Derived Functors

We are only concerned here with covariant functors.

Definition 11   Let $\mathscr{C}$ be an abelian category and $\mathscr{C}'$ be some additive catogory. For any two numbers $a$ and $b$ (which can be $\infty$ or $-\infty$), such that $a+1<b$, a covariant $\partial$ functor from $\mathscr{C}$ to $\mathscr{C}'$ of degrees $a<i<b$ is a system $T = (T^i)_{a<i<b}$ of additive covariant functors from $\mathscr{C}$ to $\mathscr{C}'$ and for every $i$, $a<i<b-1$ and exact sequence $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ in $\mathscr{C}$, a morphism

$$\partial : T^i(A'')\longrightarrow T^{i+1}(A')$$

satisfying following axioms:

1. If we have following commutative diagram with both rows exact
   $$\xymatrix{ 0\ar[rr] & & A'\ar[rr] \ar[dd] & & A \ar[rr] \ar[dd] &... ... & & & & & & \\ 0\ar[rr] & & B'\ar[rr] & & B \ar[rr] & & B'' \ar[rr] & & 0 }$$
   Then the following diagram is commutative:
   $$\xymatrix{ T^i(A'') \ar[rr]^{\partial} \ar[dd] & & T^{i+1}(A') \ar[dd] \\ & & \\ T^i(B'') \ar[rr]^{\partial} & & T^{i+1}(B') }$$

2. For every exact sequence $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$, the corresponding long sequence

   $$\ldots T^i(A') \rightarrow T^i(A) \rightarrow T^i(A'') \rightarrow T^{i+1}(A') \rightarrow \ldots$$ (5)

   
   
   is a complex (that is, composition of every two consecutive morphisms is zero.

If $\mathscr{C}'$ is abelian category and (5) is exact sequence, then corresponding $\partial-functor$ is called cohomology functor. Assume $T=(T^i)$ and $T'=(T'^i)$ are two $\partial$ functors of same degrees, then we have notion of morphism between these two $\partial$-functors, which is a system of natural transformations $f=(f^i)$, where each $f^i$ is natural transformation between $T^i$ and $T'^i$, such that for each $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$, exact sequence in $\mathscr{C}$, the following diagram commutes:

$$\xymatrix{ T^i(A'') \ar[dd] \ar[rr]^{\partial} & & T^{i+1}(A') \ar[dd]\\ & & \\ T'^i(A'') \ar[rr]_{\partial} & & T'^{i+1}(A') }$$

Now a universal $\partial$ functor is a $\partial$ functor $T=(T^i)_{0\leq i\leq a}$, such that for every $T'=(T'^i)$ of same degrees, and a natural transformation $f : T^0\rightarrow T'^0$, there is a unique morphism of $\partial$ functors $\tilde{f}$ between $T$ and $T'$ which is same as $f$ in degree 0. By this definition, given any functor $F: \mathscr{C} \rightarrow \mathscr{C}'$, there can be atmost one (upto isomorphism), universal $\partial$ functor with degrees $0\leq i\leq a$, which is same as $F$ in degree 0. Its higher degrees are denoted by $S^iF$ and called satellite functors of $F$. Of course, it is not obvious that such sequence of functors exist. It will be proved in Theorem 4.

To prove the main result, Theorem 4, we need notion of effacability. First observe that to prove the existence of satellite functors in all degrees, it is enough to prove existence of $S^1F$ and $S^{-1}F$, by virtue of the formulae $S^1(S^iF) = S^{i+1}F$ and $S^{-1}(S^iF) = S^{i-1}F$.

Definition 12   An additive functor from $\mathscr{C}$ (assumed abelian) to $\mathscr{C}'$ (assumed additive) is called effacable if for every $A\in \mathscr{C}$, one can find an injective morphism $u: A\rightarrow M$ such that $F(u)=0$.

Now we can state characterisation of univeral $\partial$ functor in terms of effacability.

