Finding example of quasi isomorphism that has no quasi inverse Between differential graded algebra $V,W$, a chain map $f\colon V\to W$ induces homomorphism between its homology. If this becomes an isomorphism between the homology of $V,W$, call this quasi isomorphism. My goal is to find quasi isomorphism such that there is no quasi inverse, i.e. chain map $g\colon W\to V$ that induces isomorphism on their homologies.
There is an easy example using the following chain complexes
$A :0\to\mathbb{Z}\to\mathbb{Z}\to0$ (map given by multiplication by $2$)
$B : 0 \to 0 \to \mathbb{Z}/2\mathbb{Z} \to 0$
I can give a chain map from $A$ to $B$, the obvious one, which induces isomorphism on the homology so is an quasi isomorphism. But there is no such map from $B$ to $A$ since $\mathbb{Z}/2\mathbb{Z}$ has torsion.  and I can give algebra structure here, though stupid structure it is (e.g. defining everything to be zero), but it works. So there exists counterexample.
But I want to find counterexample using vector spaces, and it is harder than general algebra (allowing any $R$-module). Would somebody help me with this? Thank you in advance.
 A: There are no examples in the category of vector spaces, for an elementary proof of this argue that any complex
$$\cdots \to A_i \to A_{i + 1} \to \cdots$$
is quasi-isomorphic (with quasi-inverse) to the complex
$$\cdots \to H^i(A_\bullet) \overset{0}{\to} H^{i + 1}(A_\bullet) \to \cdots$$
This is done in two steps, first argue that every complex can be written in the form
$$\cdots \to I_i \oplus K_i \oplus I_{i + 1} \to I_{i + 1} \oplus K_{i + 1} \oplus I_{i + 2} \to \cdots$$
where the map sends $I_{i + 1}$ to itself and kills $I_i$ and $K_i$.  Next write down the obvious chain maps to and from the complex
$$\cdots \to K_i \overset{0}{\to} K_{i + 1} \to \cdots$$
and check that they are indeed quasi-inverse to each other.
A: Here's a more technological argument, using model categories.
Give the category of bounded chain complexes over a field the projective model structure. Every chain complex is fibrant, and since we're over a field, they are all cofibrant (i.e. free). The "Whitehead theorem" for model categories gives what you want: a weak equivalence (i.e. a quasi-iso) between any pair of complexes is a homotopy equivalence.
