Copulas, implication Let $C$ be a copula function. Prove that $C(t,1-t)=0$ for all $t\in[0,1]$ implies that $C(u,v)=\max(u+v-1,0)$.
I think the implication other way around is easy to see, however I can't see why the "upper diagonal" part of the copula function could not be some type of a different function with $C(u,1)=u$ and $C(1,v)=v$.  
See the image below - the leftmost plot is the Frechet-Hoeffding lower bound. I need to prove that $C$ is equal to that.

 A: the condition on C implies $U \ge 1-V$.  Since $U$ and $1-V$ are both uniform equality must hold, and $U = 1-V$.   See also frechet-hoeffding lower bound.
A: You're saying the probability of being below the diagonal $u+v=1$ is zero.
Now consider a point $(u,v)$ exactly on the diagonal, so that $u+v=1$.  What probability is assigned to the closed triangle with vertices $(u,v)$, $(1,v)$, and $(1,0)$?  By the definition of copula conjoined with the fact that the probability below the diagonal is $0$, it must be $1-u$.  And the probability assigned to that triangle must also be equal to the length of its projection onto the $v$-axis, and that length is also $1-u$.  But the probability of falling within that part of the $u$-axis is equal to the probability assigned to the closed trapezoid with vertices $(u,v)=(u,1-u)$, $(1,0$), $(u,1)$, and $(1,1)$.  That means the probability assigned to the rectangle with vertices $(u,v)=(u,1-u)$, $(1,v)$, $(u,1)$ and $(1,1)$ must be zero!  This holds for every value of $u$, thus for all such rectangles.
Conclusion: The probability of being above the diagonal must be $0$.
That means all the probability is concentrated on the diagonal $u+v=1$.  And there's only one distribution on the diagonal that gives the right marginal distributions to satisfy the definition of a copula.
The way I thought of this was by thinking first of the discrete case, the set of pairs $(u,v)$ of integers in the set $\{1,\ldots,n\}$.  Just do the $2\times2$ and $3\times3$ cases and you see quickly why all the probability must be concentrated on that one diagonal, and must be uniformly distributed there.
A: Here's a rigorous proof:
Suppose $u + v - 1 \leq 0$. Then $v \leq 1-u$, so we must have
$$
0 = C(u,v) \leq C(u,1-u) = 0
$$
Now suppose $u + v - 1 > 0$.
Construct the rectangle $A = [u,1] \times [v,1]$.
Pick $z \in [1-v,u]$ and also construct the rectangle $B = [z, 1] \times [1-z, 1]$.
The volume of $B$ is given by
\begin{align*}
V_C(B) &= C(1,1) - C(z,1) - C(1,1-z) + C(z,1-z) \\
&= 1 - z - (1-z) \\
&= 0
\end{align*}
Since $A \subseteq B$, this implies $V_C(A) = 0$.
We therefore have
\begin{align*}
0 &= V_C(A)\\
& = C(1,1) - C(u,1) - C(1,v) + C(u,v) \\
&=  1 - u - v +  C(u,v) \\
\Rightarrow \qquad  C(u,v) &= u + v - 1
\end{align*}
Combining these results, we can conclude that
$$
C(u,v) = \max(0,u + v -1)
$$
