Let $U$ be a universal algebra of type $T$, and denote $\mathrm{Con}(U)\!=\!\{\text{congruence relations on }U\}$ and $\mathrm{Sub}(U)\!=\!\{\text{subalgebras of }U\}$. Let "$\leq$" mean "subalgebra".
The correspondence theorem (2.6.20, p. 54) says: $\mathrm{Con}(U/\vartheta)\!=\!\{\alpha/\vartheta\,;\, \alpha\!\in\!\mathrm{Con}(U),\vartheta\subseteq\!\alpha\},$ where $\alpha/\vartheta\!=\!\{(a/\vartheta,b/\vartheta)\!\in\!(U/\vartheta)^2;(a,b)\!\in\!\alpha\}$, and also $(\alpha\wedge\beta)/\vartheta=(\alpha/\vartheta)\wedge(\beta/\vartheta)\;\;\text{ and }\;\; (\alpha\vee\beta)/\vartheta=(\alpha/\vartheta)\vee(\beta/\vartheta).$ In particular, for a group $G$ and ring $R$ and module $M$ and algebra $A$, we have $\{\text{normal subgroups of }G/H\}\!=\!\{H'/H;\,H'\!\unlhd\!G,H\!\subseteq\!H'\},$ $\{\text{ideals of }R/I\}\!=\!\{I'/I;\,I'\!\unlhd\!R,I\!\subseteq\!I'\},$ $\{\text{submodules of }M/N\}\!=\!\{N'/N;\,N'\!\leq\!M,N\!\subseteq\!N'\},$ $\{\text{algebra ideals of }A/I\}\!=\!\{I'/I;\,I'\!\unlhd\!A,I\!\subseteq\!I'\}.$ But we know that we also have $\{\text{subgroups of }G/H\}\!=\!\{G'/H;\,G'\!\leq\!G,H\!\subseteq\!G'\},$ $\{\text{subrings of }R/I\}\!=\!\{R'/I;\,R'\!\leq\!R,I\!\subseteq\!R'\},$ $\{\text{submodules of }M/N\}\!=\!\{M'/N;\,M'\!\leq\!M,N\!\subseteq\!M'\},$ $\{\text{subalgebras of }A/I\}\!=\!\{A'/I;\,A'\!\leq\!A,I\!\subseteq\!A'\}.$
Question: Is there some nice correspondence between $\mathrm{Sub}(U/\vartheta)$ and $\mathrm{Sub}(U)$?