# The spectral sequence for hyper-derived functors

Let $F$ be a right exact functor between two abelian categories $A$ and $B$.Suppose that $C_\bullet$ is a complex in $A$,then there is a convergent spectral sequence

$$E_{p,q}^2 = ({L_p}F)({H_q}({C_ \bullet }) \Rightarrow {\mathbb{L}_{p + q}}F(C).$$

According to Weibel's book An introduction to homological algebra,p148,the proof says that we just take a Cartan-Eilenberg resolution $P_{\bullet,\bullet}$ of $C_\bullet$ and consider the double complex $F(P_{\bullet,\bullet})$.There is a spectral sequence for a double complex

$$E_{p,q}^2 = H_p^vH_q^h({C_ \bullet }) \Rightarrow {H_{p + q}}({\rm{Tot}}(C)),$$

where $H_p^v$ is the homology for vertical direction and $H_q^h$ is horizontal.

My question is how do I get the first spectral sequence from the second spectral sequence above?I have no idea why $H_p^vH_q^h(F({P_{ \bullet , \bullet }}))$ is $({L_p}F)({H_q}({C_ \bullet })$.A rough guess is that $H_q^h$ commutes with $F$,then by the definition of Cartan-Eilenberg resolution $H_q^h(P)$ is a projective resolution of $H_q(C)$ and we done.But it is so weird that $H_q^h$ can commute with $F$ .So it still get me trouble.

## 1 Answer

The horizontal complexes of a projective Cartan-Eilemberg resolution are split complexes (since the boudaries and cycles are projectives) therefore the fundamental exact sequences splits, it implies that the homology functor commutes with addtives functors.