In theta space, is the lower arc and line segment a deformation retract of punctured theta?












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Munkres Topology Example 70.1



enter image description here



enter image description here



Let X be theta-space, U = $X setminus {a}$ and V = $X setminus {b}$. Let $U cap V = X setminus {a,b}$ be doubly punctured theta-space where $a,b$ are interior points of $A$ and $B$.



Is $B cup C$ a deformation retract of $U$?










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  • Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
    – Andreas Blass
    Nov 20 '18 at 16:15
















0














Munkres Topology Example 70.1



enter image description here



enter image description here



Let X be theta-space, U = $X setminus {a}$ and V = $X setminus {b}$. Let $U cap V = X setminus {a,b}$ be doubly punctured theta-space where $a,b$ are interior points of $A$ and $B$.



Is $B cup C$ a deformation retract of $U$?










share|cite|improve this question
























  • Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
    – Andreas Blass
    Nov 20 '18 at 16:15














0












0








0







Munkres Topology Example 70.1



enter image description here



enter image description here



Let X be theta-space, U = $X setminus {a}$ and V = $X setminus {b}$. Let $U cap V = X setminus {a,b}$ be doubly punctured theta-space where $a,b$ are interior points of $A$ and $B$.



Is $B cup C$ a deformation retract of $U$?










share|cite|improve this question















Munkres Topology Example 70.1



enter image description here



enter image description here



Let X be theta-space, U = $X setminus {a}$ and V = $X setminus {b}$. Let $U cap V = X setminus {a,b}$ be doubly punctured theta-space where $a,b$ are interior points of $A$ and $B$.



Is $B cup C$ a deformation retract of $U$?







abstract-algebra algebraic-topology fundamental-groups deformation-theory retraction






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edited Nov 20 '18 at 13:42

























asked Nov 13 '18 at 12:54







user198044



















  • Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
    – Andreas Blass
    Nov 20 '18 at 16:15


















  • Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
    – Andreas Blass
    Nov 20 '18 at 16:15
















Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
– Andreas Blass
Nov 20 '18 at 16:15




Deform the part of $A$ between $P$ and $a$ toward $P$, and deform the part of $A$ between $a$ and $Q$ toward $Q$, while leaving all points in $Bcup C$ fixed.
– Andreas Blass
Nov 20 '18 at 16:15










1 Answer
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I think it's a retract by either $r(z)=z1_{B cup C} + Re(z)1_{C cup [A setminus {a}]}$ or $r(z)=z1_{B cup C} + overline{z}1_{C cup [A setminus {a}]}$ either of which is continuous by the pasting lemma because $B cup C$ and $C cup [A setminus {a}]$ are closed in $U$ because



$$B cup C = U cap {Im(z) le 0 }$$



$$C cup [A setminus {a}] = U cap {Im(z) ge 0 }$$



I think it's a deformation retract with the straight line homotopy $H(z,t)=(1-t)z+tr(z)$ because:




  1. $H(z,0)=z forall z in B cup C$

  2. $H(z,1)=r(z) forall z in B cup C$

  3. $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$






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    1 Answer
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    1 Answer
    1






    active

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    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    I think it's a retract by either $r(z)=z1_{B cup C} + Re(z)1_{C cup [A setminus {a}]}$ or $r(z)=z1_{B cup C} + overline{z}1_{C cup [A setminus {a}]}$ either of which is continuous by the pasting lemma because $B cup C$ and $C cup [A setminus {a}]$ are closed in $U$ because



    $$B cup C = U cap {Im(z) le 0 }$$



    $$C cup [A setminus {a}] = U cap {Im(z) ge 0 }$$



    I think it's a deformation retract with the straight line homotopy $H(z,t)=(1-t)z+tr(z)$ because:




    1. $H(z,0)=z forall z in B cup C$

    2. $H(z,1)=r(z) forall z in B cup C$

    3. $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$






    share|cite|improve this answer


























      1














      I think it's a retract by either $r(z)=z1_{B cup C} + Re(z)1_{C cup [A setminus {a}]}$ or $r(z)=z1_{B cup C} + overline{z}1_{C cup [A setminus {a}]}$ either of which is continuous by the pasting lemma because $B cup C$ and $C cup [A setminus {a}]$ are closed in $U$ because



      $$B cup C = U cap {Im(z) le 0 }$$



      $$C cup [A setminus {a}] = U cap {Im(z) ge 0 }$$



      I think it's a deformation retract with the straight line homotopy $H(z,t)=(1-t)z+tr(z)$ because:




      1. $H(z,0)=z forall z in B cup C$

      2. $H(z,1)=r(z) forall z in B cup C$

      3. $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$






      share|cite|improve this answer
























        1












        1








        1






        I think it's a retract by either $r(z)=z1_{B cup C} + Re(z)1_{C cup [A setminus {a}]}$ or $r(z)=z1_{B cup C} + overline{z}1_{C cup [A setminus {a}]}$ either of which is continuous by the pasting lemma because $B cup C$ and $C cup [A setminus {a}]$ are closed in $U$ because



        $$B cup C = U cap {Im(z) le 0 }$$



        $$C cup [A setminus {a}] = U cap {Im(z) ge 0 }$$



        I think it's a deformation retract with the straight line homotopy $H(z,t)=(1-t)z+tr(z)$ because:




        1. $H(z,0)=z forall z in B cup C$

        2. $H(z,1)=r(z) forall z in B cup C$

        3. $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$






        share|cite|improve this answer












        I think it's a retract by either $r(z)=z1_{B cup C} + Re(z)1_{C cup [A setminus {a}]}$ or $r(z)=z1_{B cup C} + overline{z}1_{C cup [A setminus {a}]}$ either of which is continuous by the pasting lemma because $B cup C$ and $C cup [A setminus {a}]$ are closed in $U$ because



        $$B cup C = U cap {Im(z) le 0 }$$



        $$C cup [A setminus {a}] = U cap {Im(z) ge 0 }$$



        I think it's a deformation retract with the straight line homotopy $H(z,t)=(1-t)z+tr(z)$ because:




        1. $H(z,0)=z forall z in B cup C$

        2. $H(z,1)=r(z) forall z in B cup C$

        3. $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 13:42







        user198044





































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