counterexample for $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence I want to understand a counterexample for: 
Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence,
because the statement should be wrong. 
I want to understand, why the short exact sequence $$0\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}\to 0, $$where $\mathbb{Z} \rightarrow \mathbb{Z}$ is the multiplication by 2 and $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$ is the projection, 
gives a counterexample. 
Clearly the exact sequence above can be considered as an acyclic chain complex, i.e. all homology-groups of this complex are zero. But what is $f_*$  and why is this a counterexample, can we elaborate it? 
Best.
 A: Take $C_*=0\to \mathbb{Z}\stackrel{2}\to\mathbb{Z}\to 0$ (with the $\mathbb{Z}$s in degree $0$ and $1$ and every other term in the complex $0$) and let $D_*=0\to\mathbb{Z}/2\mathbb{Z}\to 0$ (with the $\mathbb{Z}/2\mathbb{Z}$ in degree $0$ and every other term in the complex $0$).  Then there is a unique nonzero chain map $f:C_*\to D_*$, and this induces an isomorphism on homology (this is essentially equivalent to the statement that the original sequence you were given is acyclic).  But there is no nonzero chain map $g:D_*\to C_*$, so $f$ cannot have a chain homotopy inverse.
A: Seeing that no one has explained why the complex you have provided is a counterexample, maybe I should mention that you have $C_{\bullet}=0 \mapsto \mathbb{Z} \mapsto \mathbb{Z} \mapsto \mathbb{Z}/2 \mapsto 0$ is exact, so its homology is zero in all degrees. Thus, you have a chain map from $C_{\bullet}$ to the chain complex with zero in each degree, where you take all maps to be 0 maps. If you try to find a chain homotopy inverse, which factors through 0, then there is only one possibility for the chain maps, i.e. they need to be the zero maps.  If you suppose that there exists a chain homotopy, such that the zero map is chain homotopic to the identity, you need to express $id: \mathbb{Z} \mapsto \mathbb{Z}$ as $id(x) = 2\theta(x)+ \underbrace{0\circ \pi}_{0}=2\theta(x).$ Pluggin in $x=1$, we have $1=2\theta(x)$, so $\theta(x)=\frac{1}{2}$, which is not possible as that is not a $\mathbb{Z}$-module homomorphism. 