Proposition 1   Let $\mathscr{C}$ and $\mathscr{C}'$ be two abelian categories and $T=(T^i)$, $0<i<a$ be covariant, exact, $\partial$ functor in degrees $0<i<a$, where $a>0$. If $T^i$ is effacable for every $i>0$, then $T$ is universal and converse holds if every object of $\mathscr{C}$ admits injective envelop.

PROOF:
Let $T'=(T'^0, T'^1)$ be another $\partial$ functor, defined in degrees 0 and $1$. Assume that we are given $f^0$, a natural transformation of $T^0$ into $T'^0$. We will prove that there is only on way to define $f^1 : T^1\rightarrow T'^1$ (and hence $T$ is universal), assuming that $T^1$ is effacable.

Let $0\rightarrow A\rightarrow M\rightarrow A'\rightarrow 0$ be exact sequence in $\mathscr{C}$ where $T^1(A)\rightarrow T^1(M)$ is zero morphism. Consider the following diagram where first row is exact and second is a complex.

$$\xymatrix{ T^0(M) \ar[rr] \ar[dd]_{f^0(M)} & & T^0(A') \ar[rr] \a... ... 0 \\ & & & & & & \\ T'^0(M) \ar[rr] & & T'^0(A') \ar[rr] & & T'^1(A) & & }$$

Since $T^0(A')\rightarrow T^1(A)$ is surjective, $T^1(A) = T^0(A')/Z$, where $Z = Ker(T^0(A')\rightarrow T^1(A) )$. We have the composite map $T^0(A')\rightarrow T'^0(A')\rightarrow T'^1(A)$, denote this by $f$. In order to define map $T^1(A)\rightarrow T'^1(A)$, such that last square of above diagram becomes commutative, the required map must be the following factorization (proving uniqueness):

$$\xymatrix{ T^0(A') \ar[rr]^f \ar[rd] & & T'^1(A) \\ & T^1(A) = T^0(A')/Z \ar[ru] }$$

To prove that such a map can be defined, we only need to check that $f(Z)=0$, which is true since, $Z=Im(T^0(M)\rightarrow T^0(A'))$ and because last row of above commuting diagram is complex. This proves first part of proposition. The second part is essentially contained in next theorem.

QED

Theorem 4   Let $\mathscr{C}$ be an abelian category in which every object admits injective envelop ³, then for every additive covariant functor on $\mathscr{C}$, the satellite functors $S^iF$ ($i\geq 0$) exist and are effacable for each $i>0$. Moreover universal $\partial$ functor $(S^iF)_{i\geq 0}$ is exact if and only if $F$ satisfies following condition: $F$ is semi-exact and for every $P\subset Q\subset R$ in $\mathscr{C}$,

$$Ker(F(Q/P) \rightarrow F(R/P)) \subseteq Im(F(Q)\rightarrow F(Q/P))$$

(it is easy to check that these conditions hold if $F$ is left exact or right exact).

PROOF:
As always, we give construction of $S^1F$ and leave details. Let $A\in \mathscr{C}$ and consider the exact sequence: $0\rightarrow A\rightarrow M\rightarrow A'\rightarrow 0$, where $F(A)\rightarrow F(M)$ is zero morphism. Now define $S^1F(A) := F(A')/Im(F(M)\rightarrow F(A'))$.

QED

Now we are in position to define derived functors. Let $\mathscr{C}$ and $\mathscr{C}'$ be two abelian categories, where $\mathscr{C}$ has enough injectives. For any $A\in \mathscr{C}$, consider its injective resolution $0\rightarrow A\rightarrow I^0\rightarrow I^1\rightarrow \ldots$. Then derived functors of $F$ ⁴ are defined as $R^iF(A) = H^i(F(\mathbb{I}))$, where $\mathbb{I}$ stands for the complex $0\rightarrow I^0\rightarrow I^1\rightarrow \ldots$.

