the direct sum of the maps $H_i(U\cap V) \rightarrow H_i(U)$ 
and $H_i(U\cap V) \rightarrow H_i(V)$ induced by the inclusions
$U\cap V \subset U$ and $U\cap V \subset V$.

The map $H_i(U)\oplus H_i(V) \rightarrow H_i(X)$ is the difference
of the two maps $H_i(U) \rightarrow H_i(X)$ and $H_i(V) \rightarrow H_i(X)$
induced by the inclusions $U\subset X$ and $V\subset X$.

The boundary map $H_i(X) \rightarrow H_{i-1}(U\cap V)$ is the most
interesting map, since it carries the geometric information describing
exactly how $X$ has been glued together from $U$ and $V$.

Imagine that $U$ and $V$ meet along an "interface" $U\cap V$.
If a class $[c] \in H_i(X)$ is represented by an $i$-chain $c$ (say a simplicial
or singular chain), then we can cut $[c]$ into two along the interface $U\cap V$, to write it as $c = c_U + c_V,$  where $c_U$ lies entirely in $U$,
and $c_V$ entirely in $V$. Now $c$ was a cycle, i.e. $\partial c = 0,$ 
since it represents a homology class.  But neither $c_U$ nor $c_V$ 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 $\partial c = 0,$ we find that $\partial c_U = - \partial c_V$.  (This reflects
that geometrically $c_U$ and $c_V$ share a common boundary, along which
we can reglue them to obtain the cycle $c$.)  The boundary map
$H_i(X) \rightarrow H_{i-1}(U\cap V)$
is then defined via
$[c] \mapsto [\partial c_U].$
(Note that $[\partial c_U]$ makes sense, since $\partial^2 = 0$, so $\partial c_U$ 
is a cycle.  On the other hand, although $\partial c_U$ is a boundary in 
$U$, it typically won't be a boundary in $U\cap V$, and so typically
$[\partial c_U]$ is non-zero in $H_{i-1}(U\cap V)$.)

[EXAMPLE n_sphere] We can apply the Mayer-Vietoris sequence to compute
the homology of an $n$-sphere.  If we slice an $n$-sphere down the middle,
it gets decomposed into two $n$-disks glued along their common boundary,
which is an $(n-1)$-sphere.

Our computation will thus be inductive, beginning with $n = 0$.  Since
$S^0$ is just the subset $\{-1, 1\} \subset \R$, i.e. a pair of points,
we have from [ref Example #connected] that $H_0(S^0) = \Z \oplus \Z,$
while [ref Example #dimension] shows that $H_i(S^0) = 0$ for $i > 0$.

Let us now consider the case $n = 1$. If we cut the circle $S^1$ into two,
it becomes the union of two intervals (or if you prefer, $1$-disks), 
say $U$ and $V$,
glued along their common boundary $U\cap V$
(which is a pair of points, i.e.<comment thread="756" /> $S^0$).
Intervals are contractible (being convex; see [ref Example #contractible]),
and so have vanishing homology above degree $0$, and homology equal to $\Z$
in degree $0$.

The Mayer-Vietoris sequence thus becomes
[maths]
0 \rightarrow H_1(S^1) \rightarrow H_0(S^0) = \Z \oplus \Z\rightarrow 
H^0(U)\oplus H^0(V) = \Z \oplus \Z \rightarrow H_0(S^1) \rightarrow 0.
[/maths]
(All homology vanishes in degrees $i > 1$, as we see either by contemplating
the higher degree parts of the Mayer-Vietoris sequence, or by appealing to
[ref Example #dimension].)
Thus everything hinges on understanding the third map.  If this were 
an isomorphism, we would conclude that $H_i(S^1)$ vanishes for all $i$.
Of course, this is impossible, because [ref Example #connected] tells
us that $H_0(S^1) = \Z$.  We conclude that the third arrow must have an
image of rank $1$, and hence, by the rank-nullity theorem, it must also
have a rank $1$ kernel.  Consequently, $H_1(S^1)= \Z$.

We could also directly compute the structure of the third arrow.
Namely, if $P$ and $Q$ are the two points on the boundary of $U \cap V$,
its source is spanned by $[P]$ and $[Q]$ (the homology classes of these
two points).  On the other hand, each of $U$ and $V$ is path connected,
and so all points in $U$ or $V$ represent the same homology class.
Thus $[P]$ and $[Q]$ map to the same element in $H_0(U)$, and also map
to the same element in $H_0(V)$.
Thus the image does inded have rank $1$.
And the kernel has an explicit description: it is spanned by $[P] - [Q]$.

This also has a geometric meaning: if we consider the description of
the boundary map in the Mayer-Vietoris sequence given in 
[ref Example #Mayer_Vietoris], we see that $H_0(X)$ must be generated
by the class $[c]$ of some $1$-cycle $c$ such that, when we cut $c$ into two
and thus decompose it as $c = c_U + c_V,$ we have $\partial c_U = P - Q$ (or at
least, the two should be homologous).  But there is one obvious such
$1$-cycle, namely, the circle itself!  Thus $H_1(S^1)$ is generated by
$[S^1]$.

Proceeding with the induction, one finds that $H_i(S^n)$ vanishes unless
$i = 0$ or $n$.  Of course, $H_0(S^n)$ is generated by the class of any point,
while $H_n(S^n)$ is generated by $[S^n]$.

[EXAMPLE manifold_fundamental]  If $M$ is any connected, compact, orientable $n$-dimensional manifold, then $H_n(M)$ is isomorphic to $\Z$, and is generated
by $[M]$, the class of $M$ itself.  (To be precise, we have to fix an orientation
of $M$ in order for this class to be well-defined; reversing the orientation
changes its sign.)

[GENERAL DISCUSSION]
[note article incomplete] A careful, comparative, discussion of all the different methods is needed, as are
zillions more examples. [/note]

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How to compute the (co)homology of a space

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