Mixing
If $f: I to X$ is a path and $c: I to X$ the constant path $c(s)=f(1)$, how is $f cdot c$ a reparametrization...
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From Algebraic Topology by Hatcher:
Definition of reparametrization:
Paragraph in question:
If a reparametrization is defined to be a composition $f varphi$ where $varphi: I to I$ satisfies $varphi(0)=0$ and $varphi(1)=1$, how is $fcdot c$ a reparametrization if $c$ does not satisfy the definition?
general-topology proof-verification algebraic-topology definition
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
$endgroup$
add a comment |
$begingroup$
From Algebraic Topology by Hatcher:
Definition of reparametrization:
Paragraph in question:
If a reparametrization is defined to be a composition $f varphi$ where $varphi: I to I$ satisfies $varphi(0)=0$ and $varphi(1)=1$, how is $fcdot c$ a reparametrization if $c$ does not satisfy the definition?
general-topology proof-verification algebraic-topology definition
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
$endgroup$
1
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17
add a comment |
$begingroup$
From Algebraic Topology by Hatcher:
Definition of reparametrization:
Paragraph in question:
If a reparametrization is defined to be a composition $f varphi$ where $varphi: I to I$ satisfies $varphi(0)=0$ and $varphi(1)=1$, how is $fcdot c$ a reparametrization if $c$ does not satisfy the definition?
general-topology proof-verification algebraic-topology definition
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
$endgroup$
From Algebraic Topology by Hatcher:
Definition of reparametrization:
Paragraph in question:
If a reparametrization is defined to be a composition $f varphi$ where $varphi: I to I$ satisfies $varphi(0)=0$ and $varphi(1)=1$, how is $fcdot c$ a reparametrization if $c$ does not satisfy the definition?
general-topology proof-verification algebraic-topology definition
general-topology proof-verification algebraic-topology definition
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
asked Jan 21 at 22:04
WolfgangWolfgang
4,31443377
4,31443377
1
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17
add a comment |
1
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17
1
1
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17
add a comment |
1 Answer
1
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oldest
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$begingroup$
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
c(2t-1) & 1/2le tle 1
end{cases}$
so in fact
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
f(1) & 1/2le tle 1
end{cases}$
And since
$phi(t)= begin{cases}
2t & 0le tle 1/2 \
1 & 1/2le tle 1
end{cases}$,
a simple calculation gives $phicirc f=ccdot f, $ as desired.
The other case is done in exactly the same way.
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
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add a comment |
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1 Answer
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$begingroup$
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
c(2t-1) & 1/2le tle 1
end{cases}$
so in fact
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
f(1) & 1/2le tle 1
end{cases}$
And since
$phi(t)= begin{cases}
2t & 0le tle 1/2 \
1 & 1/2le tle 1
end{cases}$,
a simple calculation gives $phicirc f=ccdot f, $ as desired.
The other case is done in exactly the same way.
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
c(2t-1) & 1/2le tle 1
end{cases}$
so in fact
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
f(1) & 1/2le tle 1
end{cases}$
And since
$phi(t)= begin{cases}
2t & 0le tle 1/2 \
1 & 1/2le tle 1
end{cases}$,
a simple calculation gives $phicirc f=ccdot f, $ as desired.
The other case is done in exactly the same way.
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
c(2t-1) & 1/2le tle 1
end{cases}$
so in fact
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
f(1) & 1/2le tle 1
end{cases}$
And since
$phi(t)= begin{cases}
2t & 0le tle 1/2 \
1 & 1/2le tle 1
end{cases}$,
a simple calculation gives $phicirc f=ccdot f, $ as desired.
The other case is done in exactly the same way.
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
$endgroup$
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
c(2t-1) & 1/2le tle 1
end{cases}$
so in fact
$fcdot c(t)= begin{cases}
f(2t) & 0le tle 1/2 \
f(1) & 1/2le tle 1
end{cases}$
And since
$phi(t)= begin{cases}
2t & 0le tle 1/2 \
1 & 1/2le tle 1
end{cases}$,
a simple calculation gives $phicirc f=ccdot f, $ as desired.
The other case is done in exactly the same way.
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
answered Jan 22 at 0:09
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
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1
$begingroup$
Here $fcdot c$ is multiplication of paths, so it means follow path $f$ for half the interval, then follow path $c$ for the other half (which just sits still). It’s a reparameterization by the function $phi$ whose graph is on the left there.
$endgroup$
– csprun
Jan 21 at 22:33
$begingroup$
f$cdot$c is a reparamterization if there exists $phi$ such that (f$cdot$c)$phi$ is a continuous map such that $phi$(0)=0 and $phi$(1)=1.
$endgroup$
– Joel Pereira
Jan 21 at 22:53
$begingroup$
@JoelPereira That is not what the definition is saying. According to the definition, what you have there, $(fcdot c)phi$, is a reparametrization of $fcdot c$. We want a reparametrization of $f$.
$endgroup$
– Wolfgang
Jan 21 at 23:16
$begingroup$
@csprun How is that a reparametrization according to the definition given?
$endgroup$
– Wolfgang
Jan 21 at 23:17