1 Proof of Theorem 1
1.1 Notation
A multigraph is simple if it has no loops or multiedges. In this case, we simply refer to it as a graph.
Let be a (multi)graph. For , we denote by the subgraph of induced by , and by the subgraph of induced by . If , then we denote . For , we denote . If , then we denote . For , is the disjoint union of and a set of isolated vertices. If , then we denote . For a set of pairs of edges with , we denote . If , then we denote . Graph is called cubic if all of its degrees are exactly . Graph is called subcubic if its degrees are not bigger than .
A (multi)graph is connected if the removal of any vertices leaves the graph connected. A (multi)graph is edgeconnected if the removal of at most edges leaves the graph connected. Note that a subcubic (multi)graph with at least vertices is connected if and only if it is edgeconnected.
A (multi)graph is essentially edgeconnected if the removal of at most three edges does not yield two components with at least two vertices each. A (multi)graph is cyclically edgeconnected if the removal of at most three edges does not yield two components that both contain a cycle. For a cubic (multi)graph, these last two notions are equivalent.
1.2 Proof
We proceed by contradiction. Let be a counterexample to Theorem 1 with the fewest vertices. We use the following very convenient theorem from Munaro [Mun16]:
Theorem 2 (Theorem 3.4.10 in [Mun16]).
If is simple, then it is not cyclically 4edgeconnected.
To obtain a contradiction, we argue that is in fact, a simple graph that is essentially edgeconnected, as follows.
Lemma 1.1.
The multigraph is an essentially edgeconnected simple graph.
Proof.
While Claims 1, 2, 3 are known and easy properties of a minimum counterexample to Jones’ conjecture on subcubic graphs (see e.g. [Mun16]), we include their proofs because we believe they may constitute a useful warmup. The uninterested reader may skip them guiltfree.
Claim 1.
The multigraph is regular.
Proof.
Suppose has a vertex with degree at most . Then satisfies Jones’ Conjecture by minimality of . As no cycle of contains , we have and , therefore also satisfies Jones’ Conjecture, a contradiction.
Suppose has a vertex with degree , and let and be the two neighbors of . Then satisfies Jones’ Conjecture, so . The cycles of are in bijection with the cycles of , by exchanging the edges and and an edge when appropriate. Hence . Moreover, if is a feedback vertex set of that does not contain , then is a feedback vertex set of , and if is a feedback vertex set of that contains , then is a feedback vertex set of . Thus , and satisfies Jones’ Conjecture, a contradiction. Hence is cubic.
Claim 2.
The multigraph is connected.
Proof.
Suppose that is not connected. As is cubic, that means that is not edgeconnected. Let be a separating edge of . Both components and of verify Jones’ Conjecture by minimality of . Since is separating, it is not in any cycle of . The union of any feedback vertex set of and any feedback vertex set of is a feedback vertex set of , so . The union of any cycle packing of and any cycle packing of is a cycle packing of , so . Therefore satisfies Jones’ Conjecture, a contradiction.
In particular since is cubic, Claim 2 implies that is a simple graph.
Claim 3.
The graph is connected.
Proof.
Assume that it is not connected, and thus not edgeconnected. Let and be a edgecut, where and are in the same connected component of , which we denote . Let be the other connected component of . We write and . Note that this may lead to a double edge. See Figure 1 for an illustration. By minimality of , we know that , , and all satisfy Jones’ Conjecture.
Note that since , we have . We first argue that . Assume for a contradiction that . Note that for any feedback vertex set of , either or or and are in distinct components of , so . Thus, , a contradiction.
By symmetry, we have . Therefore, every cycle packing of contains the edge and every cycle packing of contains the edge . We can thus combine a cycle packing of and a cycle packing of by making a single cycle out of those two cycles. So . However, if is a feedback vertex set of and is a feedback vertex set of , then is a feedback vertex set of . Therefore , a contradiction. Therefore is connected.
Claim 4.
The graph is essentially edgeconnected.
Proof.
Assume that is not essentially edgeconnected, and thus not cyclically edgeconnected. Consider a nontrivial edgecut . Let and be the two components of . For , we define as the graph obtained from by contracting into a single vertex . See Figure 2 for an illustration. We define (resp. , ) as the graph obtained from by connecting with an edge vertices from incident to and (resp. to and for or to and for ). Again, this may lead to a double edge. Note that for both values of , all of , , , and have fewer vertices than , and thus satisfy Jones’ Conjecture.
First note that for both values of , . In order to prove that let us assume without loss of generality that . Then remove from . What remains from after deleting is a forest that could hypothetically create connections between vertices from incident to . However if we are given any tree and its three vertices then (in fact it is always a single vertex), where are sets of vertices on unique paths between corresponding vertices, so it is possible to break the connections between all three pairs of these vertices by removing a single vertex of . Because of that we see that what leads to . Therefore we see that . We also have , yet .
It follows that for both values of :
(1) 
(2) 
And that:
(3) 
(4) 
We are now ready for a closer analysis.

