Q:

Planar graphs and homomorphism graphs?

I'm working with graphs. I would like to please be verified with this one.
Draw, if possible, a planar representation of each graph:

I got the planar representation of the 2nd graph. This is what I got.

I think the third is not possible. Is that right?

Also, do you know how to show that the two graphs are homeomorphic? I understood how to show that graphs are isomorphic, but I'm not really getting homeomorphism. I know that graphs are said to be homeomorphic if both can be obtained from the same graph by subdivisions of edges. But how can I show that they are homeomorphic?

The vertices don't have labels.

A:

As mentionned by Damascuz, for you first question you can use the fact that any planar graph has at most $3n-6$ edges. This limits can be derived from hand-shaking lemma and Euler's formula.
You might also know Kuratowski's theorem : It states that a finite graph is planar if and only if it does not contain a subgraph that is a subdivision of $K_5$ (the complete graph on five vertices) or of $K_{3,3}$ (complete bipartite graph on six vertices, three of which connect to each of the other three).

In your third example, the graph contains a $K_{3,3}$ by grouping the verticex as $\{A,B,F\}$ and $\{C,D,E\}$ for instance.
For your second question. I disagree with Damascuz. Your graph are homeomorphic. The fact that they have the same number of vertices is not sufficient to only check for isomorphism. For the left graph, add a vertex on the diagonal. For the right graph, add a vertex on the handle, the edges that stick out of the square : you will end up with the same graph. Therefore the two graphs are homeomorphic. 

