Question about the one point compactification $mathbb{R}^2 cup infty$ of $S^2$












1












$begingroup$


Question about the one point compactification $mathbb{R}^2 cup infty$ of $S^2$.



Given $a,b in S^2$, can somebody give me the explicit homeomorphism $gamma$ from $S^2$ to $mathbb{R}^2 cup infty$ such that $gamma(a) = 0$ and $gamma(b) = infty$? For some reason in my brain I thought this was only possible if $a$ and $b$ are antipodal points, that seems like something that might be true, right? ha. Anyway apparently it's not according to the proof of Lemma 61.2 in Munkres topology, which takes arbitrary $a$ and $b$ and then maps one to $0$ and the other to $infty$.



thanks in advance :-)










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$endgroup$












  • $begingroup$
    Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
    $endgroup$
    – Guido A.
    Dec 14 '18 at 2:46










  • $begingroup$
    nope, i meant $mathbb{R}^2$ >.>
    $endgroup$
    – Math is hard
    Dec 14 '18 at 2:47






  • 2




    $begingroup$
    Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
    $endgroup$
    – Cheerful Parsnip
    Dec 14 '18 at 2:55










  • $begingroup$
    Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 9:33






  • 1




    $begingroup$
    By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 10:26
















1












$begingroup$


Question about the one point compactification $mathbb{R}^2 cup infty$ of $S^2$.



Given $a,b in S^2$, can somebody give me the explicit homeomorphism $gamma$ from $S^2$ to $mathbb{R}^2 cup infty$ such that $gamma(a) = 0$ and $gamma(b) = infty$? For some reason in my brain I thought this was only possible if $a$ and $b$ are antipodal points, that seems like something that might be true, right? ha. Anyway apparently it's not according to the proof of Lemma 61.2 in Munkres topology, which takes arbitrary $a$ and $b$ and then maps one to $0$ and the other to $infty$.



thanks in advance :-)










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
    $endgroup$
    – Guido A.
    Dec 14 '18 at 2:46










  • $begingroup$
    nope, i meant $mathbb{R}^2$ >.>
    $endgroup$
    – Math is hard
    Dec 14 '18 at 2:47






  • 2




    $begingroup$
    Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
    $endgroup$
    – Cheerful Parsnip
    Dec 14 '18 at 2:55










  • $begingroup$
    Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 9:33






  • 1




    $begingroup$
    By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 10:26














1












1








1





$begingroup$


Question about the one point compactification $mathbb{R}^2 cup infty$ of $S^2$.



Given $a,b in S^2$, can somebody give me the explicit homeomorphism $gamma$ from $S^2$ to $mathbb{R}^2 cup infty$ such that $gamma(a) = 0$ and $gamma(b) = infty$? For some reason in my brain I thought this was only possible if $a$ and $b$ are antipodal points, that seems like something that might be true, right? ha. Anyway apparently it's not according to the proof of Lemma 61.2 in Munkres topology, which takes arbitrary $a$ and $b$ and then maps one to $0$ and the other to $infty$.



thanks in advance :-)










share|cite|improve this question











$endgroup$




Question about the one point compactification $mathbb{R}^2 cup infty$ of $S^2$.



Given $a,b in S^2$, can somebody give me the explicit homeomorphism $gamma$ from $S^2$ to $mathbb{R}^2 cup infty$ such that $gamma(a) = 0$ and $gamma(b) = infty$? For some reason in my brain I thought this was only possible if $a$ and $b$ are antipodal points, that seems like something that might be true, right? ha. Anyway apparently it's not according to the proof of Lemma 61.2 in Munkres topology, which takes arbitrary $a$ and $b$ and then maps one to $0$ and the other to $infty$.



thanks in advance :-)







general-topology






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share|cite|improve this question













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share|cite|improve this question








edited Dec 14 '18 at 2:47







Math is hard

















asked Dec 14 '18 at 2:41









Math is hardMath is hard

817211




817211












  • $begingroup$
    Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
    $endgroup$
    – Guido A.
    Dec 14 '18 at 2:46










  • $begingroup$
    nope, i meant $mathbb{R}^2$ >.>
    $endgroup$
    – Math is hard
    Dec 14 '18 at 2:47






  • 2




    $begingroup$
    Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
    $endgroup$
    – Cheerful Parsnip
    Dec 14 '18 at 2:55










  • $begingroup$
    Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 9:33






  • 1




    $begingroup$
    By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 10:26


















  • $begingroup$
    Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
    $endgroup$
    – Guido A.
    Dec 14 '18 at 2:46










  • $begingroup$
    nope, i meant $mathbb{R}^2$ >.>
    $endgroup$
    – Math is hard
    Dec 14 '18 at 2:47






