Mixing
Is homotopy a one-way mapping?
Clash Royale CLAN TAG#URR8PPP
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If there exists a homotopy $f simeq g$, does this imply that a reverse homotopy $g simeq f$ exists too, or this is not true in general?
Example: a homotopy from $mathbbR^2$ to origin point $(0, 0)$.
A direct homotopy from identity map $mathbb1_mathbbR^2$ to $p_0: x mapsto 0$, which sends every point $x$ in $mathbbR^2$ to $0$, exists: $H_rightarrow(t) = tcdot p_0 + (1-t)cdotmathbb1_mathbbR^2$.
Is it ok to claim that $H_leftarrow(t) = (1-t)cdot p_0 + tcdotmathbb1_mathbbR^2$ is a reverse homotopy?
algebraic-topology
asked Jul 16 at 17:00
Entsy
284
add a comment |Â
up vote
1
down vote
favorite
If there exists a homotopy $f simeq g$, does this imply that a reverse homotopy $g simeq f$ exists too, or this is not true in general?
Example: a homotopy from $mathbbR^2$ to origin point $(0, 0)$.
A direct homotopy from identity map $mathbb1_mathbbR^2$ to $p_0: x mapsto 0$, which sends every point $x$ in $mathbbR^2$ to $0$, exists: $H_rightarrow(t) = tcdot p_0 + (1-t)cdotmathbb1_mathbbR^2$.
Is it ok to claim that $H_leftarrow(t) = (1-t)cdot p_0 + tcdotmathbb1_mathbbR^2$ is a reverse homotopy?
algebraic-topology
asked Jul 16 at 17:00
Entsy
284
1
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If there exists a homotopy $f simeq g$, does this imply that a reverse homotopy $g simeq f$ exists too, or this is not true in general?
Example: a homotopy from $mathbbR^2$ to origin point $(0, 0)$.
A direct homotopy from identity map $mathbb1_mathbbR^2$ to $p_0: x mapsto 0$, which sends every point $x$ in $mathbbR^2$ to $0$, exists: $H_rightarrow(t) = tcdot p_0 + (1-t)cdotmathbb1_mathbbR^2$.
Is it ok to claim that $H_leftarrow(t) = (1-t)cdot p_0 + tcdotmathbb1_mathbbR^2$ is a reverse homotopy?
algebraic-topology
asked Jul 16 at 17:00
Entsy
284
If there exists a homotopy $f simeq g$, does this imply that a reverse homotopy $g simeq f$ exists too, or this is not true in general?
Example: a homotopy from $mathbbR^2$ to origin point $(0, 0)$.
A direct homotopy from identity map $mathbb1_mathbbR^2$ to $p_0: x mapsto 0$, which sends every point $x$ in $mathbbR^2$ to $0$, exists: $H_rightarrow(t) = tcdot p_0 + (1-t)cdotmathbb1_mathbbR^2$.
Is it ok to claim that $H_leftarrow(t) = (1-t)cdot p_0 + tcdotmathbb1_mathbbR^2$ is a reverse homotopy?
algebraic-topology
asked Jul 16 at 17:00
Entsy
284
asked Jul 16 at 17:00
Entsy
284
asked Jul 16 at 17:00
Entsy
284
asked Jul 16 at 17:00
Entsy
284
284
1
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22
add a comment |Â
1
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22
1
1
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
A homotopy
from $f$ to $g$ is a
continuous map $H:Xtimes[0,1]to Y$ with $H(x,0)=f(x)$
and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $tilde H$
with
$$tilde H(x,t)=H(x,1-t)$$
from $g$ to $f$.
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A homotopy
from $f$ to $g$ is a
continuous map $H:Xtimes[0,1]to Y$ with $H(x,0)=f(x)$
and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $tilde H$
with
$$tilde H(x,t)=H(x,1-t)$$
from $g$ to $f$.
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
3
down vote
accepted
A homotopy
from $f$ to $g$ is a
continuous map $H:Xtimes[0,1]to Y$ with $H(x,0)=f(x)$
and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $tilde H$
with
$$tilde H(x,t)=H(x,1-t)$$
from $g$ to $f$.
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A homotopy
from $f$ to $g$ is a
continuous map $H:Xtimes[0,1]to Y$ with $H(x,0)=f(x)$
and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $tilde H$
with
$$tilde H(x,t)=H(x,1-t)$$
from $g$ to $f$.
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
A homotopy
from $f$ to $g$ is a
continuous map $H:Xtimes[0,1]to Y$ with $H(x,0)=f(x)$
and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $tilde H$
with
$$tilde H(x,t)=H(x,1-t)$$
from $g$ to $f$.
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
answered Jul 16 at 17:03
Lord Shark the Unknown
85.7k951112
85.7k951112
add a comment |Â
add a comment |Â
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1
It is correct. Homotopy between maps is an equivalence relation.
– Douglas Molin
Jul 17 at 9:59
Thanks @DouglasMolin, it is indeed is, so reflexivity, symmetry and transitivity are all included, cool!
– Entsy
Jul 17 at 16:22