# Relation between long exact sequences and Derived functors

I know that if i have a short exact sequence of chain complexes

$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$

then i can extend it to long exact sequence of homology groups as

$$\cdots\rightarrow H_n(A)\rightarrow H_n(B)\rightarrow H_n(C)\rightarrow H_{n-1}(A)\rightarrow H_{n-1}(B)\rightarrow H_{n-1}(C)\rightarrow\cdots$$

I some how see that this is kind of motivation for studying derived functors but i am not very sure about this kind of relation..

• Note that homology is a special case of derived functors: the functor $H_0$ is right exact, and $H_i$ is its $i$th left derived functor ("$H_i = L_iH_0$"). – Najib Idrissi Aug 29 '14 at 8:02
• Cross-posted on MO mathoverflow.net/questions/179648/… – Tobias Kildetoft Aug 29 '14 at 8:22
• @NajibIdrissi : I am sorry, i did not understand your point totally... can you please elaborate a bit more... – user87543 Aug 29 '14 at 9:01
• Let $Ch$ be the category of chain complexes, and $Ab$ the category of abelian groups. Then $H_0 : Ch \to Ab$ is a functor that maps a complex to its zeroth homology group. This functor is right-exact: if $0 \to A \to B \to C \to 0$ is an exact sequence, then $H_0 A \to H_0 B \to H_0 C \to 0$ is also exact. Because $Ch$ has enough projectives, we can compute the left derived functors of $H_0$, and it turns out and the $i$th left derived functor of $H_0$ is precisely $H_i$, as evidenced by the long exact sequence you wrote in your question. – Najib Idrissi Aug 29 '14 at 9:06
• But you only mentioned this on the MO post, not here. – Tobias Kildetoft Aug 29 '14 at 9:08

I will answer your question in what follows, but first it might be useful to say something about what derived functors are and what the point is.

The idea of derived functors is that if you have a functor which preserves (say) left exactness when applied to a short exact sequence, but not right exactness (e.g. $Hom(X, \text{--})$ for a module $X$ ), then its derived functors do produce an exact sequence, but now a long exact sequence.

As you probably know, the long exact sequence is $$0 \to F(A) \to F(B) \to F(C) \to R^1F(A) \to R^1F(B) \to R^1(C) \to R^2(A) \to \cdots$$

Here I am writing $R^iF$ to denote the $i$th derived functor (a right derived functor in this case, since the functor $F$ is left exact; so the letter $R$ is for "right").

Note that the derived functors are a so-called $\delta$-functor : not only do we have the sequence of functors $R^iF$, but the above long e.s.'s are themselves functorial in the s.e.s. $0 \to A \to B \to C \to 0.$ (The "$\delta$" in $\delta$-functor is a reference to the collection of connecting morphisms $R^iF(C) \to R^{i+1}F(A)$, which are commonly denoted $\delta$.)

There are a few comments to make:

• The existence of these l.e.s.'s doesn't characterize the $R^iF$ as a $\delta$-functor.

E.g. if you had the collection $R^iF,$ I could define another functor $\widetilde{R}^iF$ as follows: in all degrees $i$ except $i = 1$, set $\widetilde{R}^iF =R^iF,$ but in degree $1$, define $\widetilde{R}^iF(A) = R^iF(A) \oplus A$ (for any object $A$). Define the connecting morphisms on the extra summand we put in to be just $0$. So the new l.e.s. is the same as the old one, except that we have just direct summed in a copy of the original s.e.s. into the degree $1$ part of the l.e.s. of $R^iF$'s.

This is dealt with by the following point.

• The $R^iF$ are taken to be minimal, in some appropriate sense. Precisely, they are a universal $\delta$-functor for $F$, i.e. they map canonically to any other $\delta$-functor. (This is discussed in Hartshorne's algebraic geometry book at the beginning of Ch. III, but in many other places too, I presume.)

So, intuitively, they don't contain any extra garbage that we might have added gratuitously in some degree.

• Although it's not obvious if you're learning this for the first time, to construct the universal $\delta$-functor, you can use injective resolutions of objects, and apply $F$ (the original, underived, functor) to these resolutions. (This gives a cochain complex, whose cohomology yields the $R^iF$'s.) To get the l.e.s.'s: first prove that a s.e.s. of objects gives a s.e.s. of resolutions. Then, since s.e.s.'s of injective objects are split, and since $F$ will take a split s.e.s. to a (split) s.e.s., applying $F$ gives a s.e.s. of complexes. Passing to cohomology, and using the fact recalled in your question, gives the l.e.s. of derived functors.

So the connection between the fact you recalled in your question, and derived functors, actually doesn't come until a fairly technical part of the story, namely in the actual construction of the derived functors via injective resolutions.

Note that although sometimes you use resolutions to compute the derived functors, often you use more indirect means, and so the construction via injective resolutions is often not a key point to focus on in applications (even though it underlies the theory).

• What is the point of the whole story? Why do we go to all the trouble of constructing these derived functors, looking at the l.e.s.'s, and so on?

