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Let $\mathcal{A}$ be an abelian category and $f, g \colon X^\bullet \to Y^\bullet$ be two morphisms of complexes in $\mathcal{A}$. Suppose that the morphisms between cohomology induced from $f$ and $g$ are identical in all degrees. Then, my question is: can we always construct a homotopy between $f$ and $g$? If we can't, what is the counterexample?

Thank you!

kaede
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1 Answers1

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In the category of abelian groups, we have the following (vertical) chain map between (horizontal) complexes.

\begin{eqnarray*} 0\to \mathbb{Z}/(2)\qquad\\ \downarrow \qquad\,\,\,\,\downarrow 1_{\mathbb{Z}/(2)}\\ \mathbb{Z} \twoheadrightarrow \mathbb{Z}/(2)\qquad \end{eqnarray*}

For any abelian group $C$, the induced maps on cohomology with coefficients in $C$ are trivial. However this chain map is not null-homotopic (in fact any map $\mathbb{Z}/(2)\to \mathbb{Z}$ is trivial).

tkf
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  • Your answer is beautiful. Is the above chain map equivalent to zero in the derived category D(Ab)? – Liang Chen Jan 05 '24 at 08:44
  • @LiangChen I am not sure. Null-homotopic chain maps are equivalent to zero. However the above map is not null-homotopic. More generally, if a chain map factors through an acyclic complex, then it is equivalent to zero in the derived category: If $C$ is acyclic and $f\colon C\to 0,,, g\colon 0\to C$ are the unique chain maps then $fg=1_0$ so $$f^{-1}=f^{-1}1_0=f^{-1}fg=1_Cg$$ and $$0_C=gf=f^{-1}f=1_C.$$ Thus given chain maps $a\colon A\to C,,, b\colon C \to B$ we have $$ba=b1_Ca=b0_Ca=0.$$ – tkf Jan 09 '24 at 06:53
  • Thus $$\begin{array}[ccccccccccccc] {}\cdots&\to &0&\to & 0&\to&0&\to& \mathbb{Z}/(2)&\to &0&\to&\cdots\ &&\downarrow &&\downarrow&&\downarrow &&\downarrow 1_{\mathbb{Z}/(2)}&&\downarrow\ \cdots&\to&0&\to&\mathbb{Z}&\stackrel 2\to&\mathbb{Z}& \twoheadrightarrow &\mathbb{Z}/(2)&\to&0&\to&\cdots \end{array}$$ is equivalent to zero in the derived category, but not null-homotopic. – tkf Jan 09 '24 at 07:24