The rational Yang–Baxter equation lifts to RTT

In [F2] take $x=u$, $y=v$, $z=0$ and label the three factors $1=a$, $2=b$, $3=n$. For each site $n$ this gives $R_{ab}(u-v)R_{an}(u)R_{bn}(v)=R_{bn}(v)R_{an}(u)R_{ab}(u-v)\mathrm{.}$ This identity acts on $V_a\otimes V_b\otimes V_n$ and as the identity on every other site.

Write $R_{ab}=R_{ab}(u-v)$, $R_{an}=R_{an}(u)$ and $R_{bn}=R_{bn}(v)$. For $1\le n\le N$ set $T_a^{(n)}(u):=R_{an}(u)\cdots R_{a1}(u)$ and $T_b^{(n)}(v):=R_{bn}(v)\cdots R_{b1}(v)$. The case $n=1$ of the desired identity is step 1.1. Suppose it holds for $n-1$. Operators $T_a^{(n-1)}$ commute with $R_{bn}(v)$, and $R_{an}(u)$ commutes with $T_b^{(n-1)}$, because their auxiliary and physical tensor factors are disjoint. Therefore $\begin{array}{rl}{\displaystyle R_{ab}{T}_a^{(n)}{T}_b^{(n)}}& {\displaystyle =R_{ab}{R}_{an}{R}_{bn}{T}_a^{(n-1)}{T}_b^{(n-1)}}\\ {\displaystyle }& {\displaystyle =R_{bn}{R}_{an}{R}_{ab}{T}_a^{(n-1)}{T}_b^{(n-1)}}\\ {\displaystyle }& {\displaystyle =R_{bn}{R}_{an}{T}_b^{(n-1)}{T}_a^{(n-1)}{R}_{ab}}\\ {\displaystyle }& {\displaystyle =T_b^{(n)}{T}_a^{(n)}{R}_{ab},}\end{array}$ where the second equality is step 1.1 at site $n$, the third is the induction hypothesis, and the final equality uses the other disjoint-site commutation. The argument holds for each $n$ up to $N$, proving the claimed RTT identity.

Taking $n=N$ in step 2.1 gives the stated monodromy relation with $T_a^{(N)}=T_a$ and $T_b^{(N)}=T_b$. No invertibility assumption on $R_{ab}(u-v)$ is used, so the identity also holds at its singular spectral differences.