Proof of Borsuk-Ulam Theorem using singular Homology
$begingroup$
I have to prove the Borsuk-Ulam theorem following some specific steps.
The theorem says that:
For every continuous function $f: S^n rightarrow mathbb{R}^n$ there exists $x in S^n$ with $f(-x)=f(x)$.
I am using the cover of $mathbb{R}P^n$ by $S^n$. I showed every singular simplex $tildesigma$ of $mathbb{R}P^n$ can be lifted to $sigma$ and $tau sigma$ singular simplexes of $S^n$.
This induces an application $t:C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2), tildesigma rightarrow sigma + tausigma$ which is a chain complex morphism and induces a short exact sequence $$0 rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2) rightarrow 0$$ where the latter morphism is induced by the covering map. I managed to show all this. This implies directly existence of a long exact sequence in homology.
I should then show that $t circ p_star: C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(S^n,mathbb{F}_2)$ is zero where $p_star$ is induced by the covering map. Can you give me a hint for this?
I should then show that in the long exact sequence in hompology $$0=H_{n+1}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2) rightarrow H_{n}(S^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n-1}(mathbb{R}P^n,mathbb{F}_2)rightarrowcdots$$ the first and third morphisms are zero and the second and fourth isomorphisms. As the second is induced by $t$ and the third by $p_star$, I could already show that $p_star$ is zero, as $t$ is injective and the composition $t circ p_star$ zero. Thus $t$ is an isomorphism. Furthermore, the fourth morphism, $delta$ is injective. But how do I show its surjectivity?
Finally I should conclude that if $f: S^n rightarrow S^m$ is continuous such that, for every $x in S^n$, $f(-x)=-f(x)$, then $nleq m$, using "naturality of the long exact sequence". Could you please give me some hints?
algebraic-topology homology-cohomology
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
$endgroup$
add a comment |
$begingroup$
I have to prove the Borsuk-Ulam theorem following some specific steps.
The theorem says that:
For every continuous function $f: S^n rightarrow mathbb{R}^n$ there exists $x in S^n$ with $f(-x)=f(x)$.
I am using the cover of $mathbb{R}P^n$ by $S^n$. I showed every singular simplex $tildesigma$ of $mathbb{R}P^n$ can be lifted to $sigma$ and $tau sigma$ singular simplexes of $S^n$.
This induces an application $t:C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2), tildesigma rightarrow sigma + tausigma$ which is a chain complex morphism and induces a short exact sequence $$0 rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2) rightarrow 0$$ where the latter morphism is induced by the covering map. I managed to show all this. This implies directly existence of a long exact sequence in homology.
I should then show that $t circ p_star: C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(S^n,mathbb{F}_2)$ is zero where $p_star$ is induced by the covering map. Can you give me a hint for this?
I should then show that in the long exact sequence in hompology $$0=H_{n+1}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2) rightarrow H_{n}(S^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n-1}(mathbb{R}P^n,mathbb{F}_2)rightarrowcdots$$ the first and third morphisms are zero and the second and fourth isomorphisms. As the second is induced by $t$ and the third by $p_star$, I could already show that $p_star$ is zero, as $t$ is injective and the composition $t circ p_star$ zero. Thus $t$ is an isomorphism. Furthermore, the fourth morphism, $delta$ is injective. But how do I show its surjectivity?
Finally I should conclude that if $f: S^n rightarrow S^m$ is continuous such that, for every $x in S^n$, $f(-x)=-f(x)$, then $nleq m$, using "naturality of the long exact sequence". Could you please give me some hints?
algebraic-topology homology-cohomology
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
$endgroup$
$begingroup$
Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15
add a comment |
$begingroup$
I have to prove the Borsuk-Ulam theorem following some specific steps.
The theorem says that:
For every continuous function $f: S^n rightarrow mathbb{R}^n$ there exists $x in S^n$ with $f(-x)=f(x)$.
