A $4$-choosable graph that is not $(8:2)$-choosable

In 1980, Erd\H{o}s, Rubin and Taylor asked whether for all positive integers $a$, $b$, and $m$, every $(a:b)$-choosable graph is also $(am:bm)$-choosable. We provide a negative answer by exhibiting a $4$-choosable graph that is not $(8:2)$-choosable.

A list assignment for a graph G is a function L that to each vertex v of G assigns a set L(v) of colors. A set coloring ϕ of G is an L-set coloring if ϕ(v) ⊆ L(v) for every v ∈ V (G). For a positive integer b, we say that ϕ is an (L : b)-coloring of G if ϕ is an L-set coloring and |ϕ(v)| = b for every v ∈ V (G). If such an (L : b)-coloring exists, then G is (L : b)-colorable. For an integer a ≥ b, we say that G is (a : b)-choosable if G is (L : b)-colorable for every list assignment L such that |L(v)| = a for each v ∈ V (G). We abbreviate (L : 1)-coloring, (L : 1)-colorable, and (a : 1)-choosable to L-coloring, L-colorable, and a-choosable, respectively.
Questions and results. It is straightforward to see that if a graph is (a : b)-colorable, it is also (am : bm)colorable for every positive integer m: we can simply replace every color in an (a : b)-coloring by m new colors. However, this argument fails in the list coloring setting, leading Erdős, Rubin and Taylor [2] to ask whether every (a : b)-choosable graph is also (am : bm)-choosable whenever m ≥ 1. A positive answer to this question is sometimes referred to as "the (am : bm)-conjecture". Using the characterization of 2-choosable graphs [2], Tuza and Voigt [4] provided a positive answer when a = 2 and b = 1. In the other direction, Gutner and Tarsi [3] demonstrated that if k and m are positive integers and k is odd, then every (2mk : mk)-choosable graph is also 2m-choosable.
Formulated differently, the question is to know whether every (a : b)-choosable graph is also (c : d)choosable whenever c/d = a/b and c ≥ a. This formulation raises the same question when c/d > a/b, which was also asked by Erdős, Rubin and Taylor [2]. About ten years ago, Gutner and Tarsi [3] answered this last question negatively, by studying the kth choice number of a graph for large values of k. More precisely, the kth choice number of a graph G is ch :k (G), the least integer a for which G is (a : k)-choosable. Their result reads as follows.
As a direct corollary, one deduces that for all integers m ≥ 3 and > m, there exists a graph that is (a : b)-choosable and not ( : 1)-choosable with a b = m. (To see this, one can for example apply Theorem 1 with ε = 1 to the disjoint union of a clique of order m − 1 and a complete balanced bipartite graph with choice number + 1.) Another related result that should be mentioned here was obtained by Alon, Tuza and Voigt [1]. They proved that for every graph G, In other words, the fractional choice number of a graph equals its fractional chromatic number. The purpose of our work is to provide a negative answer to Erdős, Rubin and Taylor's question when a = 4 and b = 1. Gadgets and lemmas. A gadget is a pair (G, L 0 ), where G is a graph and L 0 is an assignment of lists of even size. Given a gadget (G, L 0 ), a half-list assignment for G is a list assignment L for G such that |L(v)| = |L 0 (v)|/2 for every v ∈ V (G). Let us start the construction by a key observation on list colorings of 5-cycles. Lemma 3. Consider the gadget (C, L 0 ), presented in Figure 1 Proof. The first statement is well known, but let us give the easy proof for completeness: since contains a color c 6 not belonging to L(v 2 ). By symmetry, we can assume that c 6 ∈ L(v 1 ). We color v 1 by c 6 and then for i = 5, 4, 3, 2 in order, we color v i by a color c i ∈ L(v i ) \ {c i+1 }. The resulting coloring is proper-we have c 2 = c 6 , since c 6 ∈ L(v 2 ).
Suppose now that C has an (L 0 : 2)-coloring, and for c ∈ {1, . . . , 6} let V c be the set of vertices of C on which the color c is used. Since two colors are used on each vertex of C, we have ∑ 6 c=1 |V c | = 10. On the other hand, V c is an independent set of a 5-cycle, and thus |V c | ≤ 2 for every color c. Furthermore, color 1 only appears in the lists of v 1 and v 2 , which are adjacent in C. It follows that |V 1 | ≤ 1. The situation is similar for color 2, which appears only in the lists of v 1 and v 5 , and also for color 3, which only appears in the lists of v 3 and v 4 . Consequently, ∑ 6 c=1 |V c | ≤ 3 · 2 + 3 · 1 = 9, which is a contradiction.
Corollary 4. Consider the gadget (G 1 , L 1 ), presented in Figure 1, where G 1 consists of a 5-cycle Proof. Let L be a half-list assignment for G 1 . First L-color y and x by colors c y ∈ L(y) and c x ∈ L(x)\{c y }, respectively. Since c x = c y and In an (L 1 : 2)-coloring, the vertex y would have to be assigned {1, 2} and x would have to be assigned {3, 4}, and thus the sets of available colors for v 1 and for v 3 would have to be {1, 2, 5, 6} and {3, 4, 5, 6}, respectively. However, no such (L 1 : 2) coloring of C exists according to Lemma 3.
