A reduced original normal pair for a least-weight essential disk has no projective-plane summand

Suppose $G=Q$ is a projective plane and put $Y=H$. Their remaining intersection is nonempty, since the full resolution is connected and both summands are nonempty. It consists of circles, since $Q$ is closed. The full resolution has boundary $\mathrm{\partial }Y$, namely the one essential boundary circle $C$ of $F$. Cutting out the annular or Mobius neighborhoods [F4] and inserting the switching strips preserves total Euler characteristic: the removed pieces, their boundary circles and the inserted interval-bundle switching pieces have Euler characteristic zero. For the latter, the finite rectangular charts glue around a circle, so their face and side counts cancel just as for an annulus or Mobius band. Hence $1=\chi (F)=\chi (Q)+\chi (Y)$ and $\chi (Y)=0$ by [F3]. By [F2], the connected surface $Y$ with its one boundary circle is a Mobius band.

An intersection circle is two-sided in $Q$ exactly when it is two-sided in $Y$. In a transverse cross-section there are two rays of each surface. A regular uncrossing pairs each $Q$ ray with one $Y$ ray. Following the rays once around the circle preserves this pairing by [F1]. Reversing the two rays of only one surface interchanges the two possible uncrossings, whereas reversing both, or neither, preserves the pairing. Thus the two surfaces' transverse intervals have the same reversal sign. This is the claimed equivalence by [F4].

Every two-sided simple circle separates $Q$. Otherwise cutting out its annular neighborhood would leave a connected boundary surface with Euler characteristic $1$ and two boundary circles. It would be a disk by [F2], a contradiction. Likewise every two-sided circle separates the Mobius band $Y$: a nonseparating annular cut would leave a connected Euler-zero surface with the three boundary circles consisting of $C$ and the two new ones, contrary to [F2]. Hence a two-sided intersection circle would separate both summands by step 2.1, contradicting reducedness [F1]. All remaining circles are one-sided in both surfaces.

A one-sided circle in $Q$ has a Mobius neighborhood with one boundary circle. Its complement is connected: only one component can meet that boundary circle, while any other component would be an open and closed component of the original connected surface. The complement has Euler characteristic $1$ and one boundary circle, so is a disk by [F2]. The neighborhood can be chosen disjoint from all other intersection circles. Any further circle would lie in that complementary disk and would be two-sided there, contradicting step 3.1. Thus there is exactly one intersection circle $c$. The same one-boundary-neighborhood argument makes the complement of its Mobius neighborhood in $Y$ connected. It has Euler characteristic zero and two boundary circles, $C$ and the new circle, so is an annulus by [F2].

Either switch along $c$ replaces the two crossing Mobius strips by one annulus connecting their boundary circles. Here is the finite local calculation. Follow the transverse cross-section once around $c$. Both labeled surface axes reverse by step 3.1. For either smoothing, its two interval pieces in opposite quadrants are therefore exchanged after one turn. After two turns each interval returns, with its endpoints still labeled $Q$ and $Y$ respectively. The two rectangles consequently glue in series into an interval product over the double circle, namely an annulus, with one boundary from $Q$ and one from $Y$. No interval reversal is possible after the two turns because its two endpoint labels are fixed. Attaching the disk remainder from $Q$ and the annulus remainder from $Y$ in step 4.1 gives a connected disk whose boundary is $C$. Since this is the only double curve, either fully resolved image is embedded.

Choose the irregular switch. Step 5.1 gives an embedded proper disk with essential boundary as a parametrizing component. The strict improvement supplier Irregular switches in original face braids strictly improve essential disks applies to the retained original braid, after its preliminary regular switches. It gives an essential normal disk of weight strictly below $w(Q)+w(Y)=w(F)$, contradicting minimality. This excludes a projective-plane $G$, and exchanging names excludes a projective-plane $H$. The argument excluded a summand using the actual switching data; it did not exclude projective planes from the ambient manifold.