Can Homotopy Extension Property Be Applied Here?

[madness]

Homework Statement

(i) Let $$A= (I\times I)/J$$ be the identification space of the unit square in which all points in the subspace $$J=(I\times \left\{1\right\})\cup(\left\{0,1\right\}\times I)$$ are identified. Use the circles $$C_t=\left\{(x,y)\in D^2|(x-t)^2+y^2=(1-t)^2 , t\in I\right\}$$ to construct a homeomorphism $$f:A \rightarrow D^2$$ such that $$f[s,0]=(\cos 2\pi s,\sin 2\pi s), f[J]=(1,0),f[I\times \left\{t\right\}]=C_t$$

(ii) Regard $$S^1$$ as the identification space of I in which the points $$\left\{ 0,1 \right\}$$ are identified via the homeomorphism [tex] I/{\lef\t{ 0,1 \right\} } \rightarrow S^1 ; \rightarrow (\cos2 \pi s, \sin 2\pi s) [/tex]. Use (i) to prove that a based loop $$\omega: (S^1,(1,0)) \rightarrow (X,x)$$ is homotopic rel{(1,0)} to the constant based loop $$e_x: (S^1,(1,0) \rightarrow (X,x)$$ if and only if $$\omega$$ extends to a based map $$\Omega : (D^2,(1,0)) \rightarrow (X,x)$$

Homework Equations

The Attempt at a Solution

(i) The space A is the square with 3 sides identified to a point. The suggested homeomorphism maps the remaining side to the circle $$S^1$$ and all other horizontal lines to circles centred at (t,0). Choose $$f[s,t] = (1-t)(\cos2 \pi s +t,\sin2\pi s)$$. This satisfies the first and third of the conditions required in (i), but I'm not sure about the second. I'm not sure if the question requires a proof that this is a homeomorphism.

(ii) I'm not sure about this part. Homeomorphic spaces are automatically homotopy equivalent. We can consider a loop $$\omega$$ as a function $$\alpha (t) = \omega (\cos 2\pi t, \sin 2 \pi t), \alpha (0) = \alpha(1) =x)$$ since the loop is based at x. So I think if I show that it has to extend to a map on the space A given above then it automatically has to extend to a map on the disk.

 

[madness]

Can no-one help with this one? I'm wondering what they mean by saying that the map "extends" to the disk. The homotopy given by considering the loop as a line with equal end-points is a map on the (identification space of) square, which I might try to relate to the disk in (i) through homeomorphism.

 

[ystael]

Looks like that is exactly what you should try. "Extends" means that there is a continuous map of pointed spaces $$\Omega: (D^2, (1, 0)) \to (X, x))$$ such that $$\Omega$$ restricts to $$\omega$$ on $$S^1$$: $$\Omega|_{S^1} = \omega$$.

 

[mrbohn1]

Your map does not satisfy the second condition in part (i): J contains the points (s,1) for all s, and your map takes all of these to (0,0). I think you are on the right track though. But you will need to show that this is a homeomorphism to answer the question fully.