Intuitive meaning of Exact Sequence

In the linear algebra of Euclidean space (i.e. $\mathbb R^n$), the consideration of subspaces and their orthogonal complements are fundamental: if $V$ is a subspace of $\mathbb R^n$ then we think of it as filling out "some of" the dimensions in $\mathbb R^n$, and then its orthogonal complement $V^{\perp}$ fills out the other directions. Together they span $\mathbb R^n$ in a minimial way (i.e. with no redundancies, i.e. $\mathbb R^n$ is the direct sum of $V$ and $V^{\perp}$).

Now in more general settings (say modules over a ring) we don't have an inner product and so we can't form orthogonal complements, but we can still talk about submodules and quotients.

So if $A$ is a submodule of $B$, then $A$ fills up "some of the directions" in $B$, and the remaining directions are encoded in $B/A$.

Now by itself this doesn't seem like anything new, or worth memorializing with new terminology, but often what happens is that one has a submodule $A \subset B$, and then a surjection $B \to C$, given without any a priori relation to each other.

However, if $A$ is precisely the kernel of the map $B \to C$, then we are (somewhat secretly) in the previous situation: $A$ fills out some of the directions in $B$, and all the complementary directions are encoded in $C$.

So we introduce the terminology "$\, \, 0 \to A \to B \to C \to 0$ is a short exact sequence" to capture this situation.

Since long (i.e. not necessarily short) exact sequences can always be broken up into a bunch of short exact sequences that are glued together, getting a feeling for short exact sequences is a good first step.

Of course, you should be coupling your study of these homological concepts with examples, e.g. short exact sequences arising from tangent and normal bundles to submanifolds of manifolds, all the important long exact sequences in homology theory (from algebraic topology), and so on; without these examples of naturally occuring set-ups of the "$A, B, C$" form described above, it won't be so easy to get a feel for why this concept was isolated as being a fundamental one.

There are many good answers here. I'd just like to add one example that made exact sequences 'click' for me, related to "Euler's Formula" relating the number of vertices ($V$), edges ($E$), and faces ($F$) of a simple non-self-intersecting polyhedron: $$ |F| - |E| + |V| = 2$$ Now what does this have to do with exact sequences, you may well ask! Well if you consider the free abelian groups generated by the set of faces, edges, and vertices separately, and create certain linear maps between them (see 'boundary maps' for simplicial homology), then you almost get an exact sequence: $$ \mathbb{Z}[F] \to \mathbb{Z}[E] \to \mathbb{Z}[V] $$ In fact, this sequence is exact at the middle term. If we append two rank $1$ groups on the left and right (one with a generator the the whole solid $S$, and one generated by the symbol $e =$ '$\emptyset$'), then you do get an exact sequence: $$ 0 \to \mathbb{Z}[S] \to \mathbb{Z}[F] \to \mathbb{Z}[E] \to \mathbb{Z}[V] \to \mathbb{Z}[e] \to 0 $$

Then Euler's Formula is the statement just that the alternating sum of ranks is $0$ (because there is no torsion to keep track of).

$$ -1 + |F| - |E| + |V| - 1 = 0, $$ or $$ |F| - |E| + |V| = 2. $$ Hope this helps!

Short version: An exact sequence gives an ingredients list using inclusion-exclusion.

High-tech version: in some cases the Grothendieck group has a quotient that it is easy to compute.

Some standard exact sequences

It helps if you are familiar with some of the basics of exact sequences first. None of the next 6 bullets points is deep. It is just a notation that allows easy book-keeping.

• $0 \to A \xrightarrow{a} B$ is exact iff $a$ is 1-1.
• $B \xrightarrow{b} C \to 0$ is exact iff $b$ is onto.
• If $A \xrightarrow{a} B \xrightarrow{b} C \to 0$ and $0 \to C \xrightarrow{c} D \xrightarrow{d} E$ are exact, then $A\xrightarrow{a} B \xrightarrow{cb} D \xrightarrow{d} E$ is exact $ \newcommand{\im}{\operatorname{im}} \newcommand{\cok}{\operatorname{cok}} $
• $0 \to \ker(f) \to A \xrightarrow{f} \im(f) \to 0$ is exact, for $f:A \to B$
• $0 \to \im(f) \to B \to \cok(f) \to 0$ is exact, for $f:A \to B$
• $0 \to \ker(f) \to A \xrightarrow{f} B \to \cok(f) \to 0$ is exact

Comparing $A$ and $B$

The last one is worth talking about a little: a homomorphism compares $A$ and $B$. The way in which they differ is captured by $\ker(f)$ and $\operatorname{cok}(f)$.

