Ext can be computed from any injective resolution of the second variable

Choose $a_N:I^{\bullet }(N)\to J^{\bullet }(N)$ extending $1_N$. Its reverse comparison is a homotopy inverse: both composites extend the identity, so [F2] compares them to the identity cochain maps.

Applying $\mathrm{Hom}(M,-)$ carries a homotopy $a-b=dh+hd$ to the homotopy $f\mapsto h\circ f$. Hence $H^n\mathrm{Hom}(M,a_N)$ is an isomorphism independent of $a_N$. The definition Ext via an injective resolution of the second variable identifies its source with ${\mathrm{Ext}}_I^n(M,N)$, which The balanced Ext bifunctor identifies with balanced Ext.

For $u:N\to N^{\mathrm{\prime }}$, choose comparison maps $I(u)$ and $J(u)$ extending $u$. Define their actions on cohomology by postcomposition. Independence follows from [F2]; identity and composition laws follow because comparison composites extend the corresponding object composites. Moreover $J(u)a_N$ and $a_{N^{\mathrm{\prime }}}I(u)$ both extend $u$, so [F2] makes them homotopic. Applying Hom and cohomology gives precisely the naturality square in $N$.

For $v:M^{\mathrm{\prime }}\to M$, precomposition by $v$ commutes exactly with postcomposition by $a_N$. This proves contravariant naturality in $M$ and therefore naturality in both variables. The same construction at a single $N$ uses only the individual resolution $J^{\bullet }(N)$, proving the final assertion without a global choice of comparison maps.