\rightarrow V$ by $R(x\oplus y)=S(x)+L(y)\ (x\in X,y\in Y)$. Then $R$ is a linear map on $E$ that extends $T$.

Finally, it is readily checked that $T$ extends uniquely to $E$ in the case that $W$ is generating.
\end{proof}

As well known, the Riesz decomposition property is employed in the proof of the classical Riesz-Kantorovich formulas \cite[Theorem~1.59]{AT}. Likewise, we will make use of a suitable Riesz decomposition property for multi-wedged spaces.

\begin{definition}\label{D:RDP}
Given $\alpha,\beta\in\N\cup\{\aleph_{0}\}$, we say that a multi-wedged space $(E,\W)$ has the $(\alpha,\beta)$-\textit{Riesz decomposition property} if for any $m<\alpha+1,n<\beta+1$, any collection $(W_{j})_{j\in J}$ of wedges in $\W$ with $|J|=n$, and any two collections $(x_{i})_{i\in I}$ and $(y_{j})_{j\in J}$ of vectors in $E$ such that $|I|=m$, $x_{i}\in\sum_{j\in J}W_{j}\ (i\in I)$, $y_{j}\in W_{j}\ (j\in J)$, and
\[
\sum_{i\in I}x_{i}=\sum_{j\in J}y_{i},
\]
there exist $z_{ij}\in W_{j}\ (i\in I,j\in J)$ such that
\[
x_{i}=\sum\limits_{j\in J}z_{ij}\ (i\in I)\ \text{and}\ y_{j}=\sum\limits_{i\in I}z_{ij}\ (j\in J).
\]
\end{definition}

It is known that every vector lattice has the classical Riesz decomposition property (see e.g.\ \cite[Theorem 1.20]{AB}), which coincides with the $(\aleph_{0},\aleph_{0})$-Riesz decomposition property on a vector lattice \cite[Theorem~1.54]{AT}. However, not every multi-lattice exhibits even the $(2,2)$-Riesz decomposition property, as the following simple example illustrates.

\begin{example}\label{E:noRDP}
Let $(E,\W)$ be the Dedekind complete multi-lattice with $E=\R^{2}$ and $\W=\{W_{1},W_{2}\}$, where $W_{1}=\{(x,y)\in\R^{2}:x,y\geq 0\}$ and $W_{2}=\{(x,x)\in\R^{2}:x\geq 0\}$. Then $(E,\{W_{1},W_{2}\})$ does not possess the $(2,2)$-Riesz decomposition property.
\end{example}

Indeed, it is readily checked using Proposition~\ref{P:geoint} that $(E,\{W_{1},W_{2}\})$ is a Dedekind complete multi-lattice. In order to verify that $(E,\{W_{1},W_{2}\})$ does not have the $(2,2)$-Riesz decomposition property, let $y_{1}=(1,0)$ and $y_{2}=(1,1)$. Then $y_{1}\in W_{1},y_{2}\in W_{2}$, and $y_{1}+y_{2}=(2,1)$. Next let $x_{1}=(2,0)$ and $x_{2}=(0,1)$. Then $x_{1},x_{2}\in W_{1}+W_{2}=W_{1}$ and $x_{1}+x_{2}=(2,1)$. Let $(z_{11},z_{11}'), (z_{21},z_{21}')\in W_{1}$ and $(z_{12},z_{12}), (z_{22},z_{22})\in W_{2}$, so that $z_{11},z_{11}',z_{21},z_{21}',z_{12},z_{22}\geq 0$. Suppose that $(z_{11},z_{11}'), (z_{21},z_{21}'),(z_{12},z_{12})$, and $(z_{22},z_{22})$ satisfy the following:
\begin{itemize}
\item[(1)] $(z_{11},z'_{11})+(z_{21},z_{21}')=(1,0)$,
\item[(2)] $(z_{12},z_{12})+(z_{22},z_{22})=(1,1)$,
\item[(3)] $(z_{11},z_{11}')+(z_{12},z_{12})=(2,0)$,
\item[(4)] $(z_{21},z_{21}')+(z_{22},z_{22})=(0,1)$.
\end{itemize}
From (1) we have $z_{11}'=z_{21}'=0$. That $z_{12}=0$ now follows from (3). From (3) again, we get $z_{11}=2$, and hence, from (1), we have $z_{21}=-1$, a contradiction.

For contrast, Example~\ref{E:F(S)} furnishes examples of multi-lattices with the $(\aleph_{0},\aleph_{0})$-Riesz decomposition property.

With the Riesz decomposition property for multi-wedged spaces introduced, we are almost ready to present the main result of this paper. We need the following lemma.

\begin{lemma}\label{L:P}
Let $(F,V)$ be a preordered vector space, and let $P_{\D(V)}$ be a projection onto $\D(V)$, so that $P_{U}=I-P_{\D(V)}$ is a projection onto an algebraic complement $U$ of $\D(V)$.
\begin{itemize}
\item[(1)] For $x\in E$ we have $x\sim_{V}P_{U}(x)$.
\item[(2)] If $A,B\subseteq F$ are nonempty and $A\sim_{V}B$ then $P_{U}(A)$ and $P_{U}(B)$ are singleton sets and also $P_{U}(A)=P_{U}(B)$.
\end{itemize}
\end{lemma}

\begin{proof}
(1) Let $x\in E$. Then $x-P_{U}(x)=(I-P_{U})(x)=P_{\D(V)}(x)\in\D(V)$.\\
(2) Let $A,B\subseteq F$ be nonempty sets such that $A\sim_{V}B$. Let $a\in A,b\in B$. Then $a\sim_{V}b$, so $a-b\in\D(V)$. Thus $P_{U}(a-b)=0$, and so $P_{U}(a)=P_{U}(b)$. This proves that $P_{U}(A)=P_{U}(B)$. Finally, note that if $a,a'\in A$ then for all $b\in B$ we have $P_{U}(a)=P_{U}(b)=P_{U}(a')$. Therefore, $P_{U}(A)$ is a singleton set.
\end{proof}

In Theorem~\ref{T:main} below, we consider singleton subsets of $F$ to be elements of $F$. 

\begin{theorem}\label{T:main}
Let $(E,\W)$ be a multi-wedged space, and let $(F,\{V\})$ be a Dedekind complete multi-lattice. Let $P_{\D(V)}$ be a projection onto $\D(V)$, so that $P_{U}=I-P_{\D(V)}$ is a projection onto an algebraic complement $U$ of $\D(V)$. Let $\kappa\in\N\cup\{\aleph_{0}\}$, and suppose $(E,\W)$ has the $(2,\kappa)$-Riesz decomposition property. The following hold.
\begin{itemize}
\item[(1)] $\bigl(\L(E,F),\L_{\W,V}(E,F)\bigr)$ is a $\kappa$-multi-lattice.
\item[(2)] If $\kappa=\aleph_{0}$ then $\big(\L(E,F),\L_{\W,V}(E,F)\big)$ is Dedekind complete.
\item[(3)] Let $\bigl(T_{i},\L_{W_{i},V}(E,F)\bigr)_{i\in I}$ in $\L(E,F)\times\L_{\W,V}(E,F)$ be multi-bounded above and suppose $|I|<\kappa+1$. For $R\in\L(E,F)$, we have $R\in\underset