Let $\mathcal{A}$ be an abelian category with enough projectives (injectives). I tried to prove that if every element $M$ of $\mathcal{D}^{b}(\mathcal{A})$ satisfies $$ M \cong \bigoplus_{i} H^{i}(M)[-i]$$ then the category must be hereditary. I think my guess is wrong but I want to find the mistake.
Let $M,N$ be objects of $\mathcal{A}$ so they can be identified by complexes on degree 0, then $\operatorname{Hom}_{\mathcal{D}(\mathcal{A})}(M,N[2])=\operatorname{Ext}^2_{\mathcal{A}}(M,N)$.
Every $f \in \operatorname{Hom}_{\mathcal{D}(\mathcal{A})}(M,N[2])$ can be represented by a class of equivalence of roofs of the form $L^{\bullet} \xrightarrow{q} M, L^{\bullet} \xrightarrow{f} N[2]$ where $q$ is a quasi-isomorphism and $L^{\bullet}$ is a chain complex. Since $L^{\bullet}$ is quasi-isomorphic to $M$ then $L^{\bullet}$ must be an element of the bounded derived category hence $$L^{\bullet} \cong \bigoplus_{i} H^{i}(L^{\bullet})[-i] \cong H^0 (L^{\bullet})$$
The last term is a chain complex on degree $0$. So is it true that then every element of $\operatorname{Hom}_{\mathcal{D}(\mathcal{A})}(M,N[2])$ can be represented by the following roof? $$H^0 (L^{\bullet}) \xrightarrow{q'} M, \enspace H^0 (L^{\bullet}) \xrightarrow{f} N[2]$$
In this case $f$ must be zero because the only non zero term in $N[2]$ is in degree $-2$. I don't believe this argument because it leads me to think that $Ext^1$ must also vanish.
What went wrong?
This doesn't work because you only know that $L^{\bullet} \cong \bigoplus_{i} H^{i}(L^{\bullet})[-i] \cong H^0 (L^{\bullet})$ in the derived category. That is, the isomorphism $L^\bullet\to H^0(L^\bullet)$ may not be given by a map of chain complexes, but only by a roof $K^\bullet\to L^\bullet$, $K^\bullet\to H^0(L^\bullet)$ of quasiisomorphisms. So you can't necessarily turn your map $f:L^\bullet\to N[2]$ into a map $H^0(L^\bullet)\to N[2]$, because the map $K^\bullet\to H^0(L^\bullet)$ is going the wrong way to compose.