Skip to main content
edited title
Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

A possible alternative proof of the Borsuk Ulam theorem without "Homology-Cohomology"

added 161 characters in body
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$ for all $x,y \in S^n$

If the answer to either of these two questions is "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the above equivalent formulation of the Borsuk Ulam theorem as. The proof is as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction. Because this situation leads to existence of an odd continuous function from $S^n$ to $S^{n-1}$. Now we apply the induction argument.

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$ for all $x,y \in S^n$

If the answer to either of these two questions is "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the equivalent formulation of the Borsuk Ulam theorem as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction.

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$ for all $x,y \in S^n$

If the answer to either of these two questions is "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the above equivalent formulation of the Borsuk Ulam theorem. The proof is as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction. Because this situation leads to existence of an odd continuous function from $S^n$ to $S^{n-1}$. Now we apply the induction argument.

added 65 characters in body
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$ for all $x,y \in S^n$

If the answer to either of these two questions is "yes" "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the equivalent formulation of the Borsuk Ulam theorem as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction.

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$

If the answer to either of these two questions is "yes", we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the equivalent formulation of the Borsuk Ulam theorem as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction.

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?

2.Is there a continuos function $h: M_n \to \mathbb{R}^n$ such that $h(x,y)=-h(y,x) $ and $h(x,-x)\neq 0$ for all $x,y \in S^n$

If the answer to either of these two questions is "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that:

There is no an odd continuous function $g:S^{n+1}\to S^n$

Assuming that the answer to either of the above two questions is affirmative, we give a proof for the equivalent formulation of the Borsuk Ulam theorem as follows:

Assume that $g:S^{n+1}\to S^n$ is an odd continuous function. then $f(g(x),g(-x))$ ( or $h(g(x),g(-x))$ ) is an odd continuous function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction.

deleted 10 characters in body
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123
Loading
added 15 characters in body
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123
Loading
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123
Loading