I've been struggling with the following question:
Prove that a cactus graph is planar
I've worked out something, but I'm not sure if I'm right or wrong:
Consider every cycle of the cactus graph separately: the number vertices is equal to the number of edges, because every edge contributes to $1$ cycle. The number of faces is equal to $2$ (inside and outside the cycle). If we fill in Eulers formula ($V-E+F=2$), we get: $2=2$. Considering every subgraph is planar, we can assume that our cactus graph is planar. Q.E.D.
Thanks in advance!
Welcome to MSE!
Your idea is natural, but doesn't quite work. It's possible for every proper subgraph of $G$ to be planar, even though $G$ itself is not planar. $K_{3,3}$ is an example of this phenomenon, for instance.
Here's a slick proof. According to wikipedia, every minor of a cactus graph is cactus. We also know by Kuratowski's theorem that a graph is nonplanar if and only if it has either $K_5$ or $K_{3,3}$ as a minor.
So if we can show that $K_5$ and $K_{3,3}$ are both noncactus, then every cactus graph avoids them. But that means every cactus graph is planar.
Of course, it's quite easy to see that $K_5$ and $K_{3,3}$ are both noncactus. Do you see why? In each case you should be able to find an edge which is contained in multiple cycles.
I hope this helps ^_^