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The following is taken from Munkres' Algebraic Topology book.enter image description here

I tried to determine which spaces (e.g. Mobius Strip, Klein bottle, etc) these complexes are, but to no avail.

I computed the Euler Characteristics (V-E+F) to be -1, -1, 1, -1 respectively. (top left, top right, bottom left, bottom right). However, I don't know any spaces with $\chi=-1$. The one with $\chi =1$ could be either a disk or a projective plane.

I am guessing that they may be trick questions (i.e. not valid triangulations). CW complexes are not discussed yet in the book, hence this rules out CW complexes or cell complexes.

Thanks for any help!

yoyostein
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    The first two of the complexes are moebius strips, the second is $\Bbb RP^2$ and the last of them is the torus. – Balarka Sen Aug 31 '15 at 12:33
  • Thanks! How did you derive at the answers? – yoyostein Aug 31 '15 at 12:44
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    I'm not sure how you calculated your Euler characteristics. I get 0, 0, 1, 0. –  Aug 31 '15 at 17:38
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    Regardless of whether the figure describes a valid triangulation, forming the quotient space makes sense and asking what space the quotient is homeomorphic to also makes sense. – Lee Mosher Aug 31 '15 at 17:42
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    @BalarkaSen : are you sure that the last one is a torus? See here or there. – Watson Jun 13 '16 at 15:58
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    @Watson Yes, the identification space is clearly a torus. The diagram is not a valid triangulation of the torus is all. You'll see answers and comments below both of the questions you linked says exactly this, if you read carefully. – Balarka Sen Jun 14 '16 at 02:55
  • Correct me if I'm wrong, but I don't think the last one is a torus. Based on what Munkres mentions in Section 3 Example 1, such figure can be considered as a way of denoting the corresponding abstract complex. By the definition of abstract complex, any simplex is uniquely determined by its vertices. Thus in Figure 3.14, for example, the two triangles with vertices $a$,$b$,$d$ should be identified as the same triangle. Based on this idea, the whole upper strip is actually identified to the lower strip. Finally, after identifying the edge $ad$ as well, I would say the the space is a cylinder. – Hopf eccentric Nov 29 '19 at 18:13
  • Of course, a cylinder and a torus have the same Euler characteristic. If we would guess the answer based on the Euler characteristic, a torus is a reasonable guess. – Hopf eccentric Nov 29 '19 at 18:19

1 Answers1

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the determined simplicial complex are homeomorphic to the following: a quotient space(three faces of a tetrahedron ), same as the first one, four faces of apyramid, cylinder

the answer of balarka didn't realize that the correponding simplicial map not only paste the edge together,but also paste the interior of the triangle together.

hhzt
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  • which one is wrong – hhzt Feb 08 '23 at 06:55
  • Sorry,I don't know whether you have read this section of the Munkres' algebraic topology,the complex in the figure is of dimention 2.You can see this in the example of this section. – hhzt Feb 08 '23 at 07:18
  • Please correct me if I am wrong.The meaning of the diagram is the following(I use figure 3.11 as an example):from the figure,we can form an abstract simplicial complex S,the collection of {a,b,c},{b,c,d},{a,c,d} and all none empty subsets of them.And S has a geometric realization(a simplicial complex K).Let f be the vertex map of L to K(we identify L as the complex in figure 3.11),the corresponding simplicial map from |L| to |K| is a quotient map,and this map paste the leftest triangle and the rightest triangle together(so,the 2-simplex is invovled). – hhzt Feb 08 '23 at 08:00
  • And I just wrote some spaces homeomorphic to |K|.However,the third one of my answer is wrong,it's four faces of a pyramid if we want to embed it into R3. – hhzt Feb 08 '23 at 08:00
  • @TedShifrin Munkres explains (on p. 18) that the identifications happen potentially for simplices of any dimension, not just dimensions 0 and 1. In figure 3.11, the two triangles with vertices $a$, $b$, and $c$ will be glued together. Same with the triangles with vertices $b$, $c$, and $d$ in figure 3.12. – John Palmieri Feb 08 '23 at 20:37
  • I was hasty. I will delete my comments. – Ted Shifrin Feb 08 '23 at 21:21