Let $G=(V,E)$ be a finite simple $k-$regular graph ($k\geq 1$). Does $G$ necessarily
contain a subset $E'\subset E$ of edges such that only isolated edges and cycles occur as connected components in $(V,E')$?


(The answer is easily yes for $k=1,2$.)


A counterexample would easily give a counterexample to question ["Antipodal" maps on regular graphs?](https://mathoverflow.net/questions/64746/antipodal-maps-on-regular-graphs) in the case $D=2$ by considering
the complementary graph of $G$ (respectively of two disjoint copies of $G$ if $G$ is 
"too small").



Existence of a nice  subset of edges in $k-$regular simple graphs?
It seems like such a subset should always exist. 


Consider the bipartite graph on $2|V(G)|$ vertices corresponding to the adjacency matrix of $G$. Since this graph is regular, by Hall's Theorem it has a perfect matching. In terms of the original $G$, this corresponds to a permutation $\sigma$ on $V(G)$ such that $v$ and $\sigma(v)$ are always adjacent. The cycles of $\sigma$ would then give you the desired decomposition. 


