Proof. Note that for any vertex $L$ in $L_r^\infty(D, D_{d,1}^\infty)$, $L|{[1,2]}$ is an admissible loop in $\oplus_r D{d,1}^\infty$. Recall the function $q$ defined in Definition 3.16 which maps an admissible loop to an edge midpoint in the tree $\mathcal{T}{d,r}$. Since each edge has a unique descendent vertex, we can instead map the loop to this vertex which lies in the forest $\mathcal{F}{d,r}$. Using this connection, we can now choose $\mathfrak{h}{r,p,N}$ to be the same as in [SW19, Lemma 3.8]. Then we have at least $N$ admissible loops of height $\mathfrak{h} \ge \mathfrak{h}{r,p,N}$ which lie in the complement of the surface corresponding to $\sigma$ in $\oplus_r D_{d,1}^\infty$. Connecting each of these admissible loops to the base point in the complement surface, we get a set of lollipops in $Lk(\sigma)$. $\square$
We now show that the complex $L_r^\infty(D, D_{d,1}^\infty)$ is in fact contractible. The idea of proof is similar to that of [SW19, Proposition 3.1] but with significantly more technical difficulty.
Proposition 4.26. The complex $L_r^\infty(D, D_{d,1}^\infty)$ is contractible for any $r \ge 1$.
Proof. The complex $L_r^\infty(D, D_{d,1}^\infty)$ is obviously non-empty. We will show by induction that for all $k \ge 0$, any map $S^k \to L_r^\infty(D, D_{d,1}^\infty)$ is null-homotopic. Assume $L_r^\infty(D, D_{d,1}^\infty)$ is $(k-1)$-connected.
Let $f: S^k \to L_r^\infty(D, D_{d,1}^\infty)$ be a map. As usual, we can assume that the sphere $S^k$ comes with a triangulation such that the map $f$ is simplicial. We first use Lemma 1.7 to make the map $S^k$-simplexwise injective. For that we need for every $p$-simplex $\sigma$ in $L_r^\infty(D, D_{d,1}^\infty)$, its link $Lk(\sigma)$ is $(k-p-2)$-connected. But by Lemma 4.23, $Lk(\sigma)$ can be identified with $L_{r_\sigma}^\infty(D, D_{d,1}^\infty)$ for some $r_\sigma \ge 1$, so we have it is $(k-1)$-connected. Thus by Lemma 1.7, we can homotope $f$ to a map that is simplexwise injective.
Now since $S^k$ is a finite simplicial complex, the free height of the vertices of $S^k$ has a maximum value. We first want to homotope $f$ to a new map such that all the vertices have free height at least $\mathfrak{h} = \mathfrak{h}{r,k,N}$ where $N = v_0 + v_1 + \dots + v_k + 2$, where $v_i$ is the number of $i$-simplices of $S^k$ and $\mathfrak{h}{r,k,N}$ is determined by Lemma 4.25. For that we use a bad simplices argument.
We call a simplex of the sphere $S^k$ bad if all of its vertices are mapped to vertices in $L_r^\infty(D, D_{d,1}^\infty)$ that have free height less than $\mathfrak{h}$. We will modify $f$ by removing the bad simplices inductively starting by those of the highest dimension. Let $\sigma$ be a bad simplex of maximal dimension $p$ among all bad simplices. We will modify $f$ and the triangulation of $S^k$ in the star of $\sigma$ in a way that does not add any new bad simplices. In the process, we will increase the number of vertices by at most 1 in each step, and not at all if $\sigma$ is a vertex. This implies that, after doing this for all bad simplices, we will have increased the number of vertices of the triangulation of $S^k$ by at most $v_1 + \dots + v_k$. As $S^k$ originally had $v_0$ vertices, at the end of the process its new triangulation will have at most $v = v_0 + v_1 + \dots + v_k$ vertices. There are two cases.
Case 1: $p=k$. If the bad simplex $\sigma$ is of the dimension $k$ of the sphere $S^k$, then its image $f(\sigma)$ has a complement loop which bounds a surface $C$ asymptotically rigidly homeomorphic to $D_{d,r_\sigma}^\infty$ for some $r_\sigma \ge 1$ by Lemma 4.23. Now we can choose a lollipop $y$ in $C$ with free height at least $\mathfrak{h}+1$. In particular $f(\sigma) \cup y$ form a $(k+1)$-simplex. We can then add a vertex $a$ in the center of $\sigma$, replacing $\sigma$ by $\partial\sigma * a$ and replacing $f$ by the map $(f|_{\partial\sigma}) * (a \mapsto y)$ on $\partial\sigma * a$. This map is homotopic to $f$ through the simplex $f(\sigma) \cup {y}$. We have added a single vertex to the triangulation. Because $L$ has free height $\mathfrak{h}+1$, we have not added any new bad simplices, and we have removed one bad simplex, namely $\sigma$. Moreover, $f$ remains simplexwise injective.
Case 2: $p < k$. If the bad simplex $\sigma$ is a $p$-simplex for some $p < k$, by maximality of its dimension, the link of $\sigma$ is mapped to vertices of free height at least $\mathfrak{h}$ in the complement of the subsurface $f(\sigma)$. The simplex $\sigma$ has $p+1$ vertices whose images are pairwise disjoint outside the based point up to based isotopy. By Lemma 4.25 and our choice of $\mathfrak{h}$, there are at least $N = v+2$ lollipops $y_1, \dots, y_N$ of free height $\mathfrak{h}$ such that each $f(\sigma) \cup {y_i}$ form a $(p+1)$-simplex. As there are fewer vertices in the link than in the whole sphere $S^k$, and $S^k$ has at most $v$ vertices, by the pigeonhole principle, the loop part of the vertices in $f(Lk(\sigma))$ are contained in at most $v$ punctured disks bounded by the corresponding admissible loops with free height $\mathfrak{h}$. As $N = v+2$, there are at least two of the above vertices $y_i$ and $y_j$ of free height $\mathfrak{h}$ such that