Hooks of a partition correspond bijectively to removable border strips

The rim is a path whose steps move left or down. Let $(p,q)\in R(\lambda )$ and suppose first that $q>1$. If ${\lambda }_{p+1}\ge q$, then $(p+1,q)\in [\lambda ]$ and it is a rim node because $(p+2,q+1)\mathrm{\notin }[\lambda ]$ follows from $(p+1,q+1)\mathrm{\notin }[\lambda ]$ and weak decrease; in this case $(p,q-1)$ is not a rim node, since $(p+1,q)\in [\lambda ]$. If instead ${\lambda }_{p+1}<q$, then $(p+1,q)\mathrm{\notin }[\lambda ]$ and $q-1\ge 1$, so $(p,q-1)\in [\lambda ]$ is a rim node. For $q=1$ the only candidate is $(p+1,1)$, which lies in $[\lambda ]$ when $p<k$ and is then a rim node because ${\lambda }_{p+1}=1$ forces $(p+2,2)\mathrm{\notin }[\lambda ]$. So every rim node other than the south-west end $(k,1)$ has exactly one rim successor among its two leftward or downward neighbours, and symmetrically every rim node other than the north-east end $(1,{\lambda }_1)$ has exactly one rim predecessor; hence $R(\lambda )$ is a single path from $(1,{\lambda }_1)$ to $(k,1)$ whose steps move left or down. In particular an edge-connected set of rim nodes is a subpath, determined by its two ends.

Fix $x=(i,j)$ and put $P:=(i,{\lambda }_i)$, $Q:=({\lambda }_j^{\mathrm{\prime }},j)$. Both are rim nodes: $(i,{\lambda }_i)$ because $(i+1,{\lambda }_i+1)\mathrm{\notin }[\lambda ]$, and $({\lambda }_j^{\mathrm{\prime }},j)$ because ${\lambda }_{j+1}^{\mathrm{\prime }}\le {\lambda }_j^{\mathrm{\prime }}$ gives $({\lambda }_j^{\mathrm{\prime }}+1,j+1)\mathrm{\notin }[\lambda ]$; and $i\le {\lambda }_j^{\mathrm{\prime }}$, $j\le {\lambda }_i$. Along the rim from $P$ the rows weakly increase and the columns weakly decrease, so the subpath from $P$ to $Q$ lies in the rectangle $\{(p,q):i\le p\le {\lambda }_j^{\mathrm{\prime }},j\le q\le {\lambda }_i\}$ and all its nodes satisfy $p\ge i$, $q\ge j$. Conversely, if a rim node satisfies $p\ge i$ and $q\ge j$, then $q\le {\lambda }_p$ gives ${\lambda }_p\ge j$, so $p\le {\lambda }_j^{\mathrm{\prime }}$, and it cannot occur strictly before $P$ (nodes preceding $P$ have row $<i$, or row $i$ and column $>{\lambda }_i$, which is impossible), nor strictly after $Q$ (nodes following $Q$ have column $<j$, because the step out of $Q$ decreases the column as ${\lambda }_{{\lambda }_j^{\mathrm{\prime }}+1}<j$); so it lies on the subpath from $P$ to $Q$. Therefore $\xi (x)=R(\lambda )\cap \{(p,q):p\ge i,q\ge j\}$ is exactly that subpath, an edge-connected rim set with ends $P$ and $Q$, and by step 1.1 it is the unique such set. Its node count is the number of leftward steps ${\lambda }_i-j=a(x)$ plus the number of downward steps ${\lambda }_j^{\mathrm{\prime }}-i=\mathrm{\ell }(x)$ plus one, so $\mathrm{\mid }\xi (x)\mathrm{\mid }=a(x)+\mathrm{\ell }(x)+1=h(x)$; its rows are exactly $i,i+1,\dots ,{\lambda }_j^{\mathrm{\prime }}$ (a left/down path meets each intervening row and visits rows in order), so its height is ${\lambda }_j^{\mathrm{\prime }}-i=\mathrm{\ell }(x)$.

