Tensoring is right exact

The map $g\otimes 1$ is surjective: by [L8], every tensor is a finite sum of elementary tensors $c\otimes n$, and [L1] supplies $b\in B$ with $g(b)=c$, so $c\otimes n=(g\otimes 1)(b\otimes n)$.

Functoriality gives $(g\otimes 1)(f\otimes 1)=(gf)\otimes 1=0$, so $\mathrm{im}(f\otimes 1)\subseteq \ker (g\otimes 1)$.

Let $Q=(B{\otimes }_{R}N)\mathrm{/}\mathrm{im}(f\otimes 1)$. The image is a submodule because [L2] and [L7] make $f\otimes 1$ $R$-linear. Since $g\otimes 1$ kills it, [L6] gives $\bar{g}:Q\to C{\otimes }_{R}N$.

For $c\in C$ and $n\in N$, choose any $b\in B$ with $g(b)=c$ and set $q(c,n)=[b\otimes n]\in Q$. If $b^{\mathrm{\prime }}$ is another lift, then $b-b^{\mathrm{\prime }}\in \ker g=\mathrm{im}f$ by [L1], so $[b\otimes n]=[b^{\mathrm{\prime }}\otimes n]$; hence $q$ is well defined.

The function $q:C\times N\to Q$ is bilinear: lifts of sums may be taken as sums of lifts, scalar multiples as scalar multiples, and the tensor relations give the required equalities. By [L3] and [L4] it induces $h:C{\otimes }_{R}N\to Q$.

For $b\otimes n$, one has $h\bar{g}([b\otimes n])=h(g(b)\otimes n)=[b\otimes n]$, while for $c\otimes n$ and a lift $b$ one has $\bar{g}h(c\otimes n)=\bar{g}([b\otimes n])=c\otimes n$; generators and uniqueness make $h$ and $\bar{g}$ inverse.

Since $\bar{g}$ is an isomorphism, the kernel of $g\otimes 1$ is exactly the submodule quotiented out in step 2.1, namely $\mathrm{im}(f\otimes 1)$. Together with step 1.1, this is right exactness.