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At page 17 of Munkres' Elements of Algebraic topology it says, referring to fig. 3.6, that "the diagram does not determine [the torus]. It does more than paste opposite edges together": enter image description here

Is it so? I tried to glue them together, but end up with a normal torus.

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If memory serves, Munkres' point is that the resulting diagram is not a triangulation: For example, triangles $abd$ and $ade$ meet at two vertices ($a$ and $d$) but do not share an edge. –  user86418 Feb 22 at 16:56
    
It may be but it is strange, since he introduces triangulation long after that example –  user130753 Feb 22 at 17:03
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Let's count vertices, edges and faces.

The vertex-set is $\{a,b,c,d,e,f\}$. So $V=6$.

The edge-set is $\{ab,ac,ad,ae,af,bc,bd,be,bf,cd,ce,cf,de,df,ef\}$. So, $E=15$.

The face-set is $\{abd,abe,acd,acf,ade,adf,bce,bcf,bde,bef,cdf,cef,\}$. So $F=12$.

This gives the Euler characteristic as $\chi=V-E+F=6-15+12=3$. It follows that this space can not be the torus as the torus has Euler characteristic equal to $0$.

The reason this doesn't work as a cellular model for the torus is that there are edges which are identified but which we would not want to identify in the usual model of the torus. For instance, the two appearances of the edge $be$ in the model mean we would identify edges which we would ordinarily consider as lying on the surface of the torus as two intervals which compose a meridional circle.

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I guess it looks like those flotational armbands kids use for swimming: thumbs2.ebaystatic.com/d/l225/m/m1hWZ-0ap0iX06N1BJ2588w.jpg –  Neal Feb 23 at 12:53
    
I don't understand, why doesn't the upper half just folds on the lower half, thus creating a rectangular foil? –  user130753 Feb 23 at 14:04
    
@user130753 for instance the face $bde$ exists in the top half, but has no corresponding face in the bottom half (in fact each face in the model only appears once). –  Daniel Rust Feb 23 at 14:09
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