1
$\begingroup$

Let $... \subset K_{-1}=0 \subset K_0\subset ...K_n \subset...$ be an arbitrary filtered chain complex with $colim_n K_n:=K$.

Let $F_{p,p+q}=im(H_{p+q}(K_p) \to H_{p+q}(K))$

Mosher and Tangora decided to write down on page 67 that $F_{n,0}:H_n(K_n)=H_n(K)$.

First this is not even true for the filtered chain complex $K_n= C_*(X_n)$ with $X_n$ the $n-$ skeleton of a CW complex $X$; the correct statement in this case is that the induced map $H_n(K_{n+1})=H_n(K)$ is an iso. Moreover, this is not true in general because I can construct a stupid filtered complex s.t. $K_i=K_{i+1}=K_{i+....}$ however many times I would like.

What is the correct statement? I am familiar with spectral sequences in that I have done many computations with them. One of them is on my stackexchange.

$\endgroup$
  • $\begingroup$ I can't find what you said on p. 67. $F_{p,q}=\operatorname{im}\left( H_{p+q}(K^p)\to H_{p+q}(K) \right)$, right? Then certainly $F_{n,0}=H_n(K)$. $\endgroup$ – iwriteonbananas Jul 5 '16 at 12:51
  • $\begingroup$ Thanks. It is where mosher and tangora says: "Therefore the following series is finite: $H_n(K) = F_{n,0} \supset F_{n-1,1} \supset ...F_{1,n-1}\supset F_{0,n}$." Now about what you said, $F_{n,0}=im(H_n(K_n) \to H_n(K))$. I don't see why this is $H_n(K)$. $\endgroup$ – user062295 Jul 5 '16 at 13:38
  • $\begingroup$ Just look at the LES of the pair $(K,K_n)$. The $n$-th relative homology is zero, so $H_n(K_n)\to H_n(K)$ is surjective $\endgroup$ – iwriteonbananas Jul 5 '16 at 15:03
  • $\begingroup$ if $K_n=C_*(X_n)$ then what you say is correct. Why is it true if you let $K_1=C_*(X_1)....K_{n-1}=C_*(X_{n-1})$, $K_n=C_*(X_{n-1})$, and $C_*(X_{n+i})=K_{n+i+1}$? $\endgroup$ – user062295 Jul 5 '16 at 15:14
  • $\begingroup$ Oh there is the #2 convergence criterion which says that $H_{p+q}(X_p,X_{p-1})=0$ for $p<0$. This criterion doesn't hold for the contrived sequence above. $\endgroup$ – user062295 Jul 5 '16 at 17:31
0
$\begingroup$

One of the convergence criterion for the spectral sequence of a filtered complex was that $E^1_{p,q}=H_{p+q}(X_p,X_{p-1})=0$ for $q<0$.

$H_p(K,K_{p})= H_p(colim_n K_n,K_p)=colim_n H_p(K_n, K_p)=0$. By LES of triple $(X_p,X_{p+n-1},X_{p+n})$ and the convergence criterion, we have the induction step showing that $colim_n H_p(K_n, K_p)=0$ for all $n>0$.

Therefore the LES for $(K,K_p)$ shows that $F_{p,0}=im (H_p(K_p) \to H_p(K))=H_p(K)$ is surjective.

Note: What I said about CW chain complexes is not correct as my answer here shows. edit:I got p and q mixed up.
I fixed it

$\endgroup$

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.