The elements in the bounded derived category of a hereditary category I am looking for a proof for the following statement.

Let $\mathcal A$ be a hereditary category, and $D^b(\mathcal A)$ be its bounded derived category. Then for any $M \in D^b(\mathcal A)$, we have
  $$M \cong \oplus_i H^iM[-i].$$

Most of the proofs I have found are similar. For example, 2.2 in this document.
The author writes an exact sequence
$$0 \rightarrow X^{n-1} \rightarrow E^n \rightarrow H^nX \rightarrow 0.$$
But what is this $E^n$, and why there is an exact sequence like this? Then how is the second commutative diagram obtained from the first?
I think the main problem for my understanding this proof is the definition of $E^n$. Maybe it is related to $\operatorname{Ext}$'s, but I need to know exactly what it is.
Thanks very much.
 A: I'll try to fill in some details for the proof from the link you provided. Here $X^*$ is a complex of objects of $\mathcal{A}$, with differential $d^*$. 
Recall the long exact sequence for the $\operatorname{Ext}$ spaces: starting from the short exact sequence 
$$0\to \ker d^{n-1} \to X^{n-1} \to \operatorname{im} d^{n-1} \to 0 $$
and applying the functor $\operatorname{Hom}(H^nX,-)$, you get an exact sequence
$$ \operatorname{Ext}^1(H^nX,\ker d^{n-1})\to \operatorname{Ext}^1(H^nX, X^{n-1}) \stackrel{\lambda}{\to} \operatorname{Ext}^1(H^nX,\operatorname{im} d^{n-1}) \to \operatorname{Ext}^2(H^nX,\ker d^{n-1}). $$
Since the category $\mathcal{A}$ is hereditary, the $\operatorname{Ext}^2$ space vanishes, so $\lambda$ is surjective.
Now, elements of the $\operatorname{Ext}^1$ spaces correspond to short exact sequences.  Unwinding the definition of $\lambda$, it being surjective precisely means that there is a short exact sequence from $X^{n-1}$ to $H^nX$ and a commutative diagram
\begin{array}{ccccccccc}
0 & \xrightarrow{} & X^{n-1} & \xrightarrow{} & E^n & \xrightarrow{} & H^nX & \xrightarrow{} & 0\\
 &  & \downarrow &  & \downarrow &  & \downarrow{=} & & \\
0 & \xrightarrow{} & \operatorname{im}d^{n-1} & \xrightarrow{} & \ker d^{n-1} & \xrightarrow{} & H^nX & \xrightarrow{} & 0.
\end{array}
(Simply view the lower exact sequence as an element of $\operatorname{Ext}^1(H^nX,\operatorname{im} d^{n-1})$, and the upper exact sequence is then a preimage of the lower one by $\lambda$.)
From there, the proof in the paper only uses the morphisms constructed above.

If you are mainly interested in the case where $\mathcal{A}$ has enough projectives, then you might be interested in Happel's original proof, which can be found in Lemma 4.1 of this paper.
