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    Let A=( I \times I ) / J be the identification space of the unit square I \times I in which all points in the subspace J= ( I \times \{ 1 \} ) \cup ( \{ 0,1 \} \times I ) \subset I \times I are identified. Use the circles
    C_t = \{ ( x,y ) \in D^2 | (x-t)^2 + y^2 = (1-t)^2 \} \quad (t \in I) 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 \{ t \}]=C_t \subset D^2

    anyway i've established that J is essentially a square with one side removed. Then if we make the identification, J essentially contracts to a point, which we call [J]. So A is a square with three of the sides essentially contracted into the point [J] i think?

    anyway i cannot see how these circles are going to be of any use whatsoever in constructing this homeomorphism but i tried

    f : A \rightarrow D^2 ; [s,t] \mapsto \begin{cases} \left( \cos{2 \pi s}, \sin{2 \pi s} \right) \text{ for } [s,t]=[s,0] \\ (1,0) \text{ for } [s,t]=[J] \\ C_t \text{ for } [s,t]=[I \times t] \end{cases}

    it's bijective if i can find an inverse which is easy enough although i'm not sure about continuity (i guess this is to do with images of open/closed sets?).

    i'm not really sure where to go now though? am i at least on the right lines?
    thanks.
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    anyone?
 
 
 
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Updated: February 15, 2010
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