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
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
asked Nov 13 at 12:54
Jack Bauer
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Munkres Topology Example 70.1
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
asked Nov 13 at 12:54
Jack Bauer
1,226531
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
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Munkres Topology Example 70.1
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
asked Nov 13 at 12:54
Jack Bauer
1,226531
Munkres Topology Example 70.1
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
abstract-algebra algebraic-topology fundamental-groups deformation-theory retraction
asked Nov 13 at 12:54
Jack Bauer
1,226531
asked Nov 13 at 12:54
Jack Bauer
1,226531
edited 2 days ago
asked Nov 13 at 12:54
Jack Bauer
1,226531
asked Nov 13 at 12:54
Jack Bauer
1,226531
asked Nov 13 at 12:54
Jack Bauer
1,226531
1,226531
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
2 days ago
add a comment |
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
2 days ago
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
2 days ago
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
2 days ago
add a comment |
1 Answer
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0
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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:
- $H(z,0)=z forall z in B cup C$
- $H(z,1)=r(z) forall z in B cup C$
- $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$
answered 2 days ago
Jack Bauer
1,226531
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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:
- $H(z,0)=z forall z in B cup C$
- $H(z,1)=r(z) forall z in B cup C$
- $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$
answered 2 days ago
Jack Bauer
1,226531
add a comment |
up vote
0
down vote
accepted
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:
- $H(z,0)=z forall z in B cup C$
- $H(z,1)=r(z) forall z in B cup C$
- $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$
answered 2 days ago
Jack Bauer
1,226531
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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:
- $H(z,0)=z forall z in B cup C$
- $H(z,1)=r(z) forall z in B cup C$
- $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$
answered 2 days ago
Jack Bauer
1,226531
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:
- $H(z,0)=z forall z in B cup C$
- $H(z,1)=r(z) forall z in B cup C$
- $H(d,t)=(1-t)(d)+tr(d)=d forall d in B cup C, t in I$
answered 2 days ago
Jack Bauer
1,226531
answered 2 days ago
Jack Bauer
1,226531
answered 2 days ago
Jack Bauer
1,226531
answered 2 days ago
Jack Bauer
1,226531
1,226531
add a comment |
add a comment |
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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
2 days ago