$\text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N)$ Let $(M_i)_{i\in I}$ be a collection of $R$-moduls. Show that for all $N\in \text{Ob}(_R\text{Mod})$ is 
$$
\text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N).
$$
My idea is to make use of the universal property of the direct sum (note that $N\rightarrow I^{\bullet}$ is an injective resolution of $N$):
$$\begin{eqnarray*}
\text{Ext}_R^n(\bigoplus_{i\in I}M_i,N) & = &\mathcal{H}^n(\text{Hom}_R(\bigoplus_{i\in I}, I^{\bullet}))\\
& \cong & \mathcal{H}^n(\prod_{i\in I}\text{Hom}_R(M_i, I^{\bullet}))\\
& \cong & \prod_{i\in I}\mathcal{H}^n(\text{Hom}_R(M_i,I^{\bullet}))\\
& = & \prod_{i\in I}\text{Ext}_R^n(M_i,N)
\end{eqnarray*}$$
Is this correct? I'm not sure if & why $\mathcal{H}^n(\prod_{i\in I}\text{Hom}_R(M_i, I^{\bullet})) \cong  \prod_{i\in I}\mathcal{H}^n(\text{Hom}_R(M_i,I^{\bullet}))$ holds....
 A: The proof is correct. If $(C_i)_{i\in I}$ is any family of cochain complexes, then by writing out the definitions you immediately see $H^n(\prod C_i)=\prod H^n(C_i)$.
For if $D$ denotes the differential on $\prod C_i$, $d_i$ the differential on $C_i$, then
$H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$
A: You can use an injective resolution for $N$: let $E$ be injective and $0\to N\to E\to E/N\to 0$ be exact. Then the long exact sequence
$$\DeclareMathOperator{\E}{Ext}\DeclareMathOperator{\H}{Hom}
0\to
\H_R(\bigoplus_{i\in I}M_i,N)\to
\H_R(\bigoplus_{i\in I}M_i,E)\to
\H_R(\bigoplus_{i\in I}M_i,E/N)\to\\
\E_R^1(\bigoplus_{i\in I}M_i,N)\to
\E_R^1(\bigoplus_{i\in I}M_i,E)=0\to
\E_R^1(\bigoplus_{i\in I}M_i,E/N)\to\\
\E_R^2(\bigoplus_{i\in I}M_i,N)\to
\E_R^2(\bigoplus_{i\in I}M_i,E)=0\to
\E_R^2(\bigoplus_{i\in I}M_i,E/N)\to
\dots
$$
allows to say that
$$
\E_R^1(\bigoplus_{i\in I}M_i,N)\cong
\prod_{i\in I}\E_R^1(M_i,N)
$$
by the universal properties of direct products and sums and by exactness of products. Then we can use dimension shifting:
$$
\prod_{i\in I}\E_R^{n}(M_i,E/N)\cong
\E_R^{n}(\bigoplus_{i\in I}M_i,E/N)\cong
\E_R^{n+1}(\bigoplus_{i\in I}M_i,N)
$$
(the first isomorphism because $N$ is arbitrary). The fact that the isomorphism is the “canonical” one is easy to check.
