Is a chain map inducing zero on all homology groups necessarily a boundary? 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 $\Delta_n$?
 A: 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.
