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.
Best Answer
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.