On 3-colorable planar graphs without short cycles

Let G be a graph. It was proved that if G is a planar graph without {4, 6, 7}-cycles and without two 5-cycles sharing exactly one edge, then G 3-colorable. We observed that the proof of this result is not correct. 1 Let G be a simple graph with vertex set G. A planar graph is one that can be drawn on a plane in such a way that there are no “edge crossings,” i.e. edges intersect only at their common vertices. A k-face is a face whose boundary has k edges. A cycle C in a planar graph G is said to be separating, if int(C) 6= ∅ and ext(C) 6= ∅, where int(C) and ext(C) denote the sets of vertices located inside and outside C, respectively. Let Ci denote an i-cycle. A k-coloring of G is a mapping c from V (G) to the set {1, . . . , k} such that c(x) 6= c(y) for any adjacent vertices x and y. The graph G is k-colorable if it has a k-coloring. Let G denote the class of planar graphs without {4, 6, 7}-cycles and without two 5-cycles sharing exactly one edge. The following theorem was proved in [1]. Theorem. Let G be a graph in G that contains 5-cycles. Then every proper 3coloring of the vertices of any 3-face or 9-face of G can be extended into a proper 3-coloring of the whole graph.