Next we construct auxiliary gadgets, which will be combined with the gadget from Corollary 4 to deal with the case where L(v 1 ) = L(v 3 ). Let G be a graph, let S be a subset of vertices of G and L a list assignment for G. An L-coloring of S is a coloring of the subgraph of G induced by S. Moreover, if S is a subset of vertices of G that contains S and ϕ is an L-coloring of S , then ϕ extends ϕ if ϕ |S = ϕ. Let (G, L 0 ) be a gadget, let v 1 and v 3 be distinct vertices of G, and let S be a set of vertices of G not containing v 1 and v 3 . The gadget is (v 1 , v 3 , S)-relaxed if every half-list assignment L satisfies at least one of the following conditions. Proof. Let L be a half-list assignment for G 2 . If not all vertices of C 2 have the same list, then choose a color c ∈ L(y 1 ) \ L(y 2 ), and observe there exists an L-coloring of G 2 [V (C 2 ) ∪ {y 1 }] such that y 1 has color c. Let ψ 0 be the restriction of this coloring to {v 1 , v 3 }. Clearly, every L-coloring of {v 1 , v 3 , y 4 } extending ψ 0 extends to an L-coloring of G 2 , and thus (i) holds.
If all the vertices of C 2 have the same list (and hence in particular L(v 1 ) = L(v 3 )), then let c be a color in L(y 1 ) \ L(v 1 ). Observe that there exists an L-coloring of G 2 [{y 1 , y 2 , y 3 , y 4 }] such that y 1 has color c. Let ψ 0 be the restriction of this coloring to y 4 . Again, every L-coloring of {v 1 , v 3 , y 4 } extending ψ 0 extends to an L-coloring of G 2 , and thus (ii) holds.
It remains to show that if ϕ is an (L 2 : 2)-coloring of G 2 then ϕ(y 4 ) ∩ {7, 8} = ∅. Suppose, on the contrary, that ϕ(y 4 ) ∩ {7, 8} = ∅. It follows that ϕ(y 4 ) ∪ ϕ(y 3 ) = {1, 2, 3, 4}, and hence ϕ(y 2 ) = {7, 8}. As a result, ϕ(y 1 ) ⊆ {1, . . . , 6} and, by symmetry, we can assume that ϕ(y 1 ) = {5, 6}. This implies that ϕ(v) ⊆ {1, 2, 3, 4} for each v ∈ V (C 2 ). In particular, ϕ|V (C 2 ) is a (4 : 2)-coloring of C 2 , which is a contradiction since the 5-cycle C 2 has fractional chromatic number 5/2. Lemma 6. Consider the gadget (G 3 , L 3 ), obtained from the gadget (G 2 , L 2 ) of Lemma 5 as follows (see Figure 3 for an illustration of G 3 ). The graph G 3 consists of G 2 and for i ∈ {1, 2}, the vertices z i,1 , . . . , z i,7 ; the edges y 4 z i,1 and y 4 z i,2 ; the edge z i, j z i,k for every j and every k such that 1 ≤ j < k ≤ 4 and ( j, k) = (1,2); the edges of the triangle z i,5 z i,6 z i,7 and the edge z i,4 z i,5 . Let L 3 2, 3, 4, 7, 8} and L 3 Proof. Let L be a half-list assignment for G 3 . The gadget (G 2 , L 3 |V (G 2 )) is (v 1 , v 3 , {y 4 })-relaxed by Lemma 5. Suppose first that (i) holds for the restriction of L to G 2 (with S = {y 4 }), and let ψ 0 be the corresponding L-coloring of {v 1 , v 3 }. For i ∈ {1, 2}, if L(z i,1 ) ∩ L(z i,2 ) = ∅, then let c i be a color in L(z i,1 ) ∩ L(z i,2 ). Otherwise, |L(z i,1 ) ∪ L(z i,2 )| = 4 > |L(z i,3 )|, and thus we can choose a color c i ∈ (L(z i,1 ) ∪ L(z i,2 )) \ L(z i,3 ). Let c be a color in L(y 4 ) \ {c 1 , c 2 }. By (i) for G 2 , we know that ψ 0 extends to an L-coloring ψ of G 2 such that ψ(y 4 ) = c. If L(z i,1 ) ∩ L(z i,2 ) = ∅, then color both z i,1 and z i,2 by c i , otherwise color one of them by c i and the other one by an arbitrary color from its list that is different from c. There are at least two colors in L(z i,4 ) distinct from the colors of z i,1 and z i,2 , choose such a color c i so that L(z i,5 ) \ {c i } = L(z i,6 ). Color z i,4 by c i and extend the coloring to z i,3 , which is possible by the choice of c i . Observe that any L-coloring of z i,7 extends to an L-coloring of the triangle z i,5 z i,6 z i,7 where the color of z i,5 is not c i . We conclude that (G 3 , L 3 ) with the half-list assignment L satisfies (i). Suppose next that (ii) holds for the restriction of L to G 2 (with S = {y 4 }), and let ψ 0 be the corresponding L-coloring of y 4 . For i ∈ {1, 2}, greedily extend ψ 0 to an L-coloring of z i,1 , . . . , z i,7 in order, and let ψ 0 be the restriction of the resulting coloring to {z 1,7 , z 2,7 }. Observe that (G 3 , L 3 ) with the half-list assignment L satisfies (ii).