Concentrated complexes determined by cohomology groups? Let $C$ be a chain complex such that $H^0(C)=E$ and $H^i(C)=0$ for all $i\neq 0$. Is it true that $C$ is quasi-isomorphic to $C':\cdots \to 0 \to E \to 0 \to \cdots$ ?
Apparently, $C$ and $C'$ have the same cohomology in each degree. But I did not find a way to construct a morphism between them.
Thanks!
 A: The answer is that there is a zigzag of quasi-isomorphisms between them:
there is a natural morphism $\tau_{\geq 0}C\to C$ where $\tau_{\geq 0}C=(\dots \to C_n\to \dots \to C_1\to \ker(\partial_0)\to 0\to ...)$.
Because of your hypothesis, this natural morphism is a quasi-isomorphism.
Then there is also a natural morphism $\tau_{\geq 0}C\to \tau_{\leq 0}\tau_{\geq 0}C$
where $\tau_{\leq 0}D = (\dots \to 0\to \mathrm{coker}(\partial_1)\to D_{-1}\to \dots \to D_{-n}\to \dots )$
Here, again, because of your hypotheses, this is a quasi-isomorphism. But now $\tau_{\leq 0}\tau_{\geq 0}C \cong H^0(C)[0]$ ($H^0(C)$ concentrated in degree $0$)
Therefore in the end you get a zigzag $C\overset\sim\leftarrow \tau_{\geq 0}C \overset\sim\to H^0(C)[0]$.
However, it is possible that you can't find a morphism $C\to H^0(C)[0]$ which is a quasi-isomorphism, or in the other direction.
Let's take for instance the following example: $C = (0\to \mathbb Z\overset{p^2}\to \mathbb Z\to \mathbb Z/p\to 0)$ where $\mathbb{Z\to Z}/p$ is the canonical projection. Then of course this complex is concentrated in degree $0$, with homology $\mathbb Z/p$.
There is obviously no quasi-isomorphism $\mathbb Z/p\to C$, and I claim that there isn't one in the opposite direction either. Indeed, in the other direction you would need a morphism $\mathbb Z\to \mathbb Z/p$, but this automatically sends $p$ (which generates the homology of $C$) to $0$, so it can't be a quasi-isomorphism.
Conclusion: you really need the zigzag, but if you allow it, then the answer is yes.
