Determining if two graphs are homeomorphic by smoothing?

Question

In graph theory two simple graphs $G,H$ are homeomorphic if repeated subdivision of $G$ and $H$ may end up with two isomorphic graphs, that is, the existence of a bijection between the vertex sets such that the number of edges joining any two vertices of first graph is equal to the number of edges joining the corresponding vertices of the other graph. Subdivision is a process of split existing edges by include vertices of degree 2. The reverse operation of subdivision is smoothing, when a vertex of degree 2 is removed and the two edges are merged into one.

Are there examples of two simple graphs, that by smoothing can be reduced into topologically homeomorphic (possibly multiple edged) graphs, that are not homeomorphic?

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I've made a preliminar IS-PLANAR that seemes to work based on the set of edges $E$:

1. For each subset $D$ of $E$ with more than $8$ edges:
2. Smooth $D$ by remove all vertices of degree $2$ and merge their edges into one edge
3. Test if $D$ is isomorphic to $K_5$ or $K_{3,3}$
4. Etc.

I've tested a lot of random graphs and visualized graphs tested to be planar. My rutin return the first find copy of $K_5$ or $K_{3,3}$ to be visualized when $E$ is tested non planar.

Answer 1

I am not an expert but wouldn’t think so. If I am not mistaken two graphs are homeomorphic precisely if their topological realizations are homeomorphic. In the latter subdivisions do not matter at all. So if you don’t want to use the realization and define homeomorphic in purely graph theoretical terms, you can either subdivide the graphs and check for isomorphism or topological reduce / smoothen the graphs and check for isomorphism. Anyway a graph is always homeomorphic to all its subdivisions and all graphs obtained from it by smoothening, so transitivity of being homeomorphic forces us to answer your question to the negative.