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I am currently reading Joseph Rotman's book "An Introduction to Homological Algebra" (2nd edition) and I'm struggling with the definition of a bounded filtration given there. According to the book,

A filtration $(F^pM)$ of a graded module $M = (M_n)$ is $\textbf{bounded}$ if, for each $n$, there exist integers $s = s(n)$ and $t = t(n)$ such that $$F^sM_n = \{0\} \quad \quad \text{and} \quad \quad F^tM_n = M_n.$$

I am not having any problem with the definition as such, but I'm confused by the following paragraph:

If $\{F^p\}$ is a bounded filtration of a complex $\textbf{C}$, then the induced filtration on homology is also bounded, and with the same bounds. More precisely, we know that if $i^p: F^p \rightarrow \textbf{C}$ is the inclusion, then $\Phi^pH_n = \text{im }i^p_*$, where $i^p_*: H_n(F^p) \rightarrow H_n(\textbf{C})$. Since $F^s = 0$ and $F^t = \textbf{C}$, we have $\Phi^sH_n = \{0\}$ and $\Phi^tH_n = H_n$.

How do we know that $F^s= 0$ and $F^t = \textbf{C}$? By definition, $s$ and $t$ depend on $n$, so I can't see any reason, why there should exist $s$ and $t$ such that $F^sM_n = \{0\}, F^tM_n = C_n$ for $\textbf{all}$ $n$.

I thought at first that this might be a mistake by the author, but this cannot be found in the errata and moreover, he uses this later in a proof.

You can find the section of the book I am refering to on http://www.math.unam.mx/javier/homologica/rotman.pdf (p. 638 of the PDF file, p .628 of the book).

Thank you very much for your help!

