Want to know more about exact sequences of homomorphisms

Question

While starting out on Lang's Algebra I have encountered the term "exact" sequence of homomorphisms. And I understand its definition and all the demonstrated examples but I would like to know more on this, like what are the things we get to know out of an exact sequence and does it aid in any kind of constructions or whatsoever. Is it possible to generalize them and how does that look like ?.

For my background, I have only been introduced to Groups, Rings and Fields in my introductory algebra course but I would be delighted in seeing a bigger picture on these structures.

Answer 1

• The zeros in an exact sequence give normal and conormal maps in an abelian category, and injective and surjective maps for the case of abelian groups. e.g. $N \rightarrow M \rightarrow 0$ for the case of modules entails that the map $N \rightarrow M$ is surjective
• The long exact sequence is an important exact sequence.
• Triangulated categories describe such triples, each of which gives a long exact sequence.
• "Exactness" works for based spaces and spectra as well.
• e.g. * ⭢ ℤ ⭢ ℝ ⭢ S¹ ⭢ *, * ⭢ S¹ ⭢ S³ ⭢ S² ⭢ *
• spectral sequences are a major topic involving exactness. Spectral sequences involve exact sequences indexed by ℤ² instead of ℤ
• "Exact couples" can be used to understand spectral sequences

Answer 2

The MO question here will give you some context for exact sequences, but to be honest it may be hard to appreciate this concept until you learn more math. If you don't yet have a reason to care, it could be hard to explain why you should care.

You could think about quotient groups in the setting of exact sequences: instead of saying there is a surjective homomorphism $G \to H$ with kernel $N$ (so $G/N \cong H$ in a definite way), we could say there is a short exact sequence $$ 1 \to K \to G \to H \to 1 $$ where the map $K \to G$ is the inclusion and the map $G \to H$ is the one given to us. The first map from $1$ and the last map to $1$ are unique since there's only one group homomorphism out of $1$ to a group and to $1$ from a group.

Conversely, given such a short exact sequence (for some homomorphisms $K \to G$ and $G \to H$ for three groups $K$, $G$, and $H$), the mapping $G \to H$ in the short exact sequence is surjective (exactness at $H$) and its kernel is the image of $K$ in $G$ (exactness at $G$) and this kernel is isomorphic to $K$ (exactness at $K$), so the information in a short exact sequence is essentially the same as the information in building a quotient group.

So what? Well, in many areas of math you can build much longer exact sequences $$ \cdots \to G_{i-1} \to G_i \to G_{i+1} \to G_{i+1} \to \cdots $$ as part of some general mathematical machinery. This first occurred in algebraic topology and then spread to many other parts of math. If we are in a lucky situation where $G_{i-1}$ and $G_{i+2}$ are both trivial, then inside that long exact sequence a piece of it is the exact sequence $$ 1 \to G_i \to G_{i+1} \to 1 $$ and this exactness says $G_i$ and $G_{i+1}$ are isomorphic groups via the map $G_i \to G_{i+1}$ that appears there. So exact sequences built in a complicated way might reveal to us that certain homomorphisms are isomorphisms if enough surrounding groups turn out to be trivial.

