Find an equivalence relation ~ on [0,1] such that [0,1]/~ is homeomorphic to [0,1] x [0,1].

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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I've thought a lot about this question and it doesn't seem possible to me. I'm assuming there is some special trick that I am not aware of. Any help is appreciated.




general-topology




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    Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
    – Steve D
    Jul 29 at 4:34














up vote
4
down vote

favorite
1












I've thought a lot about this question and it doesn't seem possible to me. I'm assuming there is some special trick that I am not aware of. Any help is appreciated.




general-topology




share|cite|improve this question















  • 1




    Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
    – Steve D
    Jul 29 at 4:34












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I've thought a lot about this question and it doesn't seem possible to me. I'm assuming there is some special trick that I am not aware of. Any help is appreciated.




general-topology




share|cite|improve this question











I've thought a lot about this question and it doesn't seem possible to me. I'm assuming there is some special trick that I am not aware of. Any help is appreciated.





general-topology





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 3:22









user8513188

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282







  • 1




    Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
    – Steve D
    Jul 29 at 4:34












  • 1




    Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
    – Steve D
    Jul 29 at 4:34







1




1




Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
– Steve D
Jul 29 at 4:34




Space-filling curve might help ( it's a quotient map, but you'd have to suss out the relation).
– Steve D
Jul 29 at 4:34










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Steve D has given 99% of the answer. It is known that there exists a continuous surjection $p : I to I^2 = I times I$. Since domain and range are compact, $p$ is an identification map. Now define $s sim t$ if $p(s) = p(t)$. This is an equivalence relation. Let $pi : I to Q = I/ sim$ denote the quotient map. Then $p$ induces a unique function $hatp : Q to I^2$ such that $hatp circ pi = p$. It is bijective by construction and continuous by the universal property of the quotient. The function $hatp^-1$ satisfies $hatp^-1 circ p = pi$ and is therefore continuous because $p$ is an identification.






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    up vote
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    Steve D has given 99% of the answer. It is known that there exists a continuous surjection $p : I to I^2 = I times I$. Since domain and range are compact, $p$ is an identification map. Now define $s sim t$ if $p(s) = p(t)$. This is an equivalence relation. Let $pi : I to Q = I/ sim$ denote the quotient map. Then $p$ induces a unique function $hatp : Q to I^2$ such that $hatp circ pi = p$. It is bijective by construction and continuous by the universal property of the quotient. The function $hatp^-1$ satisfies $hatp^-1 circ p = pi$ and is therefore continuous because $p$ is an identification.






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      up vote
      1
      down vote



      accepted










      Steve D has given 99% of the answer. It is known that there exists a continuous surjection $p : I to I^2 = I times I$. Since domain and range are compact, $p$ is an identification map. Now define $s sim t$ if $p(s) = p(t)$. This is an equivalence relation. Let $pi : I to Q = I/ sim$ denote the quotient map. Then $p$ induces a unique function $hatp : Q to I^2$ such that $hatp circ pi = p$. It is bijective by construction and continuous by the universal property of the quotient. The function $hatp^-1$ satisfies $hatp^-1 circ p = pi$ and is therefore continuous because $p$ is an identification.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Steve D has given 99% of the answer. It is known that there exists a continuous surjection $p : I to I^2 = I times I$. Since domain and range are compact, $p$ is an identification map. Now define $s sim t$ if $p(s) = p(t)$. This is an equivalence relation. Let $pi : I to Q = I/ sim$ denote the quotient map. Then $p$ induces a unique function $hatp : Q to I^2$ such that $hatp circ pi = p$. It is bijective by construction and continuous by the universal property of the quotient. The function $hatp^-1$ satisfies $hatp^-1 circ p = pi$ and is therefore continuous because $p$ is an identification.






        share|cite|improve this answer













        Steve D has given 99% of the answer. It is known that there exists a continuous surjection $p : I to I^2 = I times I$. Since domain and range are compact, $p$ is an identification map. Now define $s sim t$ if $p(s) = p(t)$. This is an equivalence relation. Let $pi : I to Q = I/ sim$ denote the quotient map. Then $p$ induces a unique function $hatp : Q to I^2$ such that $hatp circ pi = p$. It is bijective by construction and continuous by the universal property of the quotient. The function $hatp^-1$ satisfies $hatp^-1 circ p = pi$ and is therefore continuous because $p$ is an identification.







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        share|cite|improve this answer











        answered Jul 29 at 9:10









        Paul Frost

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