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For any triangulation of a compact surface, show that

$$v \leq f$$

where $v,e,f$ is vertices, edges and triangle respectively.

Since each triangle has 3 edges and each edge is the common edge of exactly two triangles, we get $3f=2e$ and $v-e+f= \chi$. Hence $v=\dfrac{1}{2}f+\chi$ but i can't continue to indicate $f \geq v$

Desunkid
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  • Shouldn't it be $v=\dfrac12,f+\chi$? Then, maybe, you could try to prove that any triangulation yields $f\geq 2\chi$ instead. – Batominovski Sep 29 '18 at 14:54
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    However, it may be very helpful to know that $\chi\leq 2$. There is a bit of a problem. There exists a triangulation of the sphere ($\chi=2$) with two triangles whose boundaries coincide. This give $v=3$, $e=3$, and $f=2$. This is the only case that violates $v\le f$. – Batominovski Sep 29 '18 at 15:07
  • oh, thank you. Should I claim that $f \geq 4$? Could you give me more detail? – Desunkid Sep 29 '18 at 15:57
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    You don't need to claim $f\geq 4$. For a nonspherical compact surface, $\chi\leq 1$ (see here, so you need $f\geq 2$, which is obvious. For a sphere, you just have to exclude the case $f=2$ since the inequality is false. Other cases work fine. (Since $f$ is even for a sphere, we end up with $f\ge 4$ anyhow.) – Batominovski Sep 29 '18 at 16:06
  • oh, thank you @Batominovski – Desunkid Sep 29 '18 at 16:14

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