Answer 3

If you are familiar with the concept of diagram chasing, that gives two ways to use exact sequences. For simplicity, let our category in the first example be (left) modules over a unital ring.
(1) For instance, if we are given a commutative diagram $\require{AMScd}$ \begin{CD} 0 @>>> A @>{i}>> B @>{p}>> C@>>> 0\\ @VVV @V{f_1}VV @V{f_2}VV @. @VVV\\ 0 @>>> A' @>{j}>> B' @>{q}>> C' @>>> 0 \end{CD} with exact rows, we may show the existence of an $f_3\colon C\to C'$, s.t. $f_3\circ p = q\circ f_2$ using the exactness as our main tool. Indeed, for every $c\in C$ there is a $b\in B$ satisfying $p(b)=c$. We then want to define $f_3(c) = (q\circ f_2)(b)$, but... What if there is a $b'\in B$, s.t. $b'\neq b$ but $p(b') = p(b)$? This could be an issue, but the exactness saves us as $b-b'\in\ker(p) = \text{im}(i)$, so that $b-b' = i(a)$ for some $a\in A$. We may then show that $(q\circ f_2)(b-b') = (q\circ f_2 \circ i)(a) = (q\circ j\circ f_1)(a) = 0$, again by exactness, from which it follows that $f_3$ is well-defined.
(2) we may chase the diagrams to prove properties of morphisms. For example, consider the diagram \begin{CD} A_1 @>{\alpha_1}>> A_2 @>{\alpha_2}>> A_3 @>{\alpha_3}>> A_4 @>{\alpha_4}>> A_5\\ @V{f_1}VV @V{f_2}VV @V{f_3}VV @V{f_4}VV @V{f_5}VV\\ B_1 @>>{\beta_1}> B_2@>>{\beta_2}> B_3@>>{\beta_3}> B_4@>>{\beta_4}> B_5, \end{CD} and assume it commutes with exact rows. If we also assume that the morphisms $f_2, f_4$ are surjective and $f_5$ is injective, we can show that $f_3$ is surjective too. Indeed, start with an arbitrary $b_3\in B_3$, and choose any $a_4\in A_4$ s.t. $f_4(a_4) = \beta_3(b_3)$. Then note that $\alpha_4(a_4)=0$ as $(f_5\circ\alpha_4)(a_4)=(\beta_4\circ f_4)(a_4) = (\beta_4\circ\beta_3)(b_3) = 0$ and $f_5$ is injective. We then obtain $a_3\in A_3$ s.t. $\alpha_3(a_3) = a_4$ and note that $\beta_3(b_3-f_3(a_3)) = 0$. From this we obtain $b_2\in B_2$ s.t. $\beta_2(b_2) = f_3(a_3) - b_3$, for which we choose any $a_2\in A_2$ s.t. $f_2(a_2) = b_2$. Finally, we have $f_3(a_3 + \alpha_2(a_2)) = b_3$ and $f_3$ is surjective. All steps except when we get an $a_i$ out of a $b_i$ via $f_i$ are based on exactness and commutativity. For more info on this, check the 5-lemma.

Once you get to resolutions, you'll see that exactness lets you prove not only the existence but also the uniqueness (upto chain homotopy) of chain maps between them. In more precision, if $P_*\xrightarrow{\varepsilon} M\to 0, Q_*\xrightarrow{\eta} N\to 0$ are projective resolutions and $f\colon M\to N$ is any morphism, then there is a chain map $f_*\colon P_*\to Q_*$, s.t. $f\circ\varepsilon=\eta\circ f_0$, and this chain map is unique upto homotopy. This fact can be used to show that one may use any projective resolution(s) to compute $\text{Tor}_*^R(M, N)$ for a right $R$-module $M$ and a left $R$-module $N$.

Another simple example of their usefulness are the long exact sequences. If I were to tell you that the $n$th homotopy group of an $n$-sphere satisfies $\pi_n(S^n)\cong\mathbb{Z}$, that $\pi_k(S^1)=0$ for all $k>1$ and $\pi_k(S^n)=0$ for all $0<k<n$, could you tell me what $\pi_3(S^2)$ is like? Indeed you could! Any (Serre) fibration $F\to E\to B$ induces a funny homotopy LES $$\cdots\to\pi_{n+1}(B)\to\pi_n(F)\to\pi_n(E)\to\pi_n(B)\to\cdots\to\pi_1(B)\to\pi_0(F)\to\pi_0(E)\to\pi_0(B)\to *,$$ in which all objects are abelian groups for $n\ge2$, are groups but not necessarily abelian for $n=1$ and need only be pointed sets for $n=0$. The Hopf fibration $S^1\to S^3\to S^2$ fits in nicely, and yields the exact sequence $$0=\pi_3(S^1)\to\pi_3(S^3)\to\pi_3(S^2)\to\pi_2(S^1)=0,$$ from which we see that $\pi_3(S^2)\cong\pi_3(S^3)\cong\mathbb{Z}$.