Finite maps equal compact surfaces

In the earlier post What is a regular map? we stated a side result without proving it. Let’s do that now.

Prop. A topological map is finite if and only if the surface is compact.

Proof. Suppose that the surface is compact and the map is infinite; in other words is infinite. Since the valency of all the vertices is finite (AG3) we conclude that is infinite. This implies that there is an accumulation point on for the set . Let be a converging sequence of vertices; (TM1) implies that for large enough,  these vertices must all be joined to each other and, what is more, they must all have valency 2 (think of the edges between them forming a single line on which they all lie). Then, for large enough every lies on two darts and (provided we label darts appropriately) it is clear that for and large enough. This is a contradiction of (TM4).

Now for the converse; we suppose that the map is finite. Let be a face of the map; (TM4) implies that there are a finite number of darts for which . We list these darts:. Now (TM3) implies that is homeomorphic to an open disc; it is clearly sufficient to show that is homeomorphic to a closed disc (since then the surface is homeomorphic to a finite union of compact surfaces and so is itself compact).

Now why is is homeomorphic to a closed disc? First of all, there is the possibility that and the edge is a loop. In this case the result is clear. Suppose this is not the case – then all edges are segments or free-edges. With a little thought it should be clear that the edges which are segments form a closed loop; the free-edges can be thought of as spikes coming off this closed loop, “poking into” the face . This union of segments clearly forms the boundary of ; the required homeomorphism can then be obtained by “pinching” the face round each of the free-edges, and extending it smoothly to the union of segments. Hopefully the principle is clear.

QED

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