Derivation of survival copula $$ \begin{align}F(x,y) &= P(X\leq x, Y\leq y) \\
 &= C(F(x), F(y)) \end{align}$$
The copula of a bivariate distribution is equal to its CDF, $C(u,v) = F(x,y)$. The density of the data's CDF therefore can be expressed as
$$f(x,y) = c(u,v) \times f_X(x) f_Y(y)$$
where $c(u,v)$ is the copula density and $f_X(x)$ is a marginal.
If $\hat{C}(u,v)$ is the survival copula for determining
$$ \begin{align}\bar{F}(x,y) &= P(X> x, Y > y) \\
 &= \hat{C}(1-F(x), 1-F(y)) \end{align}$$
how is the first equation of the ordinary copula, $C(u,v)$, linked to the survival copula as
$$\hat{C}(u,v) = u + v - 1 + C(1-u,1-v)$$
are $u$ and $v$ merely vectors containing the (ranked) observations of the uniform marginals?
A derivation in integrals and probabilities (rather than $F$s) would be good.
 A: Here is a purely algebraic and entirely rigorous demonstration.
As a matter of notation, when $\mathcal A\subset\Omega,$ the indicator function of $\mathcal A$ is defined by
$$\mathscr{I}_{\mathcal{A}}(\omega)=\left\{\begin{aligned}1 & \quad\omega\in \mathcal{A}\\0&\quad\text{otherwise.}\end{aligned}\right.$$
Indicator functions connect probabilities and expectations because when $\Omega$ is a probability space (with probability measure $\mathbb P$) and $\mathcal A$ is measurable (which for our purposes merely means it has a probability),
$$\Pr(\mathcal A) = \int_{\Omega} \mathscr{I}_{\mathcal{A}}\,\mathrm{d}\mathbb{P}.\tag{*}$$
Indicator functions also enjoy intimate, simple relationships with set operations.  In particular,

*

*$\mathscr{I}_{\Omega\setminus \mathcal{A}} = \mathscr{I}_{\mathcal{A}^\prime} =  1 - \mathscr{I}_{\mathcal A}.$ Set complements correspond to numerical complements (subtracting from $1$).


*$\mathscr{I}_{\mathcal{A} \cap \mathcal{B}} =\mathscr{I}_{\mathcal A}\mathscr{I}_{\mathcal B}.$ Intersection corresponds to multiplication.
The question concerns $(X,Y),$ a random variable with continuous marginals $F_X$ and $F_Y.$  There are two corresponding random variables in the copula, $(U,V),$ defined by $U=F_X(X)$ and $V=F_Y(Y).$  To study the distribution of $(X,Y)$ we will, as always,  consider two arbitrary numbers $x$ and $y.$  Letting $F_X(x)=u$ and $F_Y(y)=v,$ we see $u$ and $v$ are also numbers between $0$ and $1;$ that is, $(u,v)\in [0,1]^2.$ The events conventionally associated with these numbers are $$\mathcal{A} = X\le x = U \le u$$ and $$\mathcal{B} = Y\le y = V \le v.$$
With these preliminaries in place, let's turn to the question.  You ask about the probability of the event $U\gt u,V\gt v.$  This is the intersection of the complements, $ \mathcal{A}^\prime \cap \mathcal{B}^\prime.$  Employing first $(1)$ to deal with the complements and then $(2)$ to deal with the intersection we obtain
$$\mathscr{I}_{\mathcal{A}^\prime \cap \mathcal{B}^\prime} = \left(1 - \mathscr{I}_{\mathcal{A}}\right)\left(1 - \mathscr{I}_{\mathcal{B}}\right).$$
The rest is so easy as to be trivial: the only work we need do is (1) algebraically expand this product and (2) apply linearity of expectation:
$$\begin{aligned}
\hat C(u,v) &= \Pr(U\gt u,V\gt v)\\& = \Pr\left(\mathcal{A}^\prime \cap \mathcal{B}^\prime\right)\\
&= \iint_{[0,1]^2} \mathscr{I}_{\mathcal{A}^\prime \cap \mathcal{B}^\prime}(u,v)\,\mathrm{d}u\mathrm{d}v\\
&= \iint_{[0,1]^2} \left( 1 - \mathscr{I}_{\mathcal{A}} - \mathscr{I}_{\mathcal{B}} + \mathscr{I}_{\mathcal{A}}\mathscr{I}_{\mathcal{B}} \right)(u,v)\,\mathrm{d}u\mathrm{d}v\\
&= \iint_{[0,1]^2} 1\,\mathrm{d}u\mathrm{d}v - \iint_{[0,1]^2}\mathscr{I}_{\mathcal{A}}(u,v)\,\mathrm{d}u\mathrm{d}v \\&\quad\quad\quad - \iint_{[0,1]^2}\mathscr{I}_{\mathcal{B}}(u,v)\,\mathrm{d}u\mathrm{d}v + \iint_{[0,1]^2}\mathscr{I}_{\mathcal{A}}\mathscr{I}_{\mathcal{B}}(u,v)\,\mathrm{d}u\mathrm{d}v\\
&= 1 - \Pr(U\le u) - \Pr(V\le v) + \Pr(U\le u, V\le v)\\
&= 1 - C(u,1) - C(1,v) + C(u,v)\\
&= 1 - u - v + C(u,v).
\end{aligned}$$
The first appearances of the survival copula $\hat C$ and copula $C$ are their definitions.  By construction, $$C(u,1)=\Pr(U\le u)=\Pr(F_X(X)\le F_X(x)) =\Pr(X\le x)=F_X(x)=u$$ and likewise $C(1,v)=v,$ thereby justifying the final equality.
An equivalent statement can be obtained by changing $u$ to $1-u$ and $v$ to $1-v$ (both of which remain between $0$ and $1,$ as required), yielding
$$\begin{aligned}
\hat C(1-u,1-v) &= 1 - (1-u) - (1-v) + C(1-u,1-v) \\&= u+v-1 + C(1-u,1-v).\end{aligned}$$
That seems to be what was intended by the question.
You might find it instructive to compare this to the section "Intuition from Algebraic Manipulation" at https://stats.stackexchange.com/a/43075/919 and then compare it to the visualization under "Intuition from Geometry" in the same post: it's the same underlying idea.
