Given the category $\mathbf{Chain}_A$ of chain complexes over an abelian group $A$.

The $n$-th Homology functor is: $$H_n:\mathbf{Chain}_A\to\mathbf{Ab}$$

Clearly it is additive: $$H_n(\varphi+\varphi')=H_n(\varphi)+H_n(\varphi')$$

There are special chain maps: $$\Delta_n:C_n\to D_{n+1}:\quad\vartheta_n:=\partial^D_{n+1}\Delta_n+\Delta_{n-1}\partial^C_n$$ on which the functor vanishes: $$H_n(\vartheta)=0\quad(\forall n)$$

Does the converse hold? In other words, if a chain map induces zero on all homology groups, is it necessarily of the form $\vartheta$ for some sequence $\Delta_n$?


So rephrased in simpler words, you are asking if a chain map which induces zero on homology groups is necessarily a boundary. The answer is no.

Consider the two chain complexes $C_*$ and $D_*$ given respectively by

  • $C_1 = C_0 = \mathbb{Z}$, $C_i = 0$ for $i \neq 0,1$ and $d_1 : C_1 \to C_0$ is given by $d_1(x) = 2x$;
  • $D_1 = \mathbb{Z}$, $D_i = 0$ if $i \neq 1$, all the differentials are zero.

Let also $\varphi : C_* \to D_*$ be given by $\varphi_1(x) = x$. In other words we are looking at: $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z} @>{d_1 = 2 \cdot}>> \mathbb{Z} @>>> 0 \\ @. @V{\varphi_1 = \operatorname{id}}VV @VVV \\ 0 @>>> \mathbb{Z} @>>> 0 @>>> 0 \end{CD}$$

Then clearly $H_0(C) = \mathbb{Z}/2\mathbb{Z}$, $H_1(C) = 0$, whereas $H_0(D) = 0$ and $H_1(D) = \mathbb{Z}$; besides $H_i(C) = H_i(D) = 0$ for $i \neq 0,1$. So of course have $H_i(\varphi) = 0$ for all $i$.

However, $\varphi$ is not a boundary. Otherwise there should exists a map $h : C_0 = \mathbb{Z} \to D_1 = \mathbb{Z}$ such that $$x = \varphi(x) = h(dx) = h(2x) = 2 h(x)$$ for all $x$. In particular, $\varphi(1) = 1 = 2 h(1)$, which is impossible.

  • $\begingroup$ Small remark: I assume you meant $d_1(x)=x+x$ and $h(x+x)=h(x)+h(x)$ since we're working in $\mathbf{Ab}$ respectively identified $\mathbf{Ab}$ with $\mathbb{Z}-\mathbf{Mod}$. $\endgroup$ – C-Star-W-Star Dec 9 '16 at 15:53
  • $\begingroup$ If $n \in \mathbb{Z}$ and $x$ is an element in some $A$-module, then $nx$ is always well-defined. Indeed $2x = x + x$, and since a morphism of $A$-modules is in particular a morphism of abelian groups, then $h(2x) = h(x+x) = h(x) + h(x) = 2h(x)$. Here I work in $\mathbb{Z}$-modules but $2x$ would be equally well-defined over any $A$. $\endgroup$ – Najib Idrissi Dec 9 '16 at 15:55
  • $\begingroup$ Yeah, but I particularly considered chain-complexes over merely abelian groups instead of modules. Mostly, I did so in order to unravel the structure behind homology theory. Moreover, I did so as morphisms between abelian groups are, only a priori, much less restrictive than morphisms between modules. So the existence of such a boundary may fail simply due to the stronger requirement on the morphism. However, same time there may, in principle, also occur much more examples of morphisms vanishing in homology. Finally, you're example is really nice as it covers both cases. Thank you again!!! :) $\endgroup$ – C-Star-W-Star Dec 9 '16 at 16:21

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