The set $\xi (x)$ contains no $2\times 2$ block, since a rim node $(p,q)$ forbids its southeast diagonal neighbour $(p+1,q+1)$. Put $b={\lambda }_j^{\mathrm{\prime }}$ and define ${\nu }_p={\lambda }_p$ for $p<i$ or $p>b$, and ${\nu }_p=\max (j-1,{\lambda }_{p+1}-1)$ for $i\le p\le b$, with ${\lambda }_{k+1}=0$. Inside $[i,b]$ these values decrease because ${\lambda }_{p+1}$ decreases; also ${\nu }_i<{\lambda }_i\le {\lambda }_{i-1}$ when $i>1$. At the lower boundary ${\lambda }_{b+1}<j$ gives ${\nu }_b=j-1\ge {\lambda }_{b+1}={\nu }_{b+1}$. Thus $\nu$ is a partition after deleting trailing zeros. Rows outside $[i,b]$ contain no strip node: rows above $i$ fail its row condition, and rows below $b$ have width less than $j$. Within $[i,b]$, a node $(p,q)$ is outside the strip exactly when $q<j$ or $(p+1,q+1)\in [\lambda ]$, equivalently $q\le \max (j-1,{\lambda }_{p+1}-1)$. This maximum is less than ${\lambda }_p$, since ${\lambda }_p\ge j$ and ${\lambda }_{p+1}\le {\lambda }_p$. Hence the complement is exactly $[\nu ]$. Together with step 2.1, $\xi (x)$ is a removable border strip of size $h(x)$ and height $\mathrm{\ell }(x)$, with complement a partition of $n-h(x)$.

Conversely, let $\xi$ be a removable border strip and write $[\lambda ]\setminus \xi =[\nu ]$ with $\nu$ a partition. Every node of $\xi$ lies in the rim: suppose $(p,q)\in \xi$ and $(p+1,q+1)\in [\lambda ]$. If $(p+1,q+1)\in [\nu ]$, then ${\nu }_p\ge {\nu }_{p+1}\ge q+1>q$, so $(p,q)\in [\nu ]$, a contradiction; hence $(p+1,q+1)\in \xi$. Then ${\nu }_{p+1}\le q$ and equality is impossible, since ${\nu }_p\ge {\nu }_{p+1}=q$ would contradict ${\nu }_p<q$ (from $(p,q)\in \xi$); so ${\nu }_{p+1}<q$ and $q\le {\lambda }_{p+1}$ give $(p+1,q)\in \xi$, while $q+1\le {\lambda }_p$ and ${\nu }_p<q<q+1$ give $(p,q+1)\in \xi$. Thus $\xi$ would contain the $2\times 2$ block through $(p,q)$, contrary to its definition. So $\xi$ is an edge-connected set of rim nodes and, by step 1.1, a subpath of the rim path; let $P^{\mathrm{\prime }}=(i,q_1)$ be its north-east end (maximal column) and $Q^{\mathrm{\prime }}=(p_2,j)$ its south-west end (maximal row), where we have named the coordinates so that $x^{\mathrm{\prime }}:=(i,j)$. Then $q_1={\lambda }_i$: otherwise $q_1<{\lambda }_i$, so $(i,q_1+1)\in [\lambda ]$; this node is not in $\xi$ (it is not an end of the subpath and lies beyond its north-east end), hence lies in $[\nu ]$, and ${\nu }_i\ge q_1+1>q_1>{\nu }_i$, a contradiction. Likewise $p_2={\lambda }_j^{\mathrm{\prime }}$: otherwise $p_2<{\lambda }_j^{\mathrm{\prime }}$, so $(p_2+1,j)\in [\lambda ]$; this node is not in $\xi$ (it lies beyond the south-west end), hence lies in $[\nu ]$ with ${\nu }_{p_2+1}\ge j$, while $j>{\nu }_{p_2}\ge {\nu }_{p_2+1}$, a contradiction. Therefore $P^{\mathrm{\prime }}=(i,{\lambda }_i)$ and $Q^{\mathrm{\prime }}=({\lambda }_j^{\mathrm{\prime }},j)$, so $\xi$ is the unique edge-connected rim set with these ends, namely $\xi =\xi (x^{\mathrm{\prime }})$; and $x^{\mathrm{\prime }}$ is recovered from $\xi$ as (row of the north-east end, column of the south-west end), so it is unique.

Steps 3.1 and 4.1 give a map $x\mapsto \xi (x)$ from the nodes of $[\lambda ]$ onto the removable border strips, of size $h(x)$ and height $\mathrm{\ell }(x)$, whose two-sided inverse recovers $x$; so the map is a bijection. Consequently the multiset of sizes of the removable border strips of $\lambda$ is exactly the multiset of hook lengths $h(x)$, and by [F4] a node $x$ is removable if and only if $h(x)=1$, that is, if and only if $\xi (x)$ has size $1$.