I am using the cover of $mathbb{R}P^n$ by $S^n$. I showed every singular simplex $tildesigma$ of $mathbb{R}P^n$ can be lifted to $sigma$ and $tau sigma$ singular simplexes of $S^n$.
This induces an application $t:C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2), tildesigma rightarrow sigma + tausigma$ which is a chain complex morphism and induces a short exact sequence $$0 rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2) rightarrow 0$$ where the latter morphism is induced by the covering map. I managed to show all this. This implies directly existence of a long exact sequence in homology.
I should then show that $t circ p_star: C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(S^n,mathbb{F}_2)$ is zero where $p_star$ is induced by the covering map. Can you give me a hint for this?
I should then show that in the long exact sequence in hompology $$0=H_{n+1}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2) rightarrow H_{n}(S^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n-1}(mathbb{R}P^n,mathbb{F}_2)rightarrowcdots$$ the first and third morphisms are zero and the second and fourth isomorphisms. As the second is induced by $t$ and the third by $p_star$, I could already show that $p_star$ is zero, as $t$ is injective and the composition $t circ p_star$ zero. Thus $t$ is an isomorphism. Furthermore, the fourth morphism, $delta$ is injective. But how do I show its surjectivity?
Finally I should conclude that if $f: S^n rightarrow S^m$ is continuous such that, for every $x in S^n$, $f(-x)=-f(x)$, then $nleq m$, using "naturality of the long exact sequence". Could you please give me some hints?
algebraic-topology homology-cohomology
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
$endgroup$
I have to prove the Borsuk-Ulam theorem following some specific steps.
The theorem says that:
For every continuous function $f: S^n rightarrow mathbb{R}^n$ there exists $x in S^n$ with $f(-x)=f(x)$.
I am using the cover of $mathbb{R}P^n$ by $S^n$. I showed every singular simplex $tildesigma$ of $mathbb{R}P^n$ can be lifted to $sigma$ and $tau sigma$ singular simplexes of $S^n$.
This induces an application $t:C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2), tildesigma rightarrow sigma + tausigma$ which is a chain complex morphism and induces a short exact sequence $$0 rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2)rightarrow C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(mathbb{R}P^n,mathbb{F}_2) rightarrow 0$$ where the latter morphism is induced by the covering map. I managed to show all this. This implies directly existence of a long exact sequence in homology.
I should then show that $t circ p_star: C_{star}(S^n,mathbb{F}_2) rightarrow C_{star}(S^n,mathbb{F}_2)$ is zero where $p_star$ is induced by the covering map. Can you give me a hint for this?
I should then show that in the long exact sequence in hompology $$0=H_{n+1}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2) rightarrow H_{n}(S^n,mathbb{F}_2)rightarrow H_{n}(mathbb{R}P^n,mathbb{F}_2)rightarrow H_{n-1}(mathbb{R}P^n,mathbb{F}_2)rightarrowcdots$$ the first and third morphisms are zero and the second and fourth isomorphisms. As the second is induced by $t$ and the third by $p_star$, I could already show that $p_star$ is zero, as $t$ is injective and the composition $t circ p_star$ zero. Thus $t$ is an isomorphism. Furthermore, the fourth morphism, $delta$ is injective. But how do I show its surjectivity?
Finally I should conclude that if $f: S^n rightarrow S^m$ is continuous such that, for every $x in S^n$, $f(-x)=-f(x)$, then $nleq m$, using "naturality of the long exact sequence". Could you please give me some hints?
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
edited Dec 7 '18 at 20:19
J.G.
28.7k22845
28.7k22845
asked Dec 5 '18 at 19:19
C. S.C. S.
263
asked Dec 5 '18 at 19:19
C. S.C. S.
263
asked Dec 5 '18 at 19:19
C. S.C. S.
263
263
$begingroup$
Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15
add a comment |
$begingroup$
Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15
$begingroup$
Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15
$begingroup$
Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15
add a comment |
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Only very few people seem to have been reached by this post. How could I change this?
$endgroup$
– C. S.
Dec 7 '18 at 18:15