It is shown that if $\mathscr{A}$ and $\mathscr{C}$ are sets of degrees uniformly recursive in $\mathbf{0}'$ with $\mathbf{0} \nonin \mathscr{C}$ then there is a degree $\mathbf{b}$ with $\mathbf{b}' = \mathbf{0}', \mathbf{b} \cup \mathbf{c} = \mathbf{0}'$ for every $c \in \mathscr{C}$, and $\mathbf{a} \nleq \mathbf{b}$ for every $\mathbf{a} \in \mathscr{A} \sim \{0\}$. The proof is given as an oracle construction recursive in $\mathbf{0}'$. It follows that any nonrecursive degree below $\mathbf{0}'$ can be joined to $\mathbf{0}'$ by a degree strictly below $\mathbf{0}'$. Also, if $\mathbf{a < 0}$' and $\mathbf{a}'' = \mathbf{0}''$ then there is a degree $\mathbf{b}$ such that $\mathbf{a} \cup \mathbf{b} = \mathbf{0}'$ and $\mathbf{a} \cap \mathbf{b} = \mathbf{0}$.