For any and for any , we have
Indeed, take without loss of generality and . Let us consider a minimum feedback vertex set of . Note that in , there is no path between the vertex incident to and the vertex incident to or at least one of them is in . As a consequence, either vertex incident to is in or one of them, say the vertex incident to , is either in or is not in the same component as the vertex incident to . For any minimum feedback vertex set of , we observe that is a feedback vertex set of , hence the conclusion. In particular, by combining with (3), if then for some .

For every and for every , if , then . That follows from (2), since satisfies Jones’ Conjecture.

For every , we have either or . Indeed, suppose not. Then both and , for say . Then for , in every cycle packing of there is a cycle containing and . By taking a cycle packing of and a cycle packing of , we obtain a cycle packing of (combining two cycles into one). So , a contradiction with (4).
It follows from (1) and (2) that for both values of . Note that a maximum cycle packing of uses two edges out of . It follows that for some , we have .
We assume without loss of generality that . Note that from (iii), , hence , and by (ii). We assume without loss of generality . By symmetry, and . From (i) applied with and , there is such that . Note that , so and . We derive from (ii) that , hence by (iii). Again from (ii), we obtain .
Therefore we have: , , and . Now, (i) applied with and yields a contradiction.
∎
2 Conclusion
Through a nontrivial combination of elementary tricks and using a nice preliminary result of [Mun16], we were able to close the case of Jones’ Conjecture for subcubic graphs.
The obvious question is whether this can be at all used to solve the whole conjecture. The reduction we have for subcubic graphs extends easily to the general setting, in the sense that a smallest counterexample to Jones’ Conjecture is essentially edgeconnected. It is not difficult to argue in a similar way that such a graph is vertexconnected. However, a much harder question is whether it is essentially vertexconnected. While it still seems possible, such a result using our approach would require additional tricks. Note that being in the general setting also gives us more leeway regarding possible reductions (no need to shy away from increasing the maximum degree, as long as there are fewer vertices).
A second obstacle to generalization is that even assuming that a smallest counterexample is essentially vertexconnected, Theorem 2 only deals with the subcubic case. Another argument must then be devised.
A different approach would be not to aim for the conjectured bound of but simply for any bound better than the existing one of . Unfortunately, this does not seem conceptually much easier. Let us emphasize this: a simple discharging argument yields for every planar graph , while even significant effort fails to grant a factor of instead of .
To highlight how little we understand around Jones’ Conjecture, we conclude by posing the following stronger conjecture. Note that the example of many nested disjoint cycles shows that the embedding cannot be fixed. Also note that the simple discharging argument mentioned above does not imply the following conjecture with a factor of instead of .
Conjecture 2.
For every planar graph , we have
where is the maximum size of a facepacking of , i.e., a cyclepacking where, for some embedding of , every cycle bounds a face.
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