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Why is this a wrong Triangulation ?

enter image description here

I have to say, we had triangulation at the end of the topology course, so not in details. And the professor only mentioned the basic rules for the triangulation, but in this case, idon't know why it fails.

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For a triangulation, we require that triangles intersect either at one point or one edge. In the diagram, four of the triangles have the points $1$ and $4$ as vertices, and four of the six possible pairs of these triangles intersect there but not along an edge connecting $1$ and $4$.

$T_1$ and $T_2$ pictured also show that this is not a triangulation, since they intersect only at the points $1$ and $2$.

I am assuming that the only identifications being made are those shown, i.e. the points. If the boundary edges are identified then I believe what I have said in the first paragraph is still true. Triangles corresponding to $134$ and $124$ will intersect only at the points $1$ and $4$, although the intersection of $T_1$ and $T_2$ will then be the edge $12$ which is fine.

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all the triangles have the points 1 'or' 4 as vertices. and It is really hard to imagine, where exactly it fails. – derivative Feb 11 '14 at 14:25
Good point, sorry, what I should have said was that there are triangles which have $1$ and $4$ as a vertex, but do not intersect along an edge connecting $1$ and $4$. I realise now that the surface is not a sphere, sorry. See my post which should now be edited to be true. – Tom Oldfield Feb 11 '14 at 14:44

Let's write down the abstract simplicial complex associated to your drawing (note that an abstract simplicial complex is a set of finite sets and so repeats will be omitted). We have the $2$-cells, $$\{1,2,3\},\{2,3,4\},\{1,2,4\},\{1,3,4\}$$ the $1$-cells $$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$$ and the $0$-cells $$\{1\},\{2\},\{3\},\{4\}.$$

Let's suppose the geometric realisation of this simplicial set is the torus. We know that the Euler characteristic of the torus is $0$, but by using the formula $\chi=V-E+F$ we apparently get $\chi=4-6+4=2$. So something has gone wrong and indeed, the geometric realisation of the above simplicial complex is actually homeomorphic to the sphere (in fact it's the boundary of the usual $3$-simplex).

The problem is that you can't have a triangulation where any pair of the triangles (simplices) share a common boundary. Certainly $\partial\Delta_i\cap\partial\Delta_j$ can be non-empty, but we can not have $\partial\Delta_i=\partial\Delta_j$ for $i\neq j$ as otherwise we don't have a true simplicial complex. (Note, this notion of cells sharing a common boundary is only disallowed for simplicial complexes - later on, you will hopefully be introduced to the broader notion of CW-complexes which do allow cells to intersect completely and in fact individual cells can even self intersect on their boundary.)

It's helpful to remember that any vertices which share a label will be identified in the geometric realisation, but so too will the edges and higher faces which share the same labels which are purely determined by their boundary vertices. So in your picture you would be identifying edges and faces which you do not really want to identify when forming the torus.

What's the upside of this though? Well, you've found a really nice quotient map from the torus to the $2$-sphere :).

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The upper left and lower right triangles both have vertices $(1, 2, 3)$. In a triangulation, no two triangles should have the same set of vertices.

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