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The Euler characteristic of $S^1$ should be equal to that of a triangle, the two being homeomorphic to each other. But it is $1$ for the triangle and zero for $S^1$; how does it make sense?

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how are you getting that a triangle has euler characteristic 1? – Eric O. Korman Aug 14 '12 at 23:21
It depends on your definition of a triangle. How many faces does your triangle have? – M Turgeon Aug 14 '12 at 23:21

If by “triangle” you mean “three points and the segments connecting each pair of them”, then the Euler characteristic of it is $0$ (it has three edges and three vertices).

If by “triangle” you mean the same as above, plus the part of plane enclosed by the segment, then it is certainly not homeomorphic to $S^1$, but to a two-dimensional disk $D^2$, and it has Euler characteristic $1$, just like the disk.

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