Theorem B

A surface that is disconnected by every closed curve on it is topologically equivalent to a sphere.

Proof: Let S be a surface such that every closed curve on S disconnects it. We wish to show that S is a sphere.

Claim: X(S) = 2

Let M and C be as in theorem A. since X(S) = X(M) + X(C) it follows that if X(S)=/=2 , then X(C)=/=1. So C is not a tree. Thus C contains a closed loop. This loop disconnects S into 2 pieces. But each of these pieces must contain a dual vertex. These have to be joined in M, so M must cut through the loop in C. this is a contradiction as M and c are disjoint. Therefore, X(S) = 2.

It follows now that X(C)= 1 and hence C is a tree.

If you take a tree and ‘flatten it up’ a little as in the **figure**, the result is topologically equivalent to a disc. Now define two subsets of S as follows: a point of S lies in X if it is nearer to M than C. a point of S lies in Y if it is nearer to C than M. Each of X and Y is a flattening of M or C, so is topologically a disc. Further X and Y meet along their edges. So, S is topologically equivalent to two discs sewn edge to edge – which is a sphere.