This sequences says that $A$ is exactly the same as $B$, well, except for the kernel $\ker(f)$, and actually that only gives you $A/\ker(f) \cong \im(f)$, so we are also missing $B/\im(f) = \cok(f)$. Ok, so actually if you take $A$ and get rid of $\ker(f)$, it is the same as taking $B$ and getting rid of $\cok(f)$.

$$[A] - [\ker(f)] = [B] - [\cok(f)] \quad \text{or} \quad -[\ker(f)] + [A] - [B] + [\cok(f)] = 0 $$

Inclusion-exclusion

In general, an exact sequence of the form $0 \to A_1 \xrightarrow{f_1} A_2 \xrightarrow{f_2} \ldots \xrightarrow{f_{n-2}} A_{n-1} \xrightarrow{f_{n-1}} A_n \to 0$ has the nice property that for many reasonable definitions of “size”, say $A_i$ has size $d_i$, one has that $$-d_1 + d_2 \mp \ldots + (-1)^{n-1} d_{n-1} + (-1)^n d_n = 0$$

Notice that $A_k$ contains the image $\operatorname{im}(f_{k-1})$ with leftover $A_k/\operatorname{im}(f_{k-1}) = A_k/\ker(f_k) \cong \operatorname{im}(f_k)$. Symbolizing this as $$[A_k] = [\im(f_{k-1})] + [\im(f_k)]$$

Sometimes we choose (all but one of) the $A_i$ to be very very nice, and try to understand the left-over one, say $A_k$. If we understood $[\im(f_{k-1})]$ and $[\im(f_k)]$ directly, then $A_k$ would be just fine. Now $A_{k-1}$ and $A_{k+1}$ are nice, but maybe the images are not nice. So we write:

$$[A_k] = [\im(f_{k-1})] + [\im(f_k)] = -[\im(f_{k-2})] + [A_{k-1}] + [A_{k+1}] - [\im(f_{k+1})]$$

Now these $f_i$ have $i$ further from $k$, and since our sequence is bounded by $0$s, if we keep pushing away eventually the images will disappear:

$$[A_k] = [\im(f_{k-3})] - [A_{k-2}] + [A_{k-1}] + [A_{k+1}] - [A_{k+2}] + [\im(f_{k+2})]$$

Eventually we are just solving for $[A_k]$ in: $$-[A_1] +[A_2] -[A_3] \pm \ldots + (-1)^{n-1} [A_{n-1}] + (-1)^n [A_n] = 0$$

Specific measurements

For instance, if $A_i$ are finite abelian groups and $d_i = \log(|A_i|)$, then the formula works.

If $A_i$ are finite dimensionsal vector spaces and $d_i = \dim(A_i)$, then the formula works.

If $A_i$ are vector bundles and $d_i$ are the continuous functions that take a point to the dimension of the vector bundle at that point, then the formula holds.

If $A_i$ are representations of finite groups and $d_i$ are characters, then the formula holds.

If $A_i$ are finite abelian groups and $|A_i| = \prod p_j^{e_{ij}}$ and $d_i=(e_{i1}, e_{i2}, \ldots)$, then the formula holds.

Resolutions

Why would we have all these $A_i$ if we don't even understand $A_k$?

The answer is actually pretty easy: if $A$ and $B$ are very nice (say free modules), and $f:A \to B$ is given (say by a matrix) then we may want to understand $\ker(f)$ and $\cok(f)$. Without knowing lots of details of $f$, we cannot guess both $\ker(f)$ and $\cok(f)$, but inclusion-exclusion lets us calculate one if we know the other!

I often see this where $f$ is given precisely to specify $\cok(f)$, and so all we need to do is figure out $\ker(f)$. I'll conveniently label things as $f:A_{n-2} \to A_{n-1}$ and $A_n = \cok(f)$. So we find another nice $A_{n-2}$ and a homomorphism $f_{n-2}:A_{n-2} \to A_{n-1}$ whose image is exactly $\ker(f)$. Now inclusion exclusion tells us $A_n = \cok(f)$ as soon as we figure out what $\ker(f_{n-2})$ is. We find some nice $A_{n-3}$ and $f_{n-3}:A_{n-3} \to A_{n-2}$ whose image is exactly $\ker(f_{n-2})$ and inclusion-exclusion tells us about $\cok(f)$ if only we know about $\ker(f_{n-3})$.

If we are doing things so that the kernels are getting smaller or simpler, then we succeed! If the kernels are getting worse then often this has very limited utility.