E.g. why don't people just do the naive thing of looking at the four term exact sequence $$0 \to F(A) \to F(B) \to F(C) \to \text{ cokernel } \to 0.$$ The point is that this cokernel depends not just on $A$, $B$, or $C$, but on all three of them, and in fact on the whole data of the s.e.s. $0 \to A \to B \to C \to 0.$

So basically, it's not a very flexible object, and there isn't a lot of theory you can develop about it directly.

What happens in the theory of derived functors is that the information in this cokernel is diffused into the theory in a much more subtle, flexible, and useful way.

E.g. in the l.e.s. of derived functors, the terms only depend on one of the objects $A$, $B$ ,or $C$. It is only the morphisms (especially the connecting morphisms) that depend on the whole data of the s.e.s.

In particular, suppose in some particular case we want to show that $0 \to F(A) \to F(B) \to F(C) \to 0$ is exact.
The general theory shows that what we have to do is prove that the connecting morphism $F(C) \to R^1F(A)$ equals zero.

One way we can do this is just to show that $R^1F(A) = 0$ itself. This depends only on $A$, and so doesn't depend on the original s.e.s. at all. In particular, it is something we could try to check by making a computation related to $A$, without having to think at all about $B$ or $C$ or the original s.e.s. Possibly we could compute directly with injective resolutions, or perhaps we could try to use some other more indirect method: e.g. we could try to put $A$ into some other s.e.s. we know more about, and use the derived functor l.e.s. for that s.e.s. to get information about the values of the $R^iF(A)$.

Even if we can't prove that $R^1F(A)$ vanishes, we can still hope to get information (e.g. if we can bound its size, then we can bound the size of the cokernel of $F(B) \to F(C)$). By diffusing the information through the various derived functor values and the connecting morphisms, the theory of derived functors provides a much more useful tool then you get from a more naive approach to these questions.

• Thank you for spending so much time for my question.. i read that before but could not understand it properly.. I took some time.. Now it seems reasonable to me :D Thank you – user87543 Sep 1 '14 at 7:13

I'm not sure what exactly you're asking for, but here's a blurb about derived functors that I hope is useful to you. Suppose we have a left-exact functor $F$ on the category $\cal{C}$ of modules over a given commutative ring. (The usual setting is abelian categories, but there's no harm in working with something concrete for a short description. Also, I'm using left-exact functors and therefore right-derived functors, but the same approach works precisely the same with right-exact functors.) By definition, a short exact sequence of modules $$0 \to A \to B \to C \to 0$$ in $\cal{C}$ induces by $F$ a short exact sequence $$0 \to F(A) \to F(B) \to F(C).$$ If $F$ is not exact, then we can't extend the sequence above with the map $F(C) \to 0$ in general and preserve exactness. We can keep extending the sequence, though, at the expense of adding more terms to the right to kill off preceding maps in the complex. What we ultimately want is a series of objects $R^* F$ giving us an exact sequence $$\cdots \to R^{p-1} F(C)\to R^p F(A) \to R^p F(B) \to R^p F(C) \to R^{p+1} F(A) \to \cdots$$ with $F^0(M) = M$ and $F^p(M) = 0$ for $p < 0$.

So, why bother with this construction? First of all, the "common" cohomology theories are of the form $R^* F$ for some left-exact functor $F$. Ordinary CW-complex homology and cohomology can be constructed through this operation. The most common example is sheaf cohomology, where $H^*(X, \scr{F})$ is constructed as $R^*F$ for $F = \Gamma(X, \scr{F})$, the module of global sections of $\scr{F}$. This might be a fancier case than you care about, but that's part of the point; we get a complicated thing for free with this machinery. More concretely, we can look at the functor $$F(M) = M^G = \{x\in M:\, gx = x\text{ for all g\in G}\}$$ for a group $G$ and a $\mathbb{Z}G$ module $M$. This functor is left-exact, so we get a cohomology theory for free, without having to compute any chain maps or verify any axioms. I'm not going to go into their details here, but the same construction gives you the higher $\text{Ext}^*$ and $\text{Tor}^*$ functors (important in homological algebra and other places), some of the more complicated cohomology theories in algebraic geometry, and so on.

The other advantage is the power of abstract nonsense. Because of some naturality properties, the derived functors $R^*F$ are well-defined and unique. As a result, we can instantly (well, at least briefly) prove that, for example, the cohomology of $K(\pi, 1)$ is the same thing as the cohomology of the group $\pi$. These are ostensibly different things; the former is the cohomology of a (probably infinite) CW-complex, and the latter is the cohomology of a group and defined in purely algebraic terms. The point, though, is that all the machinery of the derived functors shows easily that they're both computing the same sort of injective resolution and thus equivalent. It's possible to go into even more abstract nonsense and talk about derived categories, but hopefully this gives you a flavor of the stuff involved.

• I am not familiar with CW complexes and all but your answer is helpful for me.. Thank you :) – user87543 Sep 1 '14 at 7:15