  • 2




    $begingroup$
    Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
    $endgroup$
    – Cheerful Parsnip
    Dec 14 '18 at 2:55










  • $begingroup$
    Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 9:33






  • 1




    $begingroup$
    By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
    $endgroup$
    – Paul Frost
    Dec 14 '18 at 10:26
















$begingroup$
Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
$endgroup$
– Guido A.
Dec 14 '18 at 2:46




$begingroup$
Do you mean $S^1$? Generally, $(mathbb{R}^k)^* simeq S^k$.
$endgroup$
– Guido A.
Dec 14 '18 at 2:46












$begingroup$
nope, i meant $mathbb{R}^2$ >.>
$endgroup$
– Math is hard
Dec 14 '18 at 2:47




$begingroup$
nope, i meant $mathbb{R}^2$ >.>
$endgroup$
– Math is hard
Dec 14 '18 at 2:47




2




2




$begingroup$
Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
$endgroup$
– Cheerful Parsnip
Dec 14 '18 at 2:55




$begingroup$
Start with sterographic projection $S$ that takes $b$ to $infty$ and $a$ to somewhere in the plane. Now compose this with a translation of the plane that takes $S(a)$ to the origin. You need to check that translation of $mathbb R^2$ extends to the $1$-point compactification, which it does.
$endgroup$
– Cheerful Parsnip
Dec 14 '18 at 2:55












$begingroup$
Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
$endgroup$
– Paul Frost
Dec 14 '18 at 9:33




$begingroup$
Typo: One point compactification $mathbb{R}^2 cup infty$ of $mathbb{R}^2$.
$endgroup$
– Paul Frost
Dec 14 '18 at 9:33




1




1




$begingroup$
By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
$endgroup$
– Paul Frost
Dec 14 '18 at 10:26




$begingroup$
By the way, Munkres does not need $gamma(a) = 0$. In Munkres's proof you can easily replace $0$ by an arbitrary $x_0 = gamma(a)$.
$endgroup$
– Paul Frost
Dec 14 '18 at 10:26










4 Answers
4






active

oldest

votes


















1












$begingroup$

Stereographic projection: the north pole $P$ corresponds to the point at infinity, the South pole to zero.



In Cartesian coordinates it's $(x,y,z)to (frac y{1-x},frac z{1-x})$.



As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does this map $a$ to $0$ and $b$ to $infty$?
    $endgroup$
    – Randall
    Dec 14 '18 at 3:14










  • $begingroup$
    Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 3:39










  • $begingroup$
    OP wants this to work for generic $a neq b in S^2$.
    $endgroup$
    – Randall
    Dec 14 '18 at 3:42










  • $begingroup$
    Oh I guess I overlooked that.
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 3:46



















0












$begingroup$

Well not really explicit at all, but Naturally we can consider $S^2$ to be $text{bd}(B_{mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $mathbb{R}^2$ that is ${(x,y,0)|x,yinmathbb{R}}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $mathbb{R}^2$. Complements of closed disc are neighborhoods of $infty$ as we define an open set containing $infty$ to be the complement of a compact set in $mathbb{R}^2$



Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.






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$endgroup$





















    0












    $begingroup$

    I will use the following fact: if $equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X equiv Y$ imples $X^* equiv Y^*$. The idea is to take a homeomorphism $f : X to Y$ and define $f^* :X^* to Y^*$ via $f^*(x) = f(x)$ when $x in X$ and $f^*(infty_X) = infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.



    Now, for any $k geq 1$, the map



    $$
    g(x_1, dots, x_{k+1}) = frac{1}{1-x_{k+1}}(x_1, dots, x_k)
    $$



    from $X_k := S^k setminus {(0,dots,0,1)}$ to $mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,dots,1)$ with $x in S^k$, and $g(x)$ is the intersection of the line with ${x_{k+1} = 0} subset mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* to (mathbb{R}^k)^*$. You can check that $S^k equiv (X_k)^*$ via $f(x) = x$ when $x neq N_k$, and $f(N_k) := infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $mathbb{R}^k$,



    $$
    begin{align}
    Phi : & S^k rightarrow (mathbb{R}^k)^* \
    &xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
    end{align}
    $$



    It is clear from here that $Phi(N_k) = infty$ and $Phi(0,0,dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 to S^2$ such that $a mapsto N_k$ and $b mapsto (0,0,dots,-1)$ and post-compose it with $Phi$. Can you take it from here?






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      If you know complex function theory, working on the Riemann sphere $mathbb{C} cup {infty}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $infty$.






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Stereographic projection: the north pole $P$ corresponds to the point at infinity, the South pole to zero.