Some observations worth mentioning here are:

• By assuming $R^iF=0$ for $i<0$, we get $\partial$ functor $RF=(R^iF)_{-\infty<i<\infty}$. This functor is cohomology functor.
• By universal property of satellite functors, there is unique morphism of $\partial$ functors $(S^iF)\rightarrow R^iF$ ⁵. This morphism is isomorphism if and only if $F$ is left exact.

Therefore, if we assume that $F$ is left exact, covariant functor, it make sense to identify its derived functors with universal $\partial$ functors. We say an object $A\in \mathscr{C}$ is $F$-acyclic, if for each $i>0$, $R^iF(A)=0$. The next lemma is useful tool to compute derived functors. It states that any class of objects of $\mathscr{C}$, satisfying some conditions, can be used to form resolution, which will give same derived functors, as is done by injective resolutions. Observe that if $0\rightarrow A\rightarrow J^0\rightarrow J^1\rightarrow \ldots$ is an exact sequence, where each $C^i$ are $F$-acyclic, then $H^i(F(\mathbb{J})) = R^iF(A)$. The following lemma essentially gives some properties, which imply that a certain class of objects in $\mathscr{C}$ has only $F$-acyclic objects.

Lemma 3   Let $F$ be covariant functor from an abelian category $\mathscr{C}$ to another abelian category $\mathscr{C}'$, and suppose $\mathscr{C}$ has enough injectives. Let $\mathscr{M}$ be a class of objects of $\mathscr{C}$ satisfying following conditions:

1. For every $A\in \mathscr{C}$, there exists an injective morphism from $A$ to $M$, $M\in \mathscr{M}$.
2. If $A\oplus B$ is in $\mathscr{M}$ then both $A$ and $B$ are in $\mathscr{M}$.
3. For every exact sequence $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$, with $M',M\in \mathscr{M}$, we have $M''\in \mathscr{M}$ and in this case $0\rightarrow F(M')\rightarrow F(M)\rightarrow F(M'')\rightarrow 0$ is exact sequence.

Then every injective object is in $\mathscr{M}$ and every $M\in \mathscr{M}$ is $F$-acyclic.

PROOF:
Let $I$ be some injective object in $\mathscr{C}$. Then by first assumption, there is some $M\in \mathscr{M}$ and an injective morphism $v:I\rightarrow M$. Since $I$ is injective, $v$ admits a section (that is, there is a morphism $u:M\rightarrow I$ such that $uv=1_I$) which identifies $I$ as direct factor of $M$. Then by second assumption $I$ is in $\mathscr{M}$.

Let $M\in \mathscr{M}$ and $0\rightarrow M\rightarrow I^0\rightarrow \ldots$ be injective resolution of $M$. Let $Z^i = Ker(I^i\rightarrow I^{i+1})$. To prove that $R^pF(M)=0$ for each $p$, it suffices to prove that for each $i\geq 0$, the sequence $0\rightarrow F(Z^i)\rightarrow F(I^i)\rightarrow F(Z^{i+1})\rightarrow 0$ is exact and $0\rightarrow F(M)\rightarrow F(I^0)\rightarrow F(Z^1)\rightarrow 0$ is exact. The later sequence is exact since $M,I^0\in \mathscr{M}$ and we invoke third assumption. The first claim is proved by induction on $i$. By virtue of exact sequences $0\rightarrow Z^i\rightarrow I^i\rightarrow Z^{i+1} \rightarrow 0$, we inductively prove that each $Z^i$ is in $\mathscr{M}$. This will suffice because of third assumption and the fact that each $I^i\in \mathscr{M}$. Now $Z^0\cong M$, and $Z^1$ is third term of exact sequence $0\rightarrow M\rightarrow I^0\rightarrow Z^1\rightarrow 0$ and hence is in $\mathscr{M}$. This proves the claim and hence the lemma.

QED