If one does deletions of edges on complete multigraphs: i.e. $K^{\mu}_{n} - K^{\mu}_{m}$ then, if $n$ and $m$ differ by one: i.e. $K^{\mu}_3 - K^{\mu}_2$ and $n > m$, let $\mu$ denote the multiplicity of the graph.
then this results in the claw(or at least I suspect), why is this the case? Is there a way to prove this?
Since $n=m+1$, there exists a unique vertex $v$ of $K_n^\mu$ which is not a vertex of $K_m^\mu$. Assume that $\mu\ge 1$. So when we remove from $K_n^\mu$ the edges of $K_m^\mu$, we obtain a residual complete bipartite graph $K_{1,n}^\mu$ with one bipartition part consisting of $v$ and the other of all remaining vertices of $K_n$. The residual graph $K_{1,n}^\mu$ contains a claw $K_{1,3}$ provided $n\ge 3$.