Is homotopy a one-way mapping?

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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




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    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














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




share|cite|improve this question















  • 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












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




share|cite|improve this question











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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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












  • 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










1 Answer
1






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3
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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 $f$.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    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 $f$.






    share|cite|improve this answer

























      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 $f$.






      share|cite|improve this answer























        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 $f$.






        share|cite|improve this answer













        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 $f$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 16 at 17:03









        Lord Shark the Unknown

        85.7k951112




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