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Suppose I have (semi-infinite) chain complexes $$ \cdots \rightarrow A_i \rightarrow A_{i+1}\rightarrow \cdots$$ $$ \cdots \rightarrow B_i \rightarrow B_{i+1}\rightarrow \cdots$$ over an additive category, and $A_i = B_i = 0$ for $i>0$. Suppose that $f:A\rightarrow B$ is a chain map such that, for every $k\geq 0$, $f$ is chain homotopic to a chain map which is zero in degrees $-k,\ldots, -1,0$.

Is such an $f$ always null-homotopic?

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    $\begingroup$ My feeling is that the answer is 'no' since thay would be a kind of 'phantom map'. I haven't been able to find an example, though, but I'd love to see one! $\endgroup$ Oct 27, 2014 at 11:12
  • $\begingroup$ By duality, the situation is equivalent to the situation of two cochain-complexes, zero in negative degrees, and a cochain main which is for each $k\geq 0$ chain homotopic to a chain map which is zero in degrees $0,\ldots,k$. That in turn is just equivalent to $f$ being null-homotopic when restricted to the complex truncated at $k$-th level. Now suppose we had a CW-complex of infinite dimension whose cellular chain complex is not contractible but that of each $k$-skeleton is. Then we would have a counterexample. However, I don't know whether such CW-complexes exist. $\endgroup$
    – B K
    Oct 29, 2014 at 14:59
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    $\begingroup$ If every skeleton of a CW-complex $X$ is contractible, then the map from $X$ to a point is a quasi-isomorphism and thus a homotopy equivalence by Whitehead's theorem so $X$ is contractible. $\endgroup$ Oct 29, 2014 at 15:34

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Probably there's a much less artificial example, and even more probably there's a much simpler one, but ...

Let $A=k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field of characteristic two (purely so I don't need to bother about signs).

Let $\mathcal{A}$ be the category whose objects are pairs $(M,N)$ of $A$-modules and whose maps $\alpha:(M,N)\to(M',N')$ are matrices $\begin{pmatrix}\alpha_1&0\\\alpha_2&\alpha_3\end{pmatrix}$ where $\alpha_1:M\to M'$, $\alpha_2:M\to N'$, $\alpha_3:N\to N'$ are $A$-module homomorphisms.

Let $\mathcal{A}_0$ be the (non-full) subcategory with the same objects, but with only those maps $\alpha$ where $\alpha_2=0$.

For $k\geq0$, let $X(k)$ be the chain complex $$\dots\to 0\to (A,0)\to (A,0)\to\dots\to(A,0)\to(A,0)\to0\to\dots$$ with $(A,0)$ in degrees $k+1,\dots,0$, where all the non-trivial differentials are $\begin{pmatrix}\varepsilon&0\\ 0&0\end{pmatrix}$, and let $Y(k)$ be the complex $$\dots\to 0\to (0,A)\to (0,A)\to\dots\to(0,A)\to0\to0\to\dots$$ with $(0,A)$ in degrees $k+1,\dots,1$, where all the non-trivial differentials are $\begin{pmatrix}0&0\\ 0&\varepsilon\end{pmatrix}$.

Consider the chain map (over $\mathcal{A}$) $f(k):X(k)\to Y(k)$ given by $\begin{pmatrix}0&0\\ \varepsilon&0\end{pmatrix}$ in degree $k+1$ and zero in other degrees. This is null-homotopic (for example, take the homotopy given by $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ in degrees $k,\dots,0$), but there's no choice of contracting homotopy with degree zero component in $\mathcal{A}_0$.

Now let $\mathcal{A}^{\mathbb{N}}$ be the category of $\mathbb{N}$-graded objects of $\mathcal{A}$, and consider the chain map $$(f(k))_{k\in\mathbb{N}}:(X(k))_{k\in\mathbb{N}}\to (Y(k))_{k\in\mathbb{N}}.$$

As a map of chain complexes over $\mathcal{A}^{\mathbb{N}}$ this is null-homotopic.

However, if you regard it a map of chain complexes over the (non-full) subcategory of $\mathcal{A}^{\mathbb{N}}$ with the same objects but where maps are required to have grade $d$ components in $\mathcal{A}_0$ for all but finitely many $d$, then it isn't null-homotopic but (using contracting homotopies for finitely many $f(k)$) is homotopic to a map that is zero in any given finite set of degrees.

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    $\begingroup$ On several occasions I've posted answers on MO with the feeling that they're absurdly over-complicated, and that I'll be embarrassed when somebody gives a way more elementary answer. This time the feeling is stronger than ever, but if anybody thinks of a simpler example, then please do embarrass me! $\endgroup$ Oct 29, 2014 at 19:22
  • $\begingroup$ Indeed, I hope this question is not too stupid: how did you come up with this construction? $\endgroup$
    – Aaron Chan
    Nov 2, 2014 at 18:33
  • $\begingroup$ @AaronChan Well, I was familiar with an example, for any $n$, of a map between complexes of $A$-modules (where $A$ is as in my answer) which is not null-homotopic but is homotopic to a map that is only non-zero in degree $0$ and another map that is only non-zero in degree $n$, and thought it might be relevant. After building a couple of layers of complexity onto it to fix the ways in which it didn't answer the question, I came up with this example. I've not thought much about the question since I posted this answer, but I still think there's probably a much simpler or more natural example. $\endgroup$ Nov 2, 2014 at 19:16

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