Exactness and derived functors

 Lecture topic

Let $0\to X\to Y\to Z\to 0$ be a short exact sequence of objects in a category $A$. Let $\mathcal F:A\to B$ be a covariant functor.

Definition: The functor $\mathcal F$ is right-exact if $\mathcal F(X)\to\mathcal F(Y)\to \mathcal F(Z)\to 0$ is an exact sequence. It is left-exact if $0\to \mathcal F(X)\to\mathcal F(Y)\to \mathcal F(Z)$ is an exact sequence. It is exact if it is both left- and right-exact.

Example: These are some examples of left- and right-exact functors:
    $\Hom_A(X,-)$ is covariant left-exact
    $\Hom_A(-,X)$ is contravariant left-exact
    $-\otimes_R X$ is covariant right-exact, for $X$ a left $R$-module

Recall that $X\otimes_R Y$ is naturally isomorphic to $Y\otimes_RX$.

Definition: An object $X\in \Obj(A)$ is projective if $\Hom_A(X,-)$ is an exact functor. Similarly, $X$ is injective if $\Hom_A(-,X)$ is an exact functor.

Recall that a projective resolution of an object $X$ is a sequence of projective objects $\cdots\to P_2\to P_1\to P_0$ that may or may not terminate on the left. The homology of the sequence in degree 0 is $X$, and trivial in other degrees. Similarly, an injective resolution of $X$ is a sequence of injective objects $I_0\to I_1\to I_2\to\cdots$ that may or may not terminate on the right. The cohomology is also concentrated in degree 0, and is $X$ there. A free resolution is a projective resolution where all the objects are free (whatever that means in the context).

These types of resolutions may not exist. A category "has enough injectives (projectives)" means we can always construct injective (projective) resolutions.

Definition: Let $\mathcal F:A\to B$ be a covariant right-exact functor and $\mathcal G:A\to B$ a covariant left-exact functor. Let $X\in \Obj(A)$ with $P_\bullet$ a projective resolution of $X$ and $I_\bullet$ an injective resolution of $X$. The $i$th left-derived functor of $\mathcal F$ is $L_i\mathcal F(X) = H_i(\mathcal F(P_\bullet))$. The $i$th right-derived functor of $\mathcal G$ is $R^i\mathcal G(X) = H^i(\mathcal G(I_\bullet))$.

These objects of $B$ are well-defined up to natural isomorphism. Note that $\mathcal F^{op}:A^{op}\to B^{op}$ is a contravariant right-exact functor. Moreover, if $\mathcal F$ was contravariant right-exact and $\mathcal G$ was contravariant left-exact, then $L_i\mathcal F(X)=H_i(\mathcal F(I_\bullet))$ and $R^i\mathcal G(X)=H^i(\mathcal G(P_\bullet))$.

Example: Let $R$ be a ring with $X$ and $Y$ both $R$-bimodules. Then
$$\begin{align*} \Tor_i^R(Y,X) & =  L_i(-\otimes_RX)(Y) & \Ext^i_R(X,Y) & = R^i(\Hom_R(X,-))(Y) \\ & = L_i(Y\otimes_R - )(X), && = R^i(\Hom_R(-,Y))(X). \end{align*}$$
Recall that $\Tor_i^R(Y,X)$ is canonically isomorphic to $\Tor_i^R(X,Y)$, but it is not true for $\Ext$. Also note that $\Hom_R(X,-)$ is covariant and $\Hom_R(-,Y)$ is contravariant, while $-\otimes_R X$ and $Y\otimes_R -$ are both covariant functors.
References: Weibel (An introduction to homological algebra, Chapter 2)