# simplicial complex $K$ whose underlying space is homeomorphic to the cylinder (Example 1 page 16 Munkres)

I would like to ask about the Example 1 of Munkres' Elements of Algebraic Topology: Suppose we wish to indicate a simplicial complex $K$ whose underlying space is homeomorphic to the cylinder. Two ways of doing so are to draw pictures as in figure 3.1 and 3.2, which specifies $K$ as a collection consisting of six 2-simplices and their faces. My questions are:

(1) How can one triangulate the rectangle in Figure 3.2 such way? Is there any rules? Suppose I replace the $1$-simplex spanned by $\{d,c\}$ with 1-simplex spanned by $\{b,e\}$ is that OK?

(2) Why do we have six $2$-simplices? Why can't I have, for example, four or eight $2$-simplices?

• It'll be hard to answer your question (1) about Figure 3.2 when we cannot see Figure 3.2. For question (2), the existence of examples with six 2-simplices is just an existence result. It does not rule out the existence of examples with four 2-simplices, or with eight 2-simplices, or with 8329983470 2-simplices. Those numbers of 2-simplices have to be investigated in their own right. – Lee Mosher Apr 16 '16 at 17:35
• @LeeMosher I have added the pictures. – Chen M Ling Apr 16 '16 at 17:41

The main rule in a triagulation is that each simplex has to be unqiuely determined by its vertices. So for example, you cannot triangulate a circle by taking two one-simplices and identify their endpoints. (This is OK as a cell decomposition, but not as a triangulation.) Hence the simples triangulation of a circle is as a triangle, but of course you can also triangulate it as a square or any $n$-gon. Hence there is no triangulation of $S^1$ with less than 3 one-simplices, but you can have more one-simplices without problems.
The triangulation of the cylinder you consider basically is obtained by taking the simplest triangulation of $S^1$, forming the product with a closed intervall, and then dividing each of the resulting squares into two triangles. It does not matter how you do this subdivision, so the answer to your question (1) is yes. You can do the same thing starting from the triangulation of $S^1$ as a square, thus obtaining a triangulation of the cylinder with $8$ two-simplices and so on.
There always are lots of different triangulations of a topological space, but you can get topological invariants out of them. In this example, the simplest of these would be the alternating sum of the numbers of simplices in each dimension which equals the Euler-characteristic (which happens to be zero for both $S^1$ and the cylinder.
• Just wanna make sure. "...dividing each of the resulting squares into two triangles." So there's no rules in dividing each square into two triangles as well? So, I can replace the line segment $dc$ with $be$ and replace $ea$ with $cf$ and I still have a triangulation of a cylinder? – Chen M Ling Apr 17 '16 at 12:38