Let $\mathcal A$ be an Abelian category (I am happy to assume that $\mathcal A\cong \mathrm{Mod}(R)$ for some ring $R$ if that helps). Given the following long exact sequence in $\mathcal A$ $$ 0\to F_2\to T_1\to X\to F_1\to T_0\to 0, $$ can we find a triangle $U\to X[1]\to V\to$ in the unbounded derived category $\mathbf{D}(\mathrm{Mod}(R))$ such that the following conditions hold true?
- $H^{-1}U=T_1$, $H^0U=T_0$ and $H^nU=0$ for all $n\neq 0,1$;
- $H^{-2}V=F_2$, $H^{-1}V=F_1$ and $H^mV=0$ for all $m\neq -1,-2$.
Of course that is easy to do if either $F_2=0$ or $T_0=0$.
Let $k$ be a field, and $R=k[x]/(x^2)$.
There is an exact sequence of $R$-modules $$0\to k\to R\to R\to R\to k\to0$$ (where $k$ denotes the one-dimensional $R$-module on which $x$ acts as zero).
Suppose a triangle of the required form exists. The first non-zero cohomology, in degree $-1$, of $U$ is $R$, which is injective as an $R$-module, and so splits off. The rest of the cohomology determines that $U\cong R[1]\oplus k$.
Since $\text{Hom}(k,R[1])=0$, the map $U\to R[1]$ in the triangle must be zero on restriction to the summand isomorphic to $k$, and so the cone $V$ has a summand isomorphic to $k[1]$, which is incompatible with the required cohomology of $V$.