The high-dimensional Whitney trick

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $X^m$ be a smooth manifold, let $A^a,B^b\subseteq X$ be closed connected oriented embedded complementary transverse submanifolds with $a,b\ge 3,$ so both codimensions are at least three, and let $p,q\in A\cap B$ satisfy $\epsilon (p)=-\epsilon (q)$. Here the signs use the sheet orientations and a continuous orientation of $TX$ along the specified Whitney circle; since that circle is nullhomotopic, such an orientation exists and extends over any disk filling. Reversing it changes both signs together. Suppose that the Whitney circle determined by arcs $\alpha \subseteq A$ and $\beta \subseteq B$ joining $p$ and $q$ (taken as part of the data) is null-homotopic in $X$; by the label lemma this holds exactly when the two double points carry equal fundamental-group labels, and it is automatic when $X$ is simply connected. Then there is an isotopy of $X$, the identity near $A\cap B\setminus \{p,q\}$ and supported in a compact neighbourhood of a Whitney disk, carrying $A$ to an embedded $A^{\mathrm{\prime }}$ with $A^{\mathrm{\prime }}\cap B=(A\cap B)\setminus \{p,q\}$. In particular, if $A\cap B=\{p,q\}$ then $A$ can be isotoped to meet $B$ in no point at all, and if $X$ is simply connected the only remaining hypotheses are $a,b\ge 3$ and the opposite signs.