Want a specific deformation retract of polygon with center deleted to boundary
Given a n-sided regular polygon $P$ centered at origin. I want to show that $P- {0}$ has a strong deformation retract the boundary of the polygon.
My try: Every point of $P-{0}$ is of the form $sx$ where $0< sleq 1$ and $x$ is on the boundary, moreover this representation is unique.
Define the retraction map to be $r: P-{0} to BdP $ by $r(sx)=x$.
I want to prove that the map $r$ is continuous, after which I have the map for deformation retraction as $(sx,t) to (1-t)sx + tx $.
To prove $r$ is continuous, I consider each triangle with vertices $0,v_i,v_{i+1}$ where $v_i,v_{i+1}$ are two consecutive vertices of $P$. Take an open $epsilon$ neighbourhood around $x=r(sx)$ in the line $v_iv_{i+1}$. The preimage of this (when $epsilon$ is small enough) is the triangle with vertices $0,x-epsilon,x+epsilon$ with the two sides starting at 0 deleted, which is open in $P-{0}$. Now by pasting lemma the map $r$ is continuous.
Is this proof correct?
Is there a better approach to this? One that comes to my mind is showing the polygonal region is homeomorphic to a disc where boundary goes to boundary. But writing that map seems too laborious.
Can I show the map that I wrote for $r$ is continuous using sequential techniques? Any other way to see that $r$ is continuous?
algebraic-topology retraction
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
add a comment |
Given a n-sided regular polygon $P$ centered at origin. I want to show that $P- {0}$ has a strong deformation retract the boundary of the polygon.
My try: Every point of $P-{0}$ is of the form $sx$ where $0< sleq 1$ and $x$ is on the boundary, moreover this representation is unique.
Define the retraction map to be $r: P-{0} to BdP $ by $r(sx)=x$.
I want to prove that the map $r$ is continuous, after which I have the map for deformation retraction as $(sx,t) to (1-t)sx + tx $.
To prove $r$ is continuous, I consider each triangle with vertices $0,v_i,v_{i+1}$ where $v_i,v_{i+1}$ are two consecutive vertices of $P$. Take an open $epsilon$ neighbourhood around $x=r(sx)$ in the line $v_iv_{i+1}$. The preimage of this (when $epsilon$ is small enough) is the triangle with vertices $0,x-epsilon,x+epsilon$ with the two sides starting at 0 deleted, which is open in $P-{0}$. Now by pasting lemma the map $r$ is continuous.
Is this proof correct?
Is there a better approach to this? One that comes to my mind is showing the polygonal region is homeomorphic to a disc where boundary goes to boundary. But writing that map seems too laborious.
Can I show the map that I wrote for $r$ is continuous using sequential techniques? Any other way to see that $r$ is continuous?
algebraic-topology retraction
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
add a comment |
Given a n-sided regular polygon $P$ centered at origin. I want to show that $P- {0}$ has a strong deformation retract the boundary of the polygon.
My try: Every point of $P-{0}$ is of the form $sx$ where $0< sleq 1$ and $x$ is on the boundary, moreover this representation is unique.
Define the retraction map to be $r: P-{0} to BdP $ by $r(sx)=x$.
I want to prove that the map $r$ is continuous, after which I have the map for deformation retraction as $(sx,t) to (1-t)sx + tx $.
To prove $r$ is continuous, I consider each triangle with vertices $0,v_i,v_{i+1}$ where $v_i,v_{i+1}$ are two consecutive vertices of $P$. Take an open $epsilon$ neighbourhood around $x=r(sx)$ in the line $v_iv_{i+1}$. The preimage of this (when $epsilon$ is small enough) is the triangle with vertices $0,x-epsilon,x+epsilon$ with the two sides starting at 0 deleted, which is open in $P-{0}$. Now by pasting lemma the map $r$ is continuous.
Is this proof correct?
Is there a better approach to this? One that comes to my mind is showing the polygonal region is homeomorphic to a disc where boundary goes to boundary. But writing that map seems too laborious.
Can I show the map that I wrote for $r$ is continuous using sequential techniques? Any other way to see that $r$ is continuous?
algebraic-topology retraction
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
Given a n-sided regular polygon $P$ centered at origin. I want to show that $P- {0}$ has a strong deformation retract the boundary of the polygon.
My try: Every point of $P-{0}$ is of the form $sx$ where $0< sleq 1$ and $x$ is on the boundary, moreover this representation is unique.
Define the retraction map to be $r: P-{0} to BdP $ by $r(sx)=x$.
I want to prove that the map $r$ is continuous, after which I have the map for deformation retraction as $(sx,t) to (1-t)sx + tx $.
To prove $r$ is continuous, I consider each triangle with vertices $0,v_i,v_{i+1}$ where $v_i,v_{i+1}$ are two consecutive vertices of $P$. Take an open $epsilon$ neighbourhood around $x=r(sx)$ in the line $v_iv_{i+1}$. The preimage of this (when $epsilon$ is small enough) is the triangle with vertices $0,x-epsilon,x+epsilon$ with the two sides starting at 0 deleted, which is open in $P-{0}$. Now by pasting lemma the map $r$ is continuous.
Is this proof correct?
Is there a better approach to this? One that comes to my mind is showing the polygonal region is homeomorphic to a disc where boundary goes to boundary. But writing that map seems too laborious.
Can I show the map that I wrote for $r$ is continuous using sequential techniques? Any other way to see that $r$ is continuous?
algebraic-topology retraction
algebraic-topology retraction
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
asked Nov 27 '18 at 10:03
justanothermathstudentjustanothermathstudent
1118
1118
add a comment |
add a comment |
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