The Whitney trick in the codimension-two borderline case

Form Milnor's boundary annulus in complementary corner charts and sheet collars. Along the $r$-sheet arc choose its inward normal direction and along the $s$-sheet arc choose the direction in its oriented normal bundle. At the two corners the latter is the first-sheet velocity and its negative respectively, so opposite intersection signs give matching oriented choices. For $r=1$ the normal direction along the $s$-sheet is a line: the sign computation is precisely what lets its two endpoint choices agree. This yields a clean embedded annulus whose inner loop $\lambda$ lies outside both sheets and is nullhomotopic in $X$.

If $r\ge 3$, make a nullhomotopy of $\lambda$ transverse to $B$ relative to a fixed boundary collar. Its inverse image of $B$ has expected dimension $2-r<0$ and is empty, so $\lambda$ contracts in $X\setminus B$. If $r=1$ or $2$, the assumed injectivity of ${\pi }_1(X\setminus B)\to {\pi }_1(X)$ gives exactly the same conclusion, since $\lambda$ already lies in the complement and its ambient class is trivial. Now in $X\setminus B$ make this disk transverse to $A$, with fixed collar; its expected incidence dimension $2-s<0$ clears $A$. The relative embedding supplier applies because $m\ge 5=2\cdot 2+1$, preserving the clean collar. Repeat a sufficiently small relative transversality perturbation if necessary after embedding; compact separated-pair estimates preserve embeddedness. Attach the fixed annulus to obtain a clean embedded bigon.

Choose a smooth metric adapted to the clean sheet collars and the fixed product corners: prescribe orthogonal disk-tangent and sheet-normal blocks along the arcs, extend their positive matrices in charts, and combine extensions agreeing there by [F12]. Embed $X$ in Euclidean space by [F10] and represent the disk-normal bundle by this metric's orthogonal complement to the disk tangent space inside $TX$. Let $P(z)$ be the Euclidean orthogonal projection onto that smooth subbundle in convex bigon coordinates centered at an interior point. Put $Q(t,z)=P(tz)$ and solve ${\mathrm{\partial }}_{t}U=[{\mathrm{\partial }}_{t}Q,Q]U$, $U(0,z)=I$, on $[0,1]$ by [F7]. The coefficient $K=[{\mathrm{\partial }}_{t}Q,Q]$ is skew symmetric, and differentiating $Q^2=Q$ gives $[K,Q]={\mathrm{\partial }}_{t}Q$. Consequently $U^{T}U=I$, and uniqueness gives $U(t,z)P(0)U(t,z{)}^T=Q(t,z)$. Transporting a basis at the centre and applying [F11] in the adapted metric gives a full smooth disk-normal frame. Smoothness in $z$, including local extensions at the corners, follows from [F8] and uniqueness along the compact interval.

Along the $r$-sheet arc choose an $(r-1)$-frame $E$ tangent to that sheet and along the other arc normal to the $s$-sheet. For $r\ge 2$, [F4]'s opposite-sign endpoint calculation applies in locally oriented collars; equivalently it uses the given orientation of $A$ and of the normal bundle of $B$. In the disk frame of step 3.1, $E$ is a loop in $V_{r-1}({\mathbb{R}}^{m-2})$, with complement rank $s-1\ge 2$. By [F5] it fills over the disk; attach its prescribed smooth collar to the filling and apply [F6] relative to a smaller collar to make the filling smooth. For $r=1$, $E$ is the unique empty frame and no Stiefel assertion is needed. Take its orthogonal complement inside the disk normal bundle and apply the projection transport of step 3.1 followed by [F11] to get a global $(s-1)$-frame $H$. On the $B$ arc that subspace is exactly its disk-orthogonal tangent space, so $H$ is tangent to $B$. Compatible corner values can be attained by multiplying $H$ by a smooth disk-wide $SO(s-1)$ map interpolating their two comparison matrices, constant in the corner collars; finite plane rotations provide such a path. This changes neither its subspace nor its extendibility, and no full boundary-frame class is prescribed. Thus $E,H$ is an admissible extendible frame in every case.

The local model theorem applies to this actual framed bigon with dimensions $r,s$, requiring no lower bound on $r$ once the frame is given. Apply its auxiliary ambient isotopy to $A$ and keep $B$ fixed. It removes exactly $p,q$ and fixes all other intersection neighbourhoods. This proves the borderline range, including $r=1$, without treating the codimension-one or codimension-two clearing as ordinary generic avoidance.