The crystal of the level-one Fock space is the good-node partition graph

The crystal basis. By [F1] the $A_0$-span $F_0$ is stable under the normalized string operators, and the mod-$q$ residues of the partition basis are the crystal elements; the string lengths give ${\epsilon }_i,{\phi }_i$ as the surviving minus/plus counts, and the weight formula is [F2]. Since each fixed-residue block is a finite tensor cube of two-vertex strings, all crystal axioms (the intertwining of $\epsilon ,\phi$ with the weight, the partial-inverse property of $e_i,f_i$, and the finiteness of the strings) hold blockwise, hence on the whole graph; the graph has all partitions as vertices because every partition lies in the raw $A_0$-span.

The empty component consists of regular partitions. Use decreasing-content order and the survivor rule. If removing the selected last surviving minus creates $e$ equal rows of length $L-1$, then below its removable row there is a block of $e-1$ rows of length $L-1$ whose last corner is a consecutive same-residue minus: if the upper minus survives, so does this lower one, so the upper one cannot have been selected; hence good deletion preserves regularity. Dually, if adding the selected first surviving plus creates $e$ equal rows of the new length, then a block of $e-1$ rows of that length lies immediately above it and its first addable corner is a consecutive same-residue plus which survives whenever the lower plus does, so the lower plus is not selected; hence cogood addition preserves regularity. Both operators therefore stay inside the regular labels.

Reachability of all regular partitions and the component. Every nonempty regular partition has a top block of $a<e$ equal rows; its first removable corner has residue $L-a$, while the only earlier addable corner has residue $L$, so the minus has no matching same-residue plus and survives: a good deletion exists, it remains regular and lowers the size. Induction on the size together with the inverse axiom reaches every regular partition from the empty one, and since the empty partition is regular and all edges of the component preserve regularity, the empty component is exactly the set of $e$-regular partitions. The transposed ($e$-restricted) model arises by conjugation and is not the same vertex set.

Conclusion. The crystal basis of the whole Fock space is the good-node partition graph on all partitions with the survivor $\epsilon ,\phi$, and its empty connected component is exactly the $e$-regular partition set, proved by the block argument without an imported path count or an abstract rank-one recognition.