B^+\to B^+$ by $$r_1(a,b):=\begin{cases}
(i,j) &\text{ if } a = i', b = j'\\
(i',b) &\text{ if } a = i\\
(a,j') &\text{ if } b = j\\
(a,b)  &\text{ otherwise}
\end{cases}.$$

We can think of $r_1$ as sending the rook at $P'$ to $P$, and then projecting rooks in row $i$ and column $j$ onto row $i'$ and column $j'$ respectively. We claim $r_1(a,b)\in B^+$ for all $(a,b)\in B^+$. Since $P = (i,j)$ is an out-corner of $B$, if $(i,b)\in B^+$ then $b\leq j<j'$. Because $B^+$ is a Young diagram and $P' = (i',j')\in B^+$, we have $r_1(i,b) = (i',b)\in B^+$. In this case $M\cap S = \emptyset$ so a rook of the form $(a,j)$ with $a\neq i$ must have $a<i'$. Since $B^+$ is a Young diagram and $P' = (i',j')\in B^+$, we know $r_1(a,j) = (a,j')$ is in $B^+$. Therefore, $r_1(a,b)\in B^+$ for all $(a,b)\in B^+$.

Define $f_1: A_{{\overline}{S}} \to A_{{\overline}{T}}$ by $$f_1(M) = \{r_1(a,b): (a,b)\in M\}.$$ First, we show that given $M\in A_{{\overline}{S}}$, $f_1(M)$ is in fact in $A_{{\overline}{T}}$. Sending the rook in $P'$ to $P$ causes conflicts only for rooks in row $i$ and column $j$. We have solved the problem of a rook in row 