Why is Furch's "knotted hole ball" not shellable?

Question

I'm trying to work through the example below, but I need some explanation as to why K' is not shellable. If we try to shell it, where would we get stuck?.

Here are some relevant definitions.

Answer 1

By analogy with Lemma 5.2, indeed, let's prove that pure 3D complex $G$ that contains a knotted (in $G$) curve with all edges except one on the boundary of $G$ can not be shellable.

Suppose $G$ is such a shellable complex with minimal number of facets. Let the curve be $\gamma=C\cup e$, where $C$ is a path in boundary of $G$; let $A$ and $B$ be the vertices of $e$. Then remove the last (in the shelling order) facet $F_N$ from $G$ and get a smaller shellable complex $\hat{G}$. We claim is that $\hat{G}$ also contains a a knotted (in $\hat{G}$) curve with all edges except one on the boundary of $\hat{G}$. Once this is established, we are done since $G$ was assumed to be minimal complex with such properties.

Now, why does $\hat{G}$ contain a knotted loop with all but one edges on the boundary? We claim $e$ could not have been an edge of $F_N$, for otherwise we could homotope the path $C$, keeping $A$ and $B$ fixed, to a path in boundary of $F_N$ and get a knotted curve in the boundary of $F_N$, which is not possible since that boundary is a 2-sphere. So the edge $e$ was not an edge of $F_n$, and then it is still an interior edge of $\hat{G}$. Now we can homotope $C$, keeping keeping $A$ and $B$ fixed, to a path $\hat{C}$ in the boundary of $\hat{G}$, and $\hat{\gamma}=\hat{C}\cup e$ is a knotted loop in $\hat{C}$ with exactly one edge not on the boundary.