2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.