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    I think he means $t=t(n)$ etc throughout, is that interpretation acceptable ? – 2017-02-05
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    Thank you very much for your reply! I'm sorry, but I'm not sure if I understand what you say: Do you mean that $t$ does not depend on $n$? Wouldn't then the right wording be "there exist integers $s$ and $t$, such that for each $n$ ..."? – 2017-02-06
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    I mean he is jut omitting the $n$, assuming the dependence as given. – 2017-02-06
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    Yes, I agree with you! But I can't see how he can assume that for some $t$, $F^t = \textbf{C}$ (i.e. that there is a $t$ such that $(F^t\textbf{C})_n = C_n$ for all $n$. – 2017-02-06
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    He doesnt $t=t(n)$. – 2017-02-06
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    I think that he does: He says: "Since $F^s = 0$ and $F^t = \textbf{C}$, we have $\Phi^sH_n = \{0\}$ and $\Phi^tH_n = H_n$." – 2017-02-06
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    I agree with Rene Schipperus (or maybe this is a typo mistake from the author), $F^t=\mathbf{C}$ should be read $F^{t(n)}=\mathbf{C}_n$. Note that this notion is important for the theorem 10.14 just below, and in this theorem, the author insists clearly that the bounds $s(n)$ and $t(n)$ depends on the homological degree $n$. – 2017-02-07
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    Thank you very much for your reply, Roland! If he $\textit{does}$ assume that $s(n)$ and $t(n)$ depend on $n$, then I can't see why the induced filtration on homology should be bounded with the same upper bound $t(n)$. Why should $F^{t(n)}M_n = M_n$ imply that $\Phi^{t(n)}H_n = H_n$? It is clear that $\Phi^{t(n)}H_n = H_n$ also depends on the values of $F^{t(n)}M_{n-1}$ and $F^{t(n)}M_{n+1}$, but I think that we know nothing about these values. Or is there something I am failing to see? This is the main thing I'm worrying about. – 2017-02-08
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    @Martin Why do you say it depends on $F^{t(n)}M_{n-1}$ and $F^{t(n)}M_{n+1}$ ? $\Phi^pH_n$ is the set of classes $[x]\in H_n$ which have a representative $x\in F^p$. So it is clear that, if $F^pM_n=M_n$ then $\Phi^pH_n=H_n$. – 2017-02-09
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    Oh, finally I get it! In fact, it's quite simple: If $F^{t(n)}M_n = M_n$, then the inclusion $i^{t(n)}: F^{t(n)} \rightarrow \textbf{C}$ gives us the identity $i^{t(n)}_n: M_n \rightarrow M_n$. Now $i^{t(n)}$ induces a map in homology $i^{t(n)}_{n*}: H_n(F^{t(n)}) \rightarrow H_n(\textbf{C})$ given by $[z] \mapsto [i^{t(n)}_nz] = [z]$. (The induced map comes from Proposition 6.8, Rotman.) Hence for any $[z] \in H_n(\textbf{C})$, $[z] = i^{t(n)}_{n*}[z] \in \text{im}(i^{t(n)}_{n*})$, so $\Phi^{t(n)}H_n = H_n$. – 2017-02-09
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    I was actually so confused because I wanted to compute $\Phi^{t(n)}H_n$ by computing $H_n(F^{t(n)})$ using the definition of homology applied to the complex $... \rightarrow F^{t(n)}_{n+1} \rightarrow F^{t(n)}_n = M_{n} \rightarrow F^{t(n)}_{n-1} \rightarrow ...$. Therefore I assumed that we would need that $F^{t(n)}_{n+1} = M_{n+1}$ and $F^{t(n)}_{n-1} = M_{n-1}$ as well! Thank you very much for your help and sorry for the stupid question! – 2017-02-09
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    It was also confusing that throughout the book $\textbf{C}$ has always denoted a complex and $C_n$ an object in the complex, so I thought that $F^{t} = \textbf{C}$ is an equality of complexes. I will ask the publisher to include this in the errata list of the book. – 2017-02-09
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    Well, be carefull that the map $H_n(F^{t(n)})\rightarrow H_n(C)$ is not an isomorphism in general, even if $F^{t(n)}=C_n$. But it is clearly onto, so yes on have $\Phi^{t(n)}H_n=H_n$. I agree that the statement $F^t=\mathbf{C}$ is confusing and probably needs to be corrected in the errata. – 2017-02-10
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    When you say that $H_n(F^{t(n)}) \rightarrow H_n(\textbf{C})$ is not an isomorphism (even if $F^{t(n)}_n = C_n$) in general, do you mean that (i) there may not be an isomorphism when $F^{t(n)}$ is an arbitrary complex or do you mean that (ii) there may not be an isomorphism when $F^{t(n)}$ is a subcomplex of $\textbf{C}$? I agree with (i) because we need an injective chain map from $F^{t(n)}$ to $\textbf{C}$ but I think that we have just proved that there is always an isomorphism in case (ii). – 2017-02-10
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    No even if $F^{t(n)}$ is a subcomplex. What you proved is that $H_n(F^{t(n)})\rightarrow H_n$ is onto, and this is what is needed, remember that the filtration in $H_n$ is the image of $H_n(F^p)\rightarrow H_n$. But $H_n(F^{t(n)})$ is the set of class $[x]$, $x\in F^{t(n)}\cap\ker d_n$ up to a finer equivalence : namely $[x]=[y]$ in $H_n(F^{t(n)})$ iff $x-y$ is in the image of an element in $F^{t(n)}M_{n-1}$, whereas in $H_n$, $[x]=[y]$ iff $x-y$ is the image of an element in $M_{n-1}$. But it just mean the map $H(F^{t(n)})\rightarrow H_n$ is onto. – 2017-02-10
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    Just an example, take the complex $C=F^0C=[0\rightarrow A\rightarrow A\rightarrow 0]$ and the subcomplex $F^1C=[0\rightarrow 0\rightarrow A\rightarrow 0]$ ($A$ is in degree 0). Clearly $H_0(C)=0$ but $H_0(F^1)=A$ even if $F^1C_0=C_0$. – 2017-02-10
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    Thank you very much, Roland! Now it's completely clear to me! In fact I wanted to prove something that I didn't need to prove (and which is also impossible to prove, as your example shows!). This also explains very well why we define $\Phi^{t(n)}H_n$ to be the image of the induced map and don't let it simply be $H_n(F^{t(n)})$ - which might be "bigger" than $H_n(\textbf{C})$! Thank you very much again for the enormous amount of help! – 2017-02-10
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    You are welcome. Glad I helped – 2017-02-10

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