Can you help me prove that every planar graph can be expressed as a union of (maximum) five forests?
Ι used induction on the number of vertices. A planar graph $G$ has at most $3|V|-6$ edges, so there is at least one vertex $v$ of degree 5 or less.
Let $\{w_1,\ldots, w_j\}$; $j \le 5$ be $v$'s neighbors in $G$.
Then let $F_1,\ldots, F_5$ be the 5 edge-disjoint forests (some may be empty) such that $\cup_{j=1}^5 F_i = E(G \setminus \{v\})$. How do I prove that $F_1,\ldots, F_5$ exist?
Then for each $i \le j$ set $F'_i = F_i + \{v,w_i\}$, and (if $j$ is strictly less than 5) for each $i \in \{j+1,\ldots, 5\}$ set $F'_i=F_i$. Then each $F'_i$ is also a forest, and that each edge of $G$ is in one of the $F'_i$s.
Special thanks to user Mike for the answer.
Induction on the number of vertices: Pick a vertex $a$ of degree $\le 5$ in $G$ (exists for every planar graph), solve the problem for $G'=G-a$ and assign the edges incident with $a$ to different forests. Then $a$ cannot be part of a cycle, hence any cycle in $G$ would already be a cycle in $G'$.