The clause graph is an L-reduction with constants one and one

Take the instance map $f(F):=G_F$ of [F1], which runs in polynomial time and produces a finite simple graph whose independent sets have value ${\mathrm{val}}_{\mathrm{\Gamma }}(I)=\mathrm{\mid }I\mathrm{\mid }$. Take the decoder $g(F,I)$ of [F1], which from every independent set $I$ of $G_F$ produces in polynomial time an assignment of $F$ satisfying at least $\mathrm{\mid }I\mathrm{\mid }$ clauses, of value ${\mathrm{val}}_{\mathrm{\Pi }}(g(F,I))$ equal to its satisfied-clause count. Both objectives are maximization with nonnegative values by [F4].

The first L-reduction inequality holds with $a=1$: by the exact optimum equality of [F1], ${\mathrm{OPT}}_{\mathrm{\Gamma }}(f(F))=\alpha (G_F)={\mathrm{OPT}}_{\mathrm{M}\mathrm{a}\mathrm{x}3\mathrm{S}\mathrm{A}\mathrm{T}}(F)=1\cdot {\mathrm{OPT}}_{\mathrm{\Pi }}(F)$ for every 3-CNF formula $F$ with three literal occurrences per clause.

The second L-reduction inequality holds with $b=1$: writing $t:={\mathrm{val}}_{\mathrm{\Pi }}(g(F,I))$ for the number of clauses satisfied by the decoded assignment, [F1] gives $t\ge \mathrm{\mid }I\mathrm{\mid }$, and therefore $\mathrm{\mid }{\mathrm{OPT}}_{\mathrm{\Pi }}(F)-{\mathrm{val}}_{\mathrm{\Pi }}(g(F,I))\mathrm{\mid }={\mathrm{OPT}}_{\mathrm{M}\mathrm{a}\mathrm{x}3\mathrm{S}\mathrm{A}\mathrm{T}}(F)-t\le {\mathrm{OPT}}_{\mathrm{M}\mathrm{a}\mathrm{x}3\mathrm{S}\mathrm{A}\mathrm{T}}(F)-\mathrm{\mid }I\mathrm{\mid }=\alpha (G_F)-\mathrm{\mid }I\mathrm{\mid }=\mathrm{\mid }{\mathrm{OPT}}_{\mathrm{\Gamma }}(f(F))-{\mathrm{val}}_{\mathrm{\Gamma }}(I)\mathrm{\mid }$, where the last step uses ${\mathrm{val}}_{\mathrm{\Gamma }}(I)=\mathrm{\mid }I\mathrm{\mid }$ and step 2.1.

Steps 2.1 and 3.1 exhibit the maps and constants required by [F2], so $(f,g,1,1)$ is an L-reduction from Max-3SAT to maximum independent set. By the transfer statement [F3], a feasible independent set with relative error at most $ϵ$, that is ${\mathrm{val}}_{\mathrm{\Gamma }}(I)\ge (1-ϵ)\alpha (G_F)$, decodes to a Max-3SAT assignment with relative error at most $1\cdot 1\cdot ϵ=ϵ$; when $\alpha (G_F)=0$ the decoded assignment is optimal.

A concrete formula is $F=(x\vee x\vee x)\wedge (\mathrm{\neg }x\vee \mathrm{\neg }x\vee \mathrm{\neg }x)$ with $m=2$ clauses of three literal occurrences. No assignment satisfies both clauses: $x$ true satisfies only the first and $x$ false satisfies only the second, so ${\mathrm{OPT}}_{\mathrm{M}\mathrm{a}\mathrm{x}3\mathrm{S}\mathrm{A}\mathrm{T}}(F)=1$. The graph $G_F$ has $6$ vertices in two clause clusters of three; an independent set takes at most one vertex per cluster, and every vertex of the first cluster is complementary to every vertex of the second, so no independent set has size $2$, while a single vertex is independent; hence $\alpha (G_F)=1={\mathrm{OPT}}_{\mathrm{M}\mathrm{a}\mathrm{x}3\mathrm{S}\mathrm{A}\mathrm{T}}(F)$, and an independent set of size $1$ decodes to an assignment satisfying at least $1$ clause.

The explicit clause-literal construction therefore is an L-reduction with constants $a=b=1$: the optimum values agree, and every independent set of size $k$ decodes to an assignment satisfying at least $k$ clauses, so errors transfer unchanged and an independent-set approximation ratio carries over to Max-3SAT with the same relative error.