planar graphs course homework 1

note to grader: please note that the order of the problems isnt the same as in the homework pdf, sorry about this.
mahmood sheikh, 212835466

/home/mahmooz/brain/pen/2024-11-21-Note-15-50.xopp

prove that the dual of a planar embedded graph is planar. in other words, if $\text{[math]}$ is a planar embedded graph, prove that $\text{[math]}$ satisfies euler's formula.

given the planar embedded graph $\text{[math]}$ , we need to show that $\text{[math]}$ is planar.
let $\text{[math]}$ be the number of nodes, number of edges, number of faces and the number of connected components in $\text{[math]}$ , respectively, and let $\text{[math]}$ be their "counterparts" in $\text{[math]}$ . we have that $\text{[math]}$ for $\text{[math]}$ by def-planar-emb, we need to show that $\text{[math]}$ .
we have $\text{[math]}$ (from class we know the number of nodes in the dual is equal to the number of faces in the primal) and $\text{[math]}$ for the same reason, we have $\text{[math]}$ because the dual's embedding is a permutation whose length is the same as that of the primal's embedding, albeit with the arcs reversed (by def-embgraphdual), $\text{[math]}$ (the number of connected components) remains the same by lem-graph-emb-2, this implies that Euler's formula would still hold for the dual because the total values on both sides of the equation remain the same.