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
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
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add a comment |
$begingroup$
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
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
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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
add a comment |
$begingroup$
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
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
$endgroup$
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
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
asked Jan 5 at 7:26
Aryaman GuptaAryaman Gupta
356
356
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
add a comment |
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
add a comment |
1 Answer
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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 $1$.
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
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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 $1$.
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
$endgroup$
add a comment |
$begingroup$
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 $1$.
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
$endgroup$
add a comment |
$begingroup$
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 $1$.
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
$endgroup$
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 $1$.
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
answered Jan 7 at 10:12
BabelfishBabelfish
1,128520
1,128520
add a comment |
add a comment |
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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