Proof regarding the genus of a topological space












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Intuitively, I would think that if there exists a continuous mapping $f:Xrightarrow Y$ between 2 topological spaces $X$ and $Y$, then the genus of $Y$ must be smaller that the Genus of $X$ (as during the mapping, holes in $X$ can be eliminated but no new holes may be made). However, I am finding it difficult to prove this formally. I was thinking of using the transformation of loops in $X$ to $Y$ by applying $f$, but I cannot continue from this. As such, how can one prove this formally?






proof-verification geometric-topology







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  • 1




    $begingroup$
    Holes can be made. You can wrap a line around a circle as many times as you like.
    $endgroup$
    – Ittay Weiss
    Jan 5 at 7:36
















0












$begingroup$


Intuitively, I would think that if there exists a continuous mapping $f:Xrightarrow Y$ between 2 topological spaces $X$ and $Y$, then the genus of $Y$ must be smaller that the Genus of $X$ (as during the mapping, holes in $X$ can be eliminated but no new holes may be made). However, I am finding it difficult to prove this formally. I was thinking of using the transformation of loops in $X$ to $Y$ by applying $f$, but I cannot continue from this. As such, how can one prove this formally?






proof-verification geometric-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Holes can be made. You can wrap a line around a circle as many times as you like.
    $endgroup$
    – Ittay Weiss
    Jan 5 at 7:36














0












0








0





$begingroup$


Intuitively, I would think that if there exists a continuous mapping $f:Xrightarrow Y$ between 2 topological spaces $X$ and $Y$, then the genus of $Y$ must be smaller that the Genus of $X$ (as during the mapping, holes in $X$ can be eliminated but no new holes may be made). However, I am finding it difficult to prove this formally. I was thinking of using the transformation of loops in $X$ to $Y$ by applying $f$, but I cannot continue from this. As such, how can one prove this formally?






proof-verification geometric-topology







share|cite|improve this question









$endgroup$




Intuitively, I would think that if there exists a continuous mapping $f:Xrightarrow Y$ between 2 topological spaces $X$ and $Y$, then the genus of $Y$ must be smaller that the Genus of $X$ (as during the mapping, holes in $X$ can be eliminated but no new holes may be made). However, I am finding it difficult to prove this formally. I was thinking of using the transformation of loops in $X$ to $Y$ by applying $f$, but I cannot continue from this. As such, how can one prove this formally?






proof-verification geometric-topology




proof-verification geometric-topology






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asked Jan 5 at 7:26









Aryaman GuptaAryaman Gupta

336




336








  • 1




    $begingroup$
    Holes can be made. You can wrap a line around a circle as many times as you like.
    $endgroup$
    – Ittay Weiss
    Jan 5 at 7:36














  • 1




    $begingroup$
    Holes can be made. You can wrap a line around a circle as many times as you like.
    $endgroup$
    – Ittay Weiss
    Jan 5 at 7:36








1




1




$begingroup$
Holes can be made. You can wrap a line around a circle as many times as you like.
$endgroup$
– Ittay Weiss
Jan 5 at 7:36




$begingroup$
Holes can be made. You can wrap a line around a circle as many times as you like.
$endgroup$
– Ittay Weiss
Jan 5 at 7:36










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This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map)
$fcolon mathbb{R}^2to mathbb{R}^2/mathbb{Z}^2 cong T$ to the torus, where we identify $(x,y)sim(tilde x,tilde y)$ iff $x-tilde x in mathbb Z$ and $y-tilde y in mathbb Z$.



This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.






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    $begingroup$

    This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map)
    $fcolon mathbb{R}^2to mathbb{R}^2/mathbb{Z}^2 cong T$ to the torus, where we identify $(x,y)sim(tilde x,tilde y)$ iff $x-tilde x in mathbb Z$ and $y-tilde y in mathbb Z$.



    This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map)
      $fcolon mathbb{R}^2to mathbb{R}^2/mathbb{Z}^2 cong T$ to the torus, where we identify $(x,y)sim(tilde x,tilde y)$ iff $x-tilde x in mathbb Z$ and $y-tilde y in mathbb Z$.



      This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map)
        $fcolon mathbb{R}^2to mathbb{R}^2/mathbb{Z}^2 cong T$ to the torus, where we identify $(x,y)sim(tilde x,tilde y)$ iff $x-tilde x in mathbb Z$ and $y-tilde y in mathbb Z$.



        This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.






        share|cite|improve this answer









        $endgroup$



        This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map)
        $fcolon mathbb{R}^2to mathbb{R}^2/mathbb{Z}^2 cong T$ to the torus, where we identify $(x,y)sim(tilde x,tilde y)$ iff $x-tilde x in mathbb Z$ and $y-tilde y in mathbb Z$.



        This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 10:12









        BabelfishBabelfish

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        1,128520






























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