Between-copy proper exchanges have essential diagonals and rigid cap counts

Resolve every curve except the chosen proper arc regularly. Only the pieces of $D_0,D_1$ incident to it can be affected by undoing it. Cutting their proper traces gives four disks $E_0,K_0,E_1,K_1$. In the transverse rectangle label the upper left and right rays $0,1$, and the lower left and right rays $2,3$. The full regular pairing is $(0,2),(1,3)$; the original unswitched diagonals are $(0,3),(1,2)$; the irregular pairing is $(0,1),(2,3)$. Thus the inverse diagram has two parametrizing disks $U=E_0\cup K_1$ and $V=K_0\cup E_1$, with their joining strips understood. Each is embedded: the only remaining double arc joins one branch from each component, and neither component has a self-intersection. They are normal by [F1], with $v(U)+v(V)=2d$. The irregular resolution instead gives two embedded PL disks $W=E_0\cup E_1$ and $Z=K_0\cup K_1$. All these joins are along proper boundary intervals, so the disk topology follows by gluing two disk parameters along an interval. All switching neighborhoods miss the original edges; in particular $w(W)+w(Z)=2w(D)$.

The disks $W,Z$ cannot both be essential. Indeed their irregular resolution has a return parametrizing face arc by [F2]. It lies wholly on one of these two disk components, say $W$, and is embedded there because $W$ is embedded. If both were essential, normalize $W$ strictly decreasingly by [F3] and $Z$ nonincreasingly. To spell out the strict alternative, if a preliminary move or discard loses an edge point of $W$, strict decrease is already obtained; otherwise its face curve and the return arc persist, and the edge bigon removes two intersections while retaining an essential disk. The two normal outputs then satisfy $w(W^{\ast })+w(Z^{\ast })<2w(D)$. Both must have weight at least $w(D)$, a contradiction. Consequently at least one irregular output has null-homotopic boundary. This is a joint weight argument, not the old bound of one output by the entire two-disk input weight.

Mark the exchange disk as a rectangle with frontier intervals ${\alpha }_i:p_i\to q_i$ and boundary connectors $b:p_0\to p_1$, $c:q_0\to q_1$ in $\mathrm{\partial }M$. Let $u,u^{\mathrm{\prime }}$ be the original-boundary arcs of $E_0,K_0$, both directed from $p_0$ to $q_0$. Transport the original-boundary arcs of $E_1,K_1$ by $a\mapsto ba{c}^{-1}$, obtaining paths $v,v^{\mathrm{\prime }}$ with the same endpoints. The original disk boundaries are $u(u^{\mathrm{\prime }}{)}^{-1}$ and $v(v^{\mathrm{\prime }}{)}^{-1}$, up to basepoint transport, and both are essential. The inverse diagonal boundaries are $u(v^{\mathrm{\prime }}{)}^{-1}$ and $u^{\mathrm{\prime }}{v}^{-1}$; the irregular boundaries are $u{v}^{-1}$ and $u^{\mathrm{\prime }}(v^{\mathrm{\prime }}{)}^{-1}$, up to conjugacy or inversion. A null-homotopic path loop $a_1{a}_2^{-1}$ is precisely a homotopy between $a_1,a_2$ with endpoints fixed. If either diagonal loop and either irregular loop were null, their common path would give either $u\simeq u^{\mathrm{\prime }}$ or $v\simeq v^{\mathrm{\prime }}$ with endpoints fixed. For example $u\simeq v^{\mathrm{\prime }}$ and $u\simeq v$ imply $v\simeq v^{\mathrm{\prime }}$; pairing $u\simeq v^{\mathrm{\prime }}$ with $u^{\mathrm{\prime }}\simeq v^{\mathrm{\prime }}$ instead gives $u\simeq u^{\mathrm{\prime }}$. The other two cases interchange primes. Each contradicts an essential original boundary. Step 2.1 supplies a null irregular loop, so neither diagonal loop is null: both $U,V$ are essential. The connectors are those of the actual rectangle, not arbitrary path identifications.

The two essential normal disks $U,V$ have weights summing to $2w(D)$, so least weight forces each to have weight $w(D)$. Their vectors sum to $2d$ by step 1.1. At the first coordinate differing from $d$, one would be smaller and the other larger. Lexicographic minimality among essential normal disks of this weight excludes that possibility. Hence $v(U)=v(V)=d$.

The original strips are edge-free, and the four cut caps partition the original disk edge points. Thus $N(U)=N(E_0)+N(K_1)=N(D)$, whereas $N(D_1)=N(E_1)+N(K_1)=N(D)$. Subtraction gives $N(E_0)=N(E_1)$. The complementary equalities similarly give $N(K_0)=N(K_1)$. Each of these four caps is cut from one normal disk by the selected proper trace alone, so [F4] makes its total edge weight positive. This proves all assertions without identifying any raw cap replacement with an ambiently parallel normal cap.