Intuitive meaning of Exact Sequence

Question

I'm currently learning about exact sequences in grad sch Algebra I course, but I really can't get the intuitive picture of the concept and why it is important at all.

Can anyone explain them for me? Thanks in advance.

Answer 1

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.

Answer 2

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!

Answer 3

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.

Answer 4

The answer is the same for many abstractions in mathematics - we realize something comes up often and once we have a clear way to describe them only by their essential features, it is easier to recognize their general properties and recognize them "in the wild". One might have asked "What is the intuitive meaning of the group axioms?" and one could answer that there is no intuition a priori that it is fruitful to list such conditions to form a group, our intuition comes from the special examples the new concept generalizes. To someone not used to groups yet, it might have seemed odd to make that leap when you could just work with concrete symmetry groups but we have seen that the abstraction has helped us plenty.

Though the previous question addresses the direct question of the intuition in the condition for exact sequences, it probably doesn't help you understand them much better and I'm guessing that is what you wanted. I find a good way to understand abstract constructions/objects is to understand them deeply in their most important special cases. These will depend on what subjects you are encountering exact sequences in. I recommend section 2.10 of Fulton's "Algebraic Curves" (you can do most questions in that section without knowing geometry) for some good basic examples.

Answer 5

I saw a lovely description of what a "free" resolution is that also might bear on why the property of exactness is interesting.

Lets say we have an abelian group, $A$, with a set of generators, $X$. Then we have a natural map $F[X]\to A$ which is onto, where $F[X]$ is the free abelian group on $X$. That map has a kernel, which tells us the relationships between the generators. But that kernel itself might not be free. Let $R_1$ be the set of generators of the kernel. Then we have an exact sequence $F[R_1]\to F[X]\to A\to 0$.

We can keep going, then, and we get an exact sequence ending in $A\to 0$ where all the other groups are free and in some sense, we are measuring the "freeness" of the set of relations of relations of relations... In particular, the shortest free resolution is some sort of measure of the complexity of the underlying group.

Answer 6

In the category of (left) $R$-modules, short exact sequences are crucial for the study of the functors $\mathrm{Hom}(A,-), \mathrm{Hom}(-,A), A\otimes_R-$ and other variants, as well as many other functors.

For instance, perhaps the most basic example, given a map $f:M\to N$ of $R$-modules that is injective, you might be curious to know whether the corresponding map $f\otimes 1: M\otimes_R A\to N\otimes_R A$ is injective. Sometimes it is, and sometimes it isn't. If you've already seen the tensor product, then this is a natural question to ask. It turns out that it is more natural to phrase this question in terms of a short exact sequence: the injective map $f:M\to N$ is the same thing as saying that the sequence

$0\to M\to N \to N/M\to 0$

is exact. In turns out that when we apply the tensor functor to this sequence, we get a long exact sequence

$\cdots\to\mathrm{Tor}_2(N/M,A)\to \mathrm{Tor}_1(M,A)\to\mathrm{Tor}_1(N,A)\to\mathrm{Tor}_1(N/M,A)\to M\otimes_R A\xrightarrow{f\otimes_1} N\otimes_R A\to N/M\otimes_R A\to 0$

Here, you can take it as a black box that the $\mathrm{Tor}_*(-,A)$ objects are abelian groups ($R$-modules if $R$ is commutative). So we have a long exact sequence of abelian groups. Looking at this sequence, we will know that the corresponding map $f\otimes 1$ is an injective if we can show that $\mathrm{Tor}_1(N/M,A) = 0$. Now, one of the easiest ways to calculate $\mathrm{Tor}_1(N/M,A)$ in general is to use other short exact sequences that either contain $N/M$ or $A$, since short exact sequences always give long exact sequences as above.

Even if this is a bit abstract, you can think of long exact sequences like a crossword puzzle: if you know some of the terms or maps in a long exact sequence, you can use it to figure out other terms in the long exact sequence. Moreover, to get long exact sequences, you usually need short exact sequences!

So, short and long exact sequences some up in the question: does $A\otimes_R-$ preserve a certain injective map? Dually, you can ask whether $\mathrm{Hom}(A,-)$ preserves a certain surjective map.

Here's a reason to care about those higher $\mathrm{Tor}_*$ groups like $\mathrm{Tor}_2,\mathrm{Tor}_3,...$: if $A$ is an $R$-module then the least $n$ for which $\mathrm{Tor}_n(A,B)$ vanishes for all $n$ is called the flat dimension of $A$. Taking the supremum over all $R$-modules $A$ gives the $\mathrm{Tor}$-dimension of the ring $R$. One can also do this for the $\mathrm{Hom}$-functor, which gives a concept known as global dimension (left or right variants). These invariants are useful for studying the ring $R$ itsef. Dimension theory for instance has had great success in proving results about local rings (rings for which there is just one maximal ideal).

Answer 7

It's sometimes a good idea to think of long exact sequences in the more general context of chain complexes. A chain complex $C$ of maps $d_i$ is a sequence $$\cdots\rightarrow A_{i+1}\stackrel{d_{i+1}}{\rightarrow}A_i\stackrel{d_i}{\rightarrow}A_{i-1}\rightarrow\cdots$$ such that $d_i\circ d_{i+1}=0$ for all $i$. Now, from this very mild condition, we can do quite a lot with this chain complex. In particular, we know that $\mbox{im}\, d_{i+1}\subset\ker d_i$ and so we can take a quotient.

Let $$H_n(C)=\ker d_n/\mbox{im}\,d_{n+1}.$$ We call this the $n$th homology of the chain complex $C$. The homology of a chain complex is an extremely powerful tool and motivates much of homological algebra.

We might like to ask then, what conditions on the chain complex $C$ would tell us that its homology in every degree is trivial? It turns out that the homology of $C$ is trivial in every degree if an only if $C$ is an exact sequence. This isn't hard to show as, if $C$ is an exact sequence then $\mbox{im}\, d_{n+1}=\ker d_n$ and so $H_n(C)=\ker d_n/\mbox{im}\,d_{n+1}=0$. Also, if $H_n(C)=0$ then $\ker d_n/\mbox{im}\,d_{n+1}=0$ and so $\mbox{im}\, d_{n+1}=\ker d_n$.

Answer 8

Here is a topological interpretation:

Let $X$ be a space and $Y$ its subspace. If a boundary (in $Y$) of an $n$-dimensional relative cycle $c$ of $X\setminus Y$ is a boundary of something in $Y$ then one can build a proper $n$-dimensional cycle of $X$ from $c$, gluing this "something" to $c$.