Minimal injective resolution of a module 
Let $R$ be a commutative Noetherian ring and $M$ an $R$-module. Let $0\rightarrow M\rightarrow E^{\bullet}$ be a minimal injective resolution of $M$ and $0\rightarrow M\rightarrow I^{\bullet}$ be an arbitrary injective resolution. Then how to construct an injective chain map from $E^{\bullet}$ to $I^{\bullet}$?  

If we have
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
0 &\ \ra{} & M & \ra{} & E^0& \ra{f_1} & E^1 & \ra{f_2} & E^2& \\
 & & \da{=} & & \da{g_0} & & \da{g_1} & & \da{g_2} & &  \\
0 &\ \ra{} & M & \ra{} & I^0 & \ra{h_1} & I^1 & \ra{h_2} & I^2 & \ra{h_3} & I^3 \\
\end{array}
$$
where $g_0$, $g_1$ and $g_2$ are all injections. If we can show $E^2/Im(f_2)$ can be embedded in $I^3$, then we can find $E^3\subseteq I^3$ to continue this diagram, but  how to show  $E^2/Im(f_2)$ can be embedded in $I^3$?

It is well known that we can define some chain maps $u^{\bullet}:E^{\bullet}\to I^{\bullet}$ and $v^{\bullet}:I^{\bullet}\to E^{\bullet}$ which are identity on $M$. Thus we get a chain map $w^{\bullet}:E^{\bullet}\to E^{\bullet}$ which is identity on $M$, where $w^{\bullet}=v^{\bullet}u^{\bullet}$. Now using that $E^{\bullet}$ is minimal it's easy to prove (inductively) that $w^{\bullet}$ is a chain isomorphism and therefore $u^{\bullet}$ is injective. 
