Does this simplicial complex embed into $\mathbb{R}^3$?

Let $$K$$ be a finte 2-dimensional simplicial complex that admits an embedding into $$\mathbb{R}^3$$. Further, assume that $$H_2(K, \mathbb{R})$$ is trivial, so there are no 2-dimensonal cycles in $$K$$. Let $$\gamma$$ be a 1-dimensional null-homologous cycle which is homeomorphic to a circle; by the assumption on $$H_2$$ we know there exists a unique 2-chain whose boundary is $$\gamma$$. We also assume that $$\gamma$$ is a formal sum of edges $$\gamma = \sum c_i e_i$$ such that $$c_i \in \{-1, 0, 1\}$$ for each $$i$$.

Now consider the complex $$K \cup D$$ obtained by taking a complex $$D$$ that is homeomorphic to a disk and identifying its boundary with $$\gamma$$. We can assume that $$D$$ has a nice triangulation such that the edges on its boundary are in bijection with the edges of $$\gamma$$. Does $$K \cup D$$ admit an embedding into $$\mathbb{R}^3$$?

• The boundary of a disk is a circle, so the boundary of D must be homeomorphic to a circle. I think we might be able to provide a counterexample by constructing $K$ such that $\gamma$ is not homeomorphic to a circle. Doodling pictures on my whiteboard suggests that we can pretty easily find a case where $\gamma$ is a figure-8. Jul 10 at 5:13
• Good observation. I should have specified that I assume $\gamma$ to be homeomorphic to a circle. I will edit my question to include the assumption.
– Will
Jul 10 at 5:17

Start with the complete graph on 5 vertices, $$K_5$$. Let $$X$$ denote the graph obtained by removing one edge $$[v,w]$$ from $$K_5$$. Let $$K$$ denote the cone over $$X$$ with the tip $$p$$. It is easy to see that $$X$$ is a planar graph, which implies that the cone $$K$$ embeds in $$R^3$$. Moreover, $$K$$ is clearly contractible.
Now, to construct a simple loop $$\gamma$$ in the 1-skeleton of $$K$$, let $$c$$ denote a simple arc in $$X$$ connecting $$v, w$$. Add to this arc the edges $$[p,v], [p,w]$$. The result is a simple loop $$\gamma$$.
I will leave it to you to prove that $$Y:=K\cup_\gamma D^2$$ does not embed in $$R^3$$. (Use the fact that the link of $$p$$ in $$Y$$ is $$K_5$$, which is not a planar graph.)