Edge contraction and subdivision Let $G$ be a $3$-connected graph that is not homeomorphic to $K_5$ or $K_{3, 3}$. 
Let $G'$ be the graph obtained from $G$ by contracting an edge. 
Why is it the case that $G'$ contains no subdivision of $K_5$ or $K_{3, 3}$?
Thank you  
 A: Suppose that one of the two forbidden graphs, which I'll call $K$, embeds in $G'$. The embedding consists of various vertices and edges. All of those edges exist in $G$ as the preimages of the edges of $G'$ under the "contraction" operation. Let's call the edge in $G$ $ab$, and the quotient vertex $q$. 
The collection of edges may not, however, form an embedding of $K$ in $G$, if one of the vertices of $K$ in $G'$ happens to be the "quotient" vertex $q$. In that case, an edge of $K$, say $qs$, and an another edge, say $qr$ may be joined (at $q$) in $G'$, but in $G$, they end up being $qa$ and $br$ instead, and hence no longer connected. You can fix this, however. 
Let's pick "$a$" as the "main" preimage of $q$. For any edge of $K$ that lifts to an edge $br$ in $G$, subdivide that edge of $K$ into two pieces. Send the midpoint to $b$, one end to $r$, and the other piece to $ba$. Once you do this for every edge of $K$ that lifts in $G$ to an edge meeting $b$ rather than $a$ you have a subdivision of $K$ embedded in $G$. 
A: Well, $G$ is embeddable in the plane. And when you contract an edge, you can deform the embedding to follow that contraction, so $G'$ is also embeddable, hence contains neither $K_5 $ nor $K_{3,3}$. 
(I'm assuming here that $G$ is finite, that the embedding of $G$ is chosen so that the edge maps to a smooth curve in the plane with a nice tubular neighborhood, etc., so that I can apply the isotopy extension theorem in some form, but the main gist of the proof is what I said above.)
