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Hi there,
I've been asked to prove that (R^2, S^1) has the homtopy extension property and then extend it to the general case: (R^n, S^(n-1))
here's where I've got so far,
for (R^2, S^1) Let S^1=A, R^2=X
well S^1 is contained in R^2, so by a theorem, if A contained in X, has a mapping cylinder neighbourhood N, then (X,A) has the homotopy extension property.
so, let N= {x ; 0.5 <=||x||<= 1.5}, N is closed, A is contained in N. Let b_1={x ; 0.5=||x||} and b_2= {x ; 1.5=||x||} , let B= b_1 union b_2. then B=the boundary of N, B closed implies N\B is open, and A is contained in N\B, so N is a mapping cylinder neighbourhood.
Now, let f: B->A
M_f (mapping cylinder) = [(BxI) disjoint union A]/~ where, for x belonging to B, (x,1)~f(x)
= [(b_2 x [0,0.5] union b_1 x [-0.5,0]) disjoint union A]/~
isomorphic to Ax[-1,1].
Define a homeomorphism
h: Ax[-1,1] -> N by h(e^itheta, t) =(t+1)e^itheta for t belonging to [-1,1] and theta belonging to [0, 2Pi]
Then Ax[-1,1] is isomorphic to N and h restricted to A union B = Identity.
Hence N is a mapping cylinder neighbourhood of A and therefore, (X,A) has the Homotopy extension Property.
is this correct, particularly the bold bit, and do i just have to set A= S^(n-1) and X= R^n, for it to work in the cases for other dimensons?
Thank You
I've been asked to prove that (R^2, S^1) has the homtopy extension property and then extend it to the general case: (R^n, S^(n-1))
here's where I've got so far,
for (R^2, S^1) Let S^1=A, R^2=X
well S^1 is contained in R^2, so by a theorem, if A contained in X, has a mapping cylinder neighbourhood N, then (X,A) has the homotopy extension property.
so, let N= {x ; 0.5 <=||x||<= 1.5}, N is closed, A is contained in N. Let b_1={x ; 0.5=||x||} and b_2= {x ; 1.5=||x||} , let B= b_1 union b_2. then B=the boundary of N, B closed implies N\B is open, and A is contained in N\B, so N is a mapping cylinder neighbourhood.
Now, let f: B->A
M_f (mapping cylinder) = [(BxI) disjoint union A]/~ where, for x belonging to B, (x,1)~f(x)
= [(b_2 x [0,0.5] union b_1 x [-0.5,0]) disjoint union A]/~
isomorphic to Ax[-1,1].
Define a homeomorphism
h: Ax[-1,1] -> N by h(e^itheta, t) =(t+1)e^itheta for t belonging to [-1,1] and theta belonging to [0, 2Pi]
Then Ax[-1,1] is isomorphic to N and h restricted to A union B = Identity.
Hence N is a mapping cylinder neighbourhood of A and therefore, (X,A) has the Homotopy extension Property.
is this correct, particularly the bold bit, and do i just have to set A= S^(n-1) and X= R^n, for it to work in the cases for other dimensons?
Thank You