# What is an exact sequence?

First a definition:

Definition 1 (Sequence)
Let $A,B,C$ be groups, and let $\varphi : A \rightarrow B$ and $\phi : B \rightarrow C$ be homomorphisms. Then we say that

$A \xrightarrow\varphi B \xrightarrow\phi C$

is a sequence of homomorphisms.

We’re going to begin with a brief review of the First Isomorphism Theorem for groups:

Theorem 1 (First Isomorphism Theorem).
If $\varphi : G \rightarrow H$ is a homomorphism of groups, then $\mathrm{Ker}\varphi$ is a subgroup of $G$, and $G/\mathrm{Ker}\varphi \cong \varphi(G)$.

The First Isomorphism Theorem is a light saber in group theory: it just cuts away at many, many problems.

Here’s the basic situation: suppose we have a group $A$ and another group $C$. We want to study groups $B$ which satisfy the following:

1. $A$ is isomorphic to a subgroup of $B$; that is, there exists an injective homomorphism $\alpha : A \rightarrow B$.
2. $C$ is isomorphic to $B/\varphi(A)$; that is, there exists a surjective homomorphism $\beta : B \rightarrow C$ such that $\mathrm{Ker } \beta = \varphi(A) = \mathrm{Im }\varphi$.

Well, any time we have such a $B$ and the associated homomorphisms $\alpha,\beta$, we see (by looking at the domains and ranges of these maps) that we have a sequence of homomorphisms:

$A \xrightarrow\alpha B \xrightarrow\beta C$,

However, this particular scenario carries with it a valuable relationship between our homomorphisms: by condition (2), we know that $\mathrm{Ker } \beta = \mathrm{Im } \alpha$. This leads us to our next definition:

Definition 2 (Exact sequence)
Let $A \xrightarrow\alpha B \xrightarrow\beta C$ be a sequence of homomorphisms such that $\mathrm{Ker } \beta = \mathrm{Im } \alpha$. Then we say that this sequence is exact at $B$. If $\cdots \rightarrow X_{n-1} \rightarrow X_n \rightarrow X_{n+1} \rightarrow \cdots$ is a sequence of homomorphisms, we say that it is exact if it is exact at $X_n$ for all $n$.

This language gives us a new (equivalent) way to state the so-called “basic situation”: given groups $A,C$, we want to study groups $B$ such that the sequence $A \rightarrow B \rightarrow C$ is exact.

Now, it is easy to see that if $\mathrm{Ker } \beta = \mathrm{Im } \alpha$ then $\beta \circ \alpha = 0$. In general, however, the converse is not true. The trouble is that $\beta \circ \alpha = 0$ only implies that $\mathrm{Im } \alpha \subseteq \mathrm{Ker } \beta$.

This observation motivates a new definition:

Definition 3 (Cochain complex)
Let $\mathcal{C}$ be a sequence of homomorphisms

$0 \rightarrow C^0 \xrightarrow{d_1} \cdots \rightarrow C^{n-1} \xrightarrow{d_n} C^n \xrightarrow{d_{n+1}} \cdots$.

We say that the sequence $\mathcal{C}$ is a cochain complex if the composition of any two successive maps is zero: $d_{n+1} \circ d_n = 0$ for all $n$. If $\mathcal{C}$ is a cochain complex, we define its nth cohomology group to be the quotient group $H^n(\mathcal{C}) = \mathrm{Ker }d_{n+1}/\mathrm{Im }d_n$.

Now, if $\mathcal{C}$ happens to be an exact sequence, by definition we know that $H^n(\mathcal{C})$ is the trivial group (since we’d have $\mathrm{Ker }d_{n+1} = \mathrm{Im }d_n$). However, if we instead have $\mathrm{Im }d_n \subsetneq \mathrm{Ker }d_{n+1}$, we can see that $H^n(\mathcal{C})$ can be used to “measure” the “failure” of $\mathcal{C}$‘s “exactness” at $n$.

But where does this get used?

In differential geometry, we can use these techniques to help is classify manifolds.  The discussion that follows should all be labeled as deRham Cohomology.
First, we produce a sequence of group homomorphisms out of our manifold. The groups will be differential $n$-forms on the manifold (these technically form a vector space, which is a group under addition). We can denote the group of exact $n$-forms by $\Omega^n$. The homomorphisms will be the exterior derivatives, which are a sequence of maps $d_n$ that turn $(n-1)$-forms into $n$-forms. It is a property of the exterior derivative that $d_{n+1} \circ d_n = 0$ for all $n$, which means that

$\cdots \rightarrow \Omega^{n-1} \xrightarrow{d_n} \Omega^n \xrightarrow{d_{n+1}} \Omega^{n+1} \rightarrow \cdots$

is a cochain complex.

Recall that we say $\omega$ is an exact $n$-form if there exists an $(n-1)$-form $\mu$ such that

$\omega = d_n\mu$.

Also recall that we say that $\omega$ is a closed $n$-form if $d_{n+1}\omega = 0$.

It is an easy exercise to show that every exact form is closed. This, in particular, implies that $\mathrm{Im } d_{n} \subseteq \mathrm{Ker } d_{n+1}$, as we would expect from a cochain complex.

In this language, failure of the complex to be exact at $\Omega^n$ means $\mathrm{Im } d_{n} \subsetneq \mathrm{Ker } d_{n+1}$, which means $\Omega^n$ contains $n$-forms that are not exact.

The interesting thing is that such strange forms reveal properties of the manifold’s topological structure.  Intuitively, this suggests that we should be able to use them to distinguish between manifolds, and in fact this intuition can be realized by a pleasant result: it turns out that homeomorphic manifolds have isomorphic deRham cohomology groups.  Yes, that’s right — we don’t even require the manifolds to be diffeomorphic!