$\iota_{x;0}f(x)$ are the Laurent series expansions at $x=\infty$ and at $x=0$, respectively. In particular, $$\begin{aligned}
& &\iota_{x;\infty}(x-z_{0})^{r}=
\sum_{i\ge 0}{r\choose i}(-z_{0})^{i}x^{r-i};\\
& &\iota_{x;0}(x-z_{0})^{r}=
\sum_{i\ge 0}{r\choose i}(-z_{0})^{r-i}x^{i}\end{aligned}$$ for $r\in {\Z},\;z_{0}\in {\C}^{\times}$. It is clear that $\iota_{x;\infty}$ and $\iota_{x;0}$ are ${\C}[x,x^{-1}]$-linear and one-to-one, and they are algebra homomorphisms. From Definition \[dgood\], for $v\in V,\; \alpha\in {\cal{D}}_{P(z)}(W),\; w\in W$, the formal series $\<\alpha,Y^{o}(v,x)w\>$ lies in the range of $\iota_{x;\infty}$. Then $\iota_{x;\infty}^{-1}\<\alpha,Y^{o}(v,x)w\>$ is well defined. Furthermore, $\<\alpha,Y^{o}(v,x)w\>$ absolutely converges to $\iota_{x;\infty}^{-1}\<\alpha,Y^{o}(v,x)w\>$ in the domain $|x|>|z|$. Using the defined $\iota$-maps we define a new vertex operator map $Y^{R}$.

From the definition we have $$\begin{aligned}
\label{eiota-1three}
\iota_{x;0}^{-1}\<Y^{R}(v,x)\alpha,w\>=\iota_{x;\infty}^{-1}
\<\alpha,Y^{o}(v,x)w\>
=\iota_{x;\infty}^{-1}
\<Y^{*}(v,x)\alpha,w\>.\end{aligned}$$

Let $v\in V,\; \alpha\in {\cal{D}}_{P(z)}(W)$ and let $l,k$ be as in Lemma \[lequivconds\] (b) or (c). Since $x^{l}\iota_{x;\infty}^{-1}\<\alpha,Y^{o}(v,x)w\>$ is analytic at $x=0$, $$\begin{aligned}
\<x^{l}Y^{R}(v,x)\alpha,w\>=\iota_{x;0}(x^{l}\iota_{x;\infty}^{-1}
\<\alpha,Y^{o}(v,x)w\>)\in {\C}[[x]]\end{aligned}$$ for every $w\in W$. Hence $$\begin{aligned}
x^{l}Y^{R}(v,x)\alpha\in W^{*}[[x]].\end{aligned}$$ This proves the following result:

For $v\in V,\; \alpha\in {\cal{D}}_{P(z)}(W)$, $$\begin{aligned}
Y^{R}(v,x)\alpha\in W^{*}((x)).\;\;\;\;\Box\end{aligned}$$

Let $k\in {\N}$ be as in Lemma \[lequivconds\] (b). (Note that we may take the same $k$ for (b), (c) and (d).) Noticing that $$\begin{aligned}
(x-z)^{k}\<\alpha,Y^{o}(v,x)w\>\in {\C}[x,x^{-1}]\end{aligned}$$ and that $\iota_{x;\infty}$ and $\iota_{x;0}$ are ${\C}[x,x^{-1}]$-linear, from (\[e3.13def\]) we get $$\begin{aligned}
(x-z)^{k}\<Y^{R}(v,x)\alpha,w\>
=\iota_{x;0}\iota_{x;\infty}^{-1}(x-z)^{k}\<\alpha,Y^{o}(v,x)w\>
=(x-z)^{k}\<\alpha,Y^{o}(v,x)w\>\end{aligned}$$ for all $w\in W$. Therefore we have proved 