how to show that when an edge is removed from K5, the resluting subgraph is planar. this question might be simple to others, but I'm stuck on this question. "prove that when I deleted an edge from $K5$, it has planar sub-graph . So, I know that G is planar if and only if G contains no sub-division of $K5$ or $K3,3$. So that means when I delete an edge from K5, I no loger have sub-division of K5 and therefore, the subgraph is planar. But how do I show that explicitly that when I remove edge from $K5$, it will have planar subgraph? (I think this question is somewhat related to Kuratowski's theorem)

The key observation is that all graphs of "$K_5$ with one edge removed" are isomorphic.

To this end, you can just start with a picture of $K_5$, remove any one edge, and then try to re-draw what results as a planar graph. The important component is understanding why this approach generalizes enough to prove the graph is planar for _any_ single edge-removal.

To this end, here is a picture that came up after googling **K5 graph planar** :

!enter image description here

By way of a similar argument, you can reason about $K_{3,3}$ and draw a convincing picture:

!enter image description here

(From wikipedia **here**.)

Without loss of generality, the removed edge could be one of the two that cross above.