Goursat's Lemma: Let $A$ and $B$ be groups. There is a bijective correspondence between the subgroups of $A\times B$ and the set of quintuples $(G_1,G_2,H_1,H_2,\varphi)$ where $G_2\unlhd G_1\leq A$, $H_2\unlhd H_1\leq B$ and $\varphi: G_1/G_2\to H_1/H_2$ is an isomorphism.
So assuming that $U \leq A\times B$ and letting $G_1 = \pi_A(U)$, $H_1 = \pi_B(U)$, $ G_2 = i_A(U)$ and $H_2 = i_B(U)$, where $i_X: X \to A\times B$ for $X = A, B$, the inclusion map, we obtain that $U$ determines the quintuples
Conversely, given a quintuple $(G_1,G_2,H_1,H_2,\varphi)$ where $G_2\unlhd G_1\leq A$, $H_2\unlhd H_1\leq B$ and $\varphi: G_1/G_2\to H_1/H_2$ is an isomorphism, how can I determine the converse of this bijective correspondence stated in Goursat's Lemma?
The corresponding subgroup is $\{ (x,y) : x \in G_1, y \in H_1, y \in \phi(x)H_2 \}$.