Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I refer to example 4, fig.3.6, p.17 of Munkres' Algebraic Topology. He says the given triangulation scheme "does more than paste opposite edges together".

Not clear to me. For those who don't have the book to hand, a rectangle is divided into 6 equal squares by a horizontal midline and two verticals; each square has a south-west to north-east diagonal.

enter image description here

share|cite|improve this question
Any chance you could add the picture? – Michael Albanese Jun 5 '14 at 7:10
I believe the issue is that if a simplex has vertices $\{v_0, ..., v_n\}$, then that collection of vertices should uniquely identify that $n$-simplex. In the case of the surface, subdivided as you have described, this fails. For example, if you take a vertex on the "midline" and the vertex directly above it (which is the same as the one below it by gluing), those two vertices should belong to a unique segment. But, they don't. – Sammy Black Jun 5 '14 at 7:21
@MichaelAlbanese Got to figure out how to do it. But it's really a standard sort of triangulation diagram. – InTransit Jun 6 '14 at 15:21
@SammyBlack Thanks, I'll see if this sort of argument works on other non-triangulations. – InTransit Jun 6 '14 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.