There is a single morphism from 1S to ⊥ for every world in S, so 1S◃C means all of these morphism factor through C.
A morphism from C to ⊥ is basically a column of C and a morphism from 1S to C is basically an row in C, all of whose entries are in S, and these compose to the morphism corresponding to the entry where this column meets this row.
Thus 1S◃C if and only if when you delete all rows not entirely in S, the resulting matrix has image S.
I think this equivalent to what you said. I just wrote it out myself because that was the easiest way for me to verify what you said.
Yep.
There is a single morphism from 1S to ⊥ for every world in S, so 1S◃C means all of these morphism factor through C.
A morphism from C to ⊥ is basically a column of C and a morphism from 1S to C is basically an row in C, all of whose entries are in S, and these compose to the morphism corresponding to the entry where this column meets this row.
Thus 1S◃C if and only if when you delete all rows not entirely in S, the resulting matrix has image S.
I think this equivalent to what you said. I just wrote it out myself because that was the easiest way for me to verify what you said.
Thanks! (The way you phrased the conclusion is also much clearer/cleaner than how I phrased it)