Closed orientable 2-manifold












0














Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?



I want a prove with triangulation.










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  • 2




    The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
    – Nick
    Nov 21 '18 at 22:55










  • @Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
    – hussein
    Nov 21 '18 at 23:32
















0














Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?



I want a prove with triangulation.










share|cite|improve this question




















  • 2




    The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
    – Nick
    Nov 21 '18 at 22:55










  • @Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
    – hussein
    Nov 21 '18 at 23:32














0












0








0







Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?



I want a prove with triangulation.










share|cite|improve this question















Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?



I want a prove with triangulation.







geometry differential-geometry smooth-manifolds






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edited Nov 21 '18 at 23:04







hussein

















asked Nov 21 '18 at 22:44









husseinhussein

576




576








  • 2




    The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
    – Nick
    Nov 21 '18 at 22:55










  • @Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
    – hussein
    Nov 21 '18 at 23:32














  • 2




    The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
    – Nick
    Nov 21 '18 at 22:55










  • @Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
    – hussein
    Nov 21 '18 at 23:32








2




2




The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 at 22:55




The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 at 22:55












@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 '18 at 23:32




@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 '18 at 23:32










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By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.






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    By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.






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      By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.






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        By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.






        share|cite|improve this answer












        By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.







        share|cite|improve this answer












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        answered Nov 23 '18 at 0:35









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