1. **Stating the problem:** We have the following function compositions and equalities: $$r_l \circ r_m(F) = G$$ $$r_A \circ r_m(G) = F$$ $$r_l \circ r_l(F) = B$$ We want to determine what to assign to the blue labels $A$ and $B$. 2. **Understanding the problem:** - The notation $r_x \circ r_y(Z)$ means applying function $r_y$ to $Z$ first, then applying $r_x$ to the result. - Given $r_l \circ r_m(F) = G$, applying $r_m$ to $F$ then $r_l$ gives $G$. - Given $r_A \circ r_m(G) = F$, applying $r_m$ to $G$ then $r_A$ gives back $F$. - Given $r_l \circ r_l(F) = B$, applying $r_l$ twice to $F$ gives $B$. 3. **Finding $A$:** From the first two equations: $$r_l \circ r_m(F) = G$$ $$r_A \circ r_m(G) = F$$ Apply $r_m$ to both sides of the first equation: $$r_m(G) = r_m(r_l(r_m(F)))$$ But from the second equation, applying $r_A$ to $r_m(G)$ returns $F$, so $r_A$ acts as the inverse of $r_l$ on the image of $r_m$. Therefore, $r_A = r_l^{-1}$ (the inverse function of $r_l$) on the relevant domain. 4. **Finding $B$:** Given: $$r_l \circ r_l(F) = B$$ This means applying $r_l$ twice to $F$ results in $B$. So $B = r_l(r_l(F))$. 5. **Summary:** - The blue $A$ corresponds to the inverse function of $r_l$, i.e., $A = l^{-1}$. - The blue $B$ corresponds to the double application of $r_l$ to $F$, i.e., $B = r_l(r_l(F))$. Hence, the blue labels represent: - $A$: the inverse of $r_l$ - $B$: the result of applying $r_l$ twice to $F$