# Why is $F_{n,0}=H_n(K)$ for an arbitrary filtered complex?

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.

• 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 $F_{n,0}=H_n(K)$. – iwriteonbananas Jul 5 '16 at 12:51
• 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 $H_n(K)$. – user062295 Jul 5 '16 at 13:38
• 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 – iwriteonbananas Jul 5 '16 at 15:03
• 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 $C_*(X_{n+i})=K_{n+i+1}$? – user062295 Jul 5 '16 at 15:14
• 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. – user062295 Jul 5 '16 at 17:31

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.