The homotopical proof of the fundamental theorem of algebra
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I am reading a homotopical proof of the fundamental theorem of algebra, and the end of the proof is as follows:
... the map $zmapsto r^nz^n$ (where $r$ is a positive real number and $zin S^1$) is homotopic to a constant map $zmapsto a_0$ as maps from $S^1$ to $mathbb R^2-0$. But $mathbb R^2-0cong S^1$,...
So does it follow from here that the maps from $S^1$ to itself given by $zmapsto z^n$ and $zmapsto c$ for a constant $c$ are homotopic, and since the fundamental group of $S^1$ is $mathbb Z$, the first map induces a nonzero map while the constant map induces the zero map which is contradictory?
proof-verification algebraic-topology homotopy-theory
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add a comment |
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I am reading a homotopical proof of the fundamental theorem of algebra, and the end of the proof is as follows:
... the map $zmapsto r^nz^n$ (where $r$ is a positive real number and $zin S^1$) is homotopic to a constant map $zmapsto a_0$ as maps from $S^1$ to $mathbb R^2-0$. But $mathbb R^2-0cong S^1$,...
So does it follow from here that the maps from $S^1$ to itself given by $zmapsto z^n$ and $zmapsto c$ for a constant $c$ are homotopic, and since the fundamental group of $S^1$ is $mathbb Z$, the first map induces a nonzero map while the constant map induces the zero map which is contradictory?
proof-verification algebraic-topology homotopy-theory
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Yes. (filler characters)
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– Pig
Jan 12 at 1:06
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Thanks (fill character too) @Pig
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– User12239
Jan 12 at 10:31
add a comment |
$begingroup$
I am reading a homotopical proof of the fundamental theorem of algebra, and the end of the proof is as follows:
... the map $zmapsto r^nz^n$ (where $r$ is a positive real number and $zin S^1$) is homotopic to a constant map $zmapsto a_0$ as maps from $S^1$ to $mathbb R^2-0$. But $mathbb R^2-0cong S^1$,...
So does it follow from here that the maps from $S^1$ to itself given by $zmapsto z^n$ and $zmapsto c$ for a constant $c$ are homotopic, and since the fundamental group of $S^1$ is $mathbb Z$, the first map induces a nonzero map while the constant map induces the zero map which is contradictory?
proof-verification algebraic-topology homotopy-theory
$endgroup$
I am reading a homotopical proof of the fundamental theorem of algebra, and the end of the proof is as follows:
... the map $zmapsto r^nz^n$ (where $r$ is a positive real number and $zin S^1$) is homotopic to a constant map $zmapsto a_0$ as maps from $S^1$ to $mathbb R^2-0$. But $mathbb R^2-0cong S^1$,...
So does it follow from here that the maps from $S^1$ to itself given by $zmapsto z^n$ and $zmapsto c$ for a constant $c$ are homotopic, and since the fundamental group of $S^1$ is $mathbb Z$, the first map induces a nonzero map while the constant map induces the zero map which is contradictory?
proof-verification algebraic-topology homotopy-theory
proof-verification algebraic-topology homotopy-theory
asked Jan 11 at 23:37
User12239User12239
453216
453216
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Yes. (filler characters)
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– Pig
Jan 12 at 1:06
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Thanks (fill character too) @Pig
$endgroup$
– User12239
Jan 12 at 10:31
add a comment |
$begingroup$
Yes. (filler characters)
$endgroup$
– Pig
Jan 12 at 1:06
$begingroup$
Thanks (fill character too) @Pig
$endgroup$
– User12239
Jan 12 at 10:31
$begingroup$
Yes. (filler characters)
$endgroup$
– Pig
Jan 12 at 1:06
$begingroup$
Yes. (filler characters)
$endgroup$
– Pig
Jan 12 at 1:06
$begingroup$
Thanks (fill character too) @Pig
$endgroup$
– User12239
Jan 12 at 10:31
$begingroup$
Thanks (fill character too) @Pig
$endgroup$
– User12239
Jan 12 at 10:31
add a comment |
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Yes. (filler characters)
$endgroup$
– Pig
Jan 12 at 1:06
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Thanks (fill character too) @Pig
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– User12239
Jan 12 at 10:31