Simplicial Structures in Topology

186 V Triangulable Manifolds
we take the cylinder that gives rise to K and glue it to M in such a manner that the
circles that bound the cylinder are glued in opposite directions (one clockwise, the
other anticlockwise, as seen in Fig. V.14). Now, the spaces obtained by the addition
of a cylinder to M by these two procedures are homeomorphic to each other.
(V.2.5) Theorem. Any closed surface is homeomorphic to one of the following
surfaces:
(i) S 2
(ii) nT := T 2 #...#T 2 (n times)
(iii) nRP 2 := RP 2 #...#RP 2 (n times)
Proof. Let S be a surface and K one of its triangulations. The complex K has
a finite number of vertices, 1-simplexes, and 2-simplexes; for simplicity of notation,
we denote by Ti (respectively, ℓi) the geometric realizations of the 2-simplexes
(respectively, the 1-simplexes) of K; these are called triangles and sides. Let us go
through some preliminary considerations.
We begin by observing that each side of |K| is only a side of two triangles (see
Theorem (V.1.5)). Now, let v be a vertex of K; then the triangles with the vertex v
may be arranged in cyclic order T1,T2,...,Tr = T0 so that for each i ∈{1,...,r},the
intersection Ti−1 ∩ Ti is the side ℓi. In fact, we note that, if there were two such sets
of triangles around v, they would have only the point v in common; therefore, by
removing v from their union, we would have a non-connected space; on the other
hand, since S is a surface, there would be an open set U ⊂ S that would contain v
and be homeomorphic to an open disk of R 2 ; hence U � v would be connected, a
contradiction. It follows that the neighbourhood of every simple cycle of sides in
|K| is homeomorphic to the cylinder I × S 1 or the Möbius band.
We now consider the one-dimensional simplicial complex G (namely, the graph)
defined as follows: the vertices of G are all triangles of K, and for each pair of
adjacent triangles (that is to say, with one side in common) T1 and T2 of K, we
add the 1-simplex {T1,T2} to G. ThecomplexG may be embedded in S ∼ = |K| by
sending the triangle Ti (viewed as a vertex of G) to its barycentre b(Ti) (viewed as a
point of |K|). Through a barycentric subdivision of K, the 1-simplexes of G will be
represented by a broken line that joins the two barycentres and crosses the common
side of T1 and T2.
We now recall the concept of spanning tree (or maximal tree). A tree of |K| is
a contractible unidimensional subpolyhedron |L| ⊂|K|; the existence of trees in a
polyhedron |K| is assured by the fact that the geometric realization of any 1-simplex
of K, being homeomorphic to the disk D 1 , is contractible. The trees of a complex
are partially ordered by inclusion; a tree is called spanning if it is not contained in a
tree strictly larger; since K is finite, the existence of a spanning tree is also certain.
(V.2.6) Lemma. If |K| is path-connected and |L| is a spanning tree, then L contains
all vertices of K.
Proof. We suppose a to be a vertex of K but not of L. Let us take any vertex b of
L; since|K| is path-connected, there is a path γ : I →|K| joining a and b. Bythe