Gruber P. Convex and Discrete Geometry

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34 Koebe’s Representation Theorem for Planar Graphs 503
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Fig. 34.3. Country circle and vertex circles
is open. Let w be one of its points. Then there are points x, y ∈ C 2 , x �= y,
with f (x) = f (y) = w. Choose disjoint neighbourhoods U of x and V of y
and a neighbourhood W of w such that f maps each of U, V homeomorphically
onto W . Then, clearly, W ⊆ � z : # f −1 (z) ≥ 2 � . Secondly, it will be shown that
the set � z : # f −1 (z) = 1 � is also open. For suppose not. Then there are points
z, zn, n = 1, 2,..., with zn → z and # f −1 (z) = 1, # f −1 (zn) ≥ 2 for all n.
Let {x} = f −1 (z) and choose neighbourhoods U of x and W of z such that f
maps U homeomorphically onto W . By omitting finitely many indices and renumbering,
if necessary, we may assume that there are xn ∈ U, yn ∈ C\U such that
f (xn) = f (yn) = zn and xn → x. Since � y : # f −1� f (y) � ≥ 2 � is bounded by
assumption, by considering a suitable subsequence and renumbering, if necessary,
we may assume that yn → y ∈ C\U, say. The continuity of f then implies that
f (x) = f (y) = z, while x �= y. This is a contradiction. C is thus the disjoint union
of the two open sets � z : # f −1 (z) ≥ 2 � and � z : # f −1 (z) = 1 � . By assumption,
� z : # f −1 (z) ≥ 2 � is bounded. Thus � z : # f −1 (z) = 1 � �=∅. Since C is connected,
it follows that � z : # f −1 (z) ≥ 2 � =∅. Hence f is one-to-one, concluding the proof
of (2).
The third step is to show the following topological result on graphs.
(3) Let H be the image of G ′ in C under a (graph) isomorphism f , possibly
with edge crossings, which has the following properties:
(i) All edges of H are polygonal arcs.
(ii) For each vertex v of G ′ the images in H of the edges that leave v are
pairwise non-crossing and leave f (v) in the same clockwise order as
their originals in G ′ .
(iii) The image of each country cycle in G ′ is a closed Jordan curve in H.
(iv) If C is the boundary cycle of a bounded country of G ′ and E an edge
of G ′ leaving C, then the first segment of f (E) is in the exterior of
f (C).