Is a chain map inducing zero on all homology groups necessarily a boundary?

Question

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 $\Delta_n$?

Answer 1

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.