Every $p$-restricted partition is reached from the empty partition by cogood additions

Restriction is faithful on nonzero modules. Let $M\mathrm{\ne }0$ be a finite-dimensional $k{S}_d$-module and consider the multiplication map $\phi :k{S}_d{\otimes }_{k{S}_{d-1}}M\to M$, $\phi (g\otimes m)=gm$, where $M$ is regarded as a $k{S}_{d-1}$-module by restriction. It is $k{S}_d$-linear and surjective, since the image of $1\otimes m$ is $m$ and the elements $m$ span $M$; hence if the restriction of $M$ to $k{S}_{d-1}$ were zero, the tensor product would be zero and $M=0$, a contradiction. Therefore restriction sends no nonzero module to zero.

Nonzero finite-dimensional modules have nonzero socle: choosing a nonzero submodule of least positive dimension gives a simple submodule, so the socle, which is the sum of all simple submodules, is nonzero.

$D(\mu )$ is a nonzero simple $k{S}_d$-module by [F3], because $\mu$ is $p$-restricted.

Applying step 1.1 to $M=D(\mu )$ with $d\ge 1$: since $D(\mu )\mathrm{\ne }0$ by step 1.3, its restriction to $S_{d-1}$ is nonzero.

By step 1.2 applied to the nonzero finite-dimensional $k{S}_{d-1}$-module $\mathrm{Res}D(\mu )$ from step 2.1, the socle ${\mathrm{soc}}_{S_{d-1}}(D(\mu ){\downarrow }_{S_{d-1}})$ is nonzero.

By the residue-refined socle theorem [F4], the socle is the multiplicity-free sum of the simple modules $D(\mu \setminus A_i)$ over the residues $i$ with ${\epsilon }_i(\mu )>0$. Since this socle is nonzero by step 3.1, at least one residue satisfies ${\epsilon }_i(\mu )>0$.

Fix such a residue $i$. By [F5] the good $i$-node $A_i$ exists, $\nu :=\mu \setminus A_i$ is a $p$-restricted partition of $d-1$, and $\mu ={\tilde{f}}_i\nu$ with ${\tilde{f}}_i\nu =\nu \cup A_i$.

Induction on $d$. We prove that every $p$-restricted partition of $d$ is reachable from $\mathrm{\varnothing }$ by $d$ cogood additions. For $d=0$ the empty partition is the only partition and it is reached by the empty sequence. For $d\ge 1$ and a $p$-restricted partition $\mu$ of $d$, if $\mu =\mathrm{\varnothing }$ there is nothing to prove, and otherwise step 5.1 supplies a residue $i$ and a $p$-restricted partition $\nu$ of $d-1$ with $\mu ={\tilde{f}}_i\nu$; by the inductive hypothesis $\nu ={\tilde{f}}_{i_{d-1}}\cdots {\tilde{f}}_{i_1}\mathrm{\varnothing }$ for suitable residues, whence $\mu ={\tilde{f}}_i{\tilde{f}}_{i_{d-1}}\cdots {\tilde{f}}_{i_1}\mathrm{\varnothing }$. The residue $i$ is selected as, say, the least residue with ${\epsilon }_i(\mu )>0$, so no choice principle is used.

Good deletions carry every $p$-restricted partition to $\mathrm{\varnothing }$. At each nonempty stage step 5.1 produces a good node and, by [F5], the partition obtained by deleting it is again $p$-restricted of rank one less; iterating $d$ times reaches the only partition of rank $0$, which is $\mathrm{\varnothing }$; equivalently, by the inverse property in [F5], the two procedures are mutually inverse step by step.

Consequences for the graph. By definition the edges of the graph are the partial operations $\nu \mapsto {\tilde{f}}_i\nu$, so step 6.1 says exactly that every $p$-restricted partition is reached from $\mathrm{\varnothing }$; step 7.1 says every vertex reaches $\mathrm{\varnothing }$ by good deletions ${\tilde{e}}_i$. The empty partition is the unique partition of $0$, so it is the unique rank-$0$ vertex, and it has no incoming edge from a $p$-restricted partition because no partition of $-1$ exists; all ranks are finite. No step used the characteristic of $k$ beyond the existence of the socle theorem [F4], and no choice principle was used.