## Extensions and homological algebra

If you ever followed a course in algebra, you may have heard about extensions of “things”. Most “things” fit in what is called a category, that is, an abstract structure remembering how to compose morphisms between these “things”, and sometimes what is the sum of such “things”, the kernel of a morphism: we are interested in the structure of abelian category, which is the framework of vector spaces, modules over a ring, sheaves of modules… An extension of A by B is an exact sequence

$0 \to B \to E \to A \to 0$

where $B$ is the kernel, and $A$ is the quotient (cokernel). The group of extensions $\mathrm{Ext}^1(A,B)$ is the set of isomorphism classes of such exact sequences. It is a group by means of the Baer sum, which takes two extensions E and E’, and considers the diagram :

Here S is the fiber product of E and E’ over B. Then the Baer sum of E and E’ is the quotient of S where the two copies of A are identified. But this approach doesn’t really help knowing what the Ext group looks like. The machinery of homological algebra tells us that Ext is a derived functor for Hom. This essentially means that an exact sequence

$0 \to A \to B \to C \to 0$

determines an exact sequence

$0 \to Hom(M,A) \to Hom(M,B) \to Hom(M,C)$
$\to Ext(M,A) \to Ext(M,B) \to Ext(M,C)$

and

$0 \to Hom(C,M) \to Hom(B,M) \to Hom(A,M)$
$\to Ext(C,M) \to Ext(B,M) \to Ext(A,M)$

Thus if A and B are projective (a thing A is projective if $Hom(A,M) \to Hom(A,N)$ is surjective for any surjective morphism $M \to N$), $Ext(A,M) = Ext(B,M) = 0$, and it is not hard to compute $Ext(C,M)$. More generally, it is possible to define groups $\mathrm{Ext}^n(A,B)$ (see MacLane, Homology for an elementary definition), so that an exact sequence as above determines a long exact sequence

$0 \to Hom(M,A) \to Hom(M,B) \to Hom(M,C)$
$\to Ext^1(M,A) \to Ext^1(M,B) \to Ext^1(M,C)$
$\to Ext^2(M,A) \to Ext^2(M,B) \cdots$

We say that the $Ext^n$ are the derived functors of $Hom$. A fundamental result is that if $(P_\bullet)$ is a projective resolution of $A$ i.e. a sequence of projective things with an exact sequence

$\cdots \to P_n \to P_{n-1} \to \cdots \to P_1 \to P_0 \to A \to 0,$

the cohomology groups of the complex $\mathrm{Hom}(P_\bullet, M)$ are exactly the $\mathrm{Ext}^n(A,M)$.