A nonvertex fundamental disk has a connected two-copy normal sum

There is an integer $n\ge 1$ and two nonzero admissible integer vectors $x,y$, neither proportional to $d$, with $nd=x+y$. Here is the rational linear algebra behind this assertion. The smallest face of the nonnegative matching cone containing $d$ is obtained by setting its zero coordinates to zero. If that face has dimension at least two, its matching nullspace, restricted to the positive coordinates of $d$, has dimension at least two. Adding the equation that coordinate sum is zero leaves a nonzero null vector $h$. The coefficient matrix is integral, so ordinary finite rational row reduction gives such an $h$ rational; clear denominators to make it integral. It is not proportional to $d$, since its coordinate sum is zero while that of $d$ is positive. For a sufficiently large integer $N$, both $Nd+h$ and $Nd-h$ are nonnegative and supported on $d$. They are admissible because that support permits only $d$'s quad type in each tetrahedron, and they sum to $2Nd$. This proves the assertion. Conversely the face could have dimension one only if $d$ were on an extreme ray, contrary to the hypothesis.

For every positive integer $m$, every component of a normal realization of $md$ has vector $d$. First form $m$ disjoint parallel normal copies of $D$. Local transverse intervals to its normal disk pieces fit across their sides; a choice of transverse side continues consistently over $D$, since every loop in a disk contracts and continuation around a small surface chart returns to the same side. Thus finitely many sufficiently close levels give $m$ copies, each with vector $d$, also along the proper boundary collar. Compare any realization of $md$ with this one using the ordered matching construction in [F1], proof 1.2: match edge points in order, then face arcs, then their tetrahedral boundary circles and disk pieces. That homeomorphism maps each normal disk piece to a piece of the same type in the same tetrahedron. It therefore preserves the normal vector of each component, proving the assertion; abstract disk type alone would not suffice.

Choose $n$ least among these decompositions. It is at least two, since $D$ is fundamental. Also $d$ has coordinate gcd one: otherwise division by that gcd would split it into two nonzero admissible integer vectors. A connected normal component with vector proportional to $d$ must therefore have vector $jd$ for a positive integer $j$. Indeed a Bezout integer combination of the coordinates of $d$ is one, so the same combination of the proportional integral vector gives its proportionality scalar as an integer. Step 1.2 applied with $m=j$ shows that each component has vector $d$, so connectedness forces $j=1$. Thus any such component has vector $d$.

Among regular diagrams for the least $n$, choose one with the least number of double curves, allowing the diagrams obtained by partial regular absorption as in [F1]. Neither family has a component with vector proportional to $d$. Such a component would have vector $d$ by step 2.1; subtracting it from $nd=x+y$ gives a nonproportional decomposition of $(n-1)d$, contrary to minimality. The remaining vector on its side is nonzero and not proportional, since otherwise the entire side would have been proportional. Every component of either family intersects the other family: a disjoint component survives unchanged into the full resolution of vector $nd$, and hence has vector $d$ by step 1.2, just excluded.

Both families are connected. Suppose, for example, that $Y$ has distinct components, and choose one $Z$. Regularly absorb all the other $Y$ components into $X$, giving a surface $W$. It is embedded and normal: the only remaining old double curves join its sheets to $Z$, so it has no self-intersection, and all tetrahedral pieces remain normal by [F1]. Now $nd=v(W)+v(Z)$. The vector of $Z$ is not proportional by step 3.1, and that of $W$ cannot be proportional either, since then $nd-v(W)=v(Z)$ would be. Thus this is another eligible diagram for the same $n$. Its remaining double curves are exactly the former curves between $X$ and $Z$. There are strictly fewer: every absorbed $Y$ component intersected $X$ by step 3.1. This contradicts the chosen minimum. The same argument interchanging families proves connectedness of $X$.

Their Euler characteristics sum to $n$ by [F3]. They cannot both be closed, since their resolution has nonempty boundary. If exactly one is closed, every double curve is a circle, so the other surface's boundary is unchanged by resolution and consists of $n$ circles. Since $n\ge 2$, that connected boundary surface has Euler characteristic at most zero by [F2]. The closed summand has Euler characteristic at most two. Hence $n=2$, the closed summand has characteristic two and is a sphere, and the other has characteristic zero and two boundary circles and is an annulus. If both have boundary, each has Euler characteristic at most one, so again $n=2$ and both are disks. The full-resolution surface has vector $2d$, so [F1] gives its two disk components and their boundary homotopy classes. In the sphere case those are exactly the unchanged annulus boundary curves.

Finally assume the extra weight and essentiality hypotheses and that both disk summands were essential. Their positive weights add to $2w(D)$, while each is at least $w(D)$; hence both equal $w(D)$. Their vectors satisfy $x+y=2d$. If either differs from $d$, at the first differing coordinate the two differences have opposite nonzero signs. One vector would be lexicographically smaller than $d$, contrary to the hypothesis. Thus both equal $d$, contrary to their nonproportionality. At least one disk summand must therefore have null-homotopic boundary, as asserted. No conclusion excluding that case or the sphere-annulus case has been used.