Cohomologies.

Yoshifumi Tsuchimoto

DEFINITION 05.1   A (covariant) functor $F$ from a category $\mathcal C$ to a category $\mathcal D$ consists of the following data:

1. An function which assigns to each object $C$ of $\mathcal C$ an object $F(C)$ of $\mathcal D$ .
2. An function which assigns to each morphism $f$ of $\mathcal C$ an morphism $F(f)$ of $\mathcal D$ .

The data must satisfy the following axioms:

functor-1.

$F(1_C)=1_{F(C)}$ for any object $C$ of $\mathcal C$ .

functor-2.

$F(f\circ g)=F(f)\circ F(g)$ for any composable morphisms $f,g$ of $\mathcal C$ .

By employing the following axiom instead of the axiom (functor-2) above, we obtain a definition of a contravariant functor:

(functor-$2'$ ) $F(f\circ g)=F(g)\circ F(f)$ for any composable morphisms

DEFINITION 05.2   Let $F: \mathcal{C}_1 \to \mathcal{C}_2$ be a functor between additive categories. We call $F$ additive if for any objects $M, N$ in $\mathcal{C}_1$ ,

$$\operatorname{Hom}(M,N)\to \operatorname{Hom}(F(M),F(N))$$

is additive.

DEFINITION 05.3   Let $F$ be an additive functor from an abelian category $\mathcal{C}_1$ to $\mathcal{C}_2$ .

1. $F$ is said to be left exact (respectively, right exact ) if for any exact sequence
   $$0 \to L\to M\to N\to 0,$$
   the corresponding map
   $$0\to F(L)\to F(M)\to F(N)$$
   (respectively,
   $$F(L)\to F(M)\to F(N) \to 0)$$
   is exact
2. $F$ is said to be exact if it is both left exact and right exact.

LEMMA 05.4   Let $R$ be a (unital associative but not necessarily commutative) ring. Then for any $R$ -module $M$ , the following conditions are equivalent.

1. $M$ is a direct summand of free modules.
2. $M$ is projective

COROLLARY 05.5   For any ring $R$ , the category $(R \operatorname{-modules})$ of $R$ -modules have enough projectives. That means, for any object $M \in (R\operatorname{-modules})$ , there exists a projective object $P$ and a surjective morphism $f: P \to M$ .

DEFINITION 05.6   Let $R$ be a commutative ring. We assume $R$ is a domain (that means, $R$ has no zero-divisors except for 0 .)

An $R$ -module $M$ is said to be divisible if for any $r \in R\setminus \{0\}$ , the multplication map

$$M \overset{r \times }{\to} M$$

is surjective.

DEFINITION 05.7   Let $R$ be a commutative ring. We assume $R$ is a domain (that means, $R$ has no zero-divisors except for 0 .)

An $R$ -module $M$ is said to be divisible if for any $r \in R\setminus \{0\}$ , the multplication map

$$M \overset{r \times }{\to} M$$

is epic.

DEFINITION 05.8   Let $(K^\bullet,d_K)$ , $(L^\bullet, d_L)$ be complexes of objects of an additive category $\mathcal{C}$ .

1. A morphism of complex $u:K^\bullet \to L^\bullet$ is a family
   $$u^j: K^j \to L^j$$
   of morphisms in $\mathcal{C}$ such that $u$ commutes with $d$ . That means,
   $$u^{j+1} \circ d^j_K = d^j_K \circ u^j$$
   holds.
2. A homotopy between two morphisms $u,v: K^\bullet \to L\bullet$ of complexes is a family of morphisms
   $$h^j: K^j \to L^{j-1}$$
   such that $u-v = d \circ h + h \circ d$ holds.

LEMMA 05.9   Let $\mathcal{C}$ be an abelian category that has enough injectives. Then:

1. For any object $M$ in $\mathcal{C}$ , there exists an injective resolution of $M$ . That means, there exists an complex $I^\bullet$ and a morphism $\iota_M:M \to I^0$ such that
   $$% latex2html id marker 1010H^j(I^\bullet) = \begin{cases} ... ...iota_M) &\text{ if } j=0 \\ 0 &\text{ if } j\neq 0 \end{cases}$$

2. For any morphism $f:M\to N$ of $\mathcal{C}$ , and for any injective resolutions $(I^\bullet,\iota_M)$ , $(J^\bullet,\iota_N)$ of $M$ and $N$ (respectively), There exists a morphism $\bar f:I^\bullet \to J^\bullet$ of complexes which commutes with $f$ . Forthermore, if there are two such morphisms $\bar f$ and $f'$ , then the two are homotopic.

DEFINITION 05.10   Let $\mathcal{C}_1$ be an abelian category which has enough injectives. Let $F: \mathcal{C}_1 \to \mathcal{C}_2$ be a left exact functor to an abelian category. Then for any object $M$ of $\mathcal{C}_1$ we take an injective resolution $I^\bullet_M$ of $M$ and define

$$R^i F(M)=H^i(I^\bullet_M).$$

and call it the derived functor of $F$ .

LEMMA 05.11   The derived functor is indeed a functor.