This question is about the extension of subcopulas to copulas, shown  in Sklar, A. (1996), "Random variables, distribution functions, and copulas: A personal look backward and forward." Institute of Mathematical Statistics Lecture Notes-Monograph Series, 28, 1–14. [106, 133]. A proof for the 2-D case is in Nelsen (2006) "An introduction to copulas" (see p.16-17 [here][1]). 
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**Let me first review the result** For simplicity, let $n=2$.  
Let $\mathcal{S}\equiv \mathcal{S}_1\times \mathcal{S}_2  \subseteq [0,1]^2$, where  each $\mathcal{S}_i$ is such that $\{0,1\}\in \mathcal{S}_i$ for each $i=1,2$. 

A  $2$-D subcopula is a function $\bar{C}:\mathcal{S}\rightarrow [0,1]$ such that:

**1.** $\bar{C}$ is non decreasing. That is, the volume under $\bar{C}$ of each $2$-D box whose vertices are elements of $\mathcal{S}$ is $\geq 0$. 

- For example,   take $(a_1,b_1)\in \mathcal{S}$  and $(a_2,b_2)\in \mathcal{S}$ with $a_1\leq a_2$ and $b_1\leq b_2$. These 2 elements of $\mathcal{S}$ form the box with 4 vertices  $$(a_1,b_1)\in \mathcal{S}, (a_2,b_1)\in \mathcal{S}, (a_1,b_2)\in \mathcal{S}, (a_2,b_2)\in \mathcal{S}$$  Condition 1 requires that $$\bar{C}(a_1,b_1)+\bar{C}(a_2,b_2)-\bar{C}(a_1,b_2)-\bar{C}(a_2,b_1)\geq 0$$

- As another example, take $(0,0)\in \mathcal{S}$ and $(a,b)\in  \mathcal{S}$. These 2 elements of $\mathcal{S}$ form the box with 4 vertices  $$(0,0)\in \mathcal{S}, (a,0)\in \mathcal{S}, (0,b)\in \mathcal{S}, (a,b)\in \mathcal{S}$$  Condition 1 requires that $$\bar{C}(a,b)+\bar{C}(0,0)-\bar{C}(a,0)-\bar{C}(0,b)\underbrace{=}_{\text{See condition 2 below}} C(a,b)\geq 0$$

**2.** $\bar{C}(u) = 0$ for any $u \in \mathcal{S}$ that has at least one component equal to 0.

**3.** $\bar{C}(u) = u_i$ for any $u \in  \mathcal{S}$ that has all components, except the $i$-th, equal to 1.

A $2$-D copula  is a $2$-D subcopula for which $\mathcal{S}=[0,1]^2$.

"Extension Lemma": Let $\bar{C}:\mathcal{S}\rightarrow [0,1]$ be a $2$-D subcopula with domain $\mathcal{S}$. Then, there exists a   proper $2$-D copula $C$ such that $C(u) = \bar{C}(u)$ for all $u\in \mathcal{S}$.
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**My question** is about whether some specific constraints can be enforced while constructing   a copula $C$  extending a subcopula $\bar{C}$. 

More precisely, let us consider a subcopula $\bar{C}: \underbrace{\mathcal{S}_1\times \mathcal{S}_2}_{\equiv \mathcal{S}}\rightarrow[0,1]$. Let us consider the simple setting where $\mathcal{S}_1$ and $\mathcal{S}_2$ are **finite**.

Consider the  box with vertices $(a,b),(c,b), (a,d), (c,d)$, which we call $B$.

Assume that **at least one** of the following four conditions holds: $a\notin \mathcal{S}_1$; $b\notin \mathcal{S}_2$; $c\notin \mathcal{S}_1$; $d\notin \mathcal{S}_2$. That is, the vertices of $B$ are not all elements of $\mathcal{S}$.

Further, assume that if $B$ contains any box $D$ whose 4 vertices    are all elements of $\mathcal{S}$, then the volume of $D$ under $\bar{C}$ is equal to 0.

Can we construct a copula $C$ that extends $\bar{C}$ and such that 
$$
C(a,b)+C(c,d)-C(a,d)-C(c,b)=0 \quad ?
$$


I've tried to tweak the traditional proof of the above "Extension Lemma" to achieve my result (see p.16-17 [here][1] with $n=2$), but I haven't been successful. The proof of the lemma is based on a bilinear interpolation procedure which makes it hard to introduce variations. Do you have suggestions on how to proceed?


  [1]: https://rady.ucsd.edu/faculty/directory/valkanov/pub/classes/mfe/docs/copula_ch2.pdf