        In Cartesian coordinates it's $(x,y,z)to (frac y{1-x},frac z{1-x})$.



        As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          How does this map $a$ to $0$ and $b$ to $infty$?
          $endgroup$
          – Randall
          Dec 14 '18 at 3:14










        • $begingroup$
          Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:39










        • $begingroup$
          OP wants this to work for generic $a neq b in S^2$.
          $endgroup$
          – Randall
          Dec 14 '18 at 3:42










        • $begingroup$
          Oh I guess I overlooked that.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:46
















        1












        $begingroup$

        Stereographic projection: the north pole $P$ corresponds to the point at infinity, the South pole to zero.



        In Cartesian coordinates it's $(x,y,z)to (frac y{1-x},frac z{1-x})$.



        As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          How does this map $a$ to $0$ and $b$ to $infty$?
          $endgroup$
          – Randall
          Dec 14 '18 at 3:14










        • $begingroup$
          Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:39










        • $begingroup$
          OP wants this to work for generic $a neq b in S^2$.
          $endgroup$
          – Randall
          Dec 14 '18 at 3:42










        • $begingroup$
          Oh I guess I overlooked that.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:46














        1












        1








        1





        $begingroup$

        Stereographic projection: the north pole $P$ corresponds to the point at infinity, the South pole to zero.



        In Cartesian coordinates it's $(x,y,z)to (frac y{1-x},frac z{1-x})$.



        As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.






        share|cite|improve this answer











        $endgroup$



        Stereographic projection: the north pole $P$ corresponds to the point at infinity, the South pole to zero.



        In Cartesian coordinates it's $(x,y,z)to (frac y{1-x},frac z{1-x})$.



        As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 4:18

























        answered Dec 14 '18 at 3:05









        Chris CusterChris Custer

        12.9k3827




        12.9k3827












        • $begingroup$
          How does this map $a$ to $0$ and $b$ to $infty$?
          $endgroup$
          – Randall
          Dec 14 '18 at 3:14










        • $begingroup$
          Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:39










        • $begingroup$
          OP wants this to work for generic $a neq b in S^2$.
          $endgroup$
          – Randall
          Dec 14 '18 at 3:42










        • $begingroup$
          Oh I guess I overlooked that.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:46


















        • $begingroup$
          How does this map $a$ to $0$ and $b$ to $infty$?
          $endgroup$
          – Randall
          Dec 14 '18 at 3:14










        • $begingroup$
          Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:39










        • $begingroup$
          OP wants this to work for generic $a neq b in S^2$.
          $endgroup$
          – Randall
          Dec 14 '18 at 3:42










        • $begingroup$
          Oh I guess I overlooked that.
          $endgroup$
          – Chris Custer
          Dec 14 '18 at 3:46
















        $begingroup$
        How does this map $a$ to $0$ and $b$ to $infty$?
        $endgroup$
        – Randall
        Dec 14 '18 at 3:14




        $begingroup$
        How does this map $a$ to $0$ and $b$ to $infty$?
        $endgroup$
        – Randall
        Dec 14 '18 at 3:14












        $begingroup$
        Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
        $endgroup$
        – Chris Custer
        Dec 14 '18 at 3:39




        $begingroup$
        Put in $(1,0,0)$ and $(-1,0,0)$, respectively. You get $"infty "$ and $(0,0)$.
        $endgroup$
        – Chris Custer
        Dec 14 '18 at 3:39












        $begingroup$
        OP wants this to work for generic $a neq b in S^2$.
        $endgroup$
        – Randall
        Dec 14 '18 at 3:42




        $begingroup$
        OP wants this to work for generic $a neq b in S^2$.
        $endgroup$
        – Randall
        Dec 14 '18 at 3:42












        $begingroup$
        Oh I guess I overlooked that.
        $endgroup$
        – Chris Custer
        Dec 14 '18 at 3:46




        $begingroup$
        Oh I guess I overlooked that.
        $endgroup$
        – Chris Custer
        Dec 14 '18 at 3:46











        0












        $begingroup$

        Well not really explicit at all, but Naturally we can consider $S^2$ to be $text{bd}(B_{mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $mathbb{R}^2$ that is ${(x,y,0)|x,yinmathbb{R}}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $mathbb{R}^2$. Complements of closed disc are neighborhoods of $infty$ as we define an open set containing $infty$ to be the complement of a compact set in $mathbb{R}^2$



        Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Well not really explicit at all, but Naturally we can consider $S^2$ to be $text{bd}(B_{mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $mathbb{R}^2$ that is ${(x,y,0)|x,yinmathbb{R}}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $mathbb{R}^2$. Complements of closed disc are neighborhoods of $infty$ as we define an open set containing $infty$ to be the complement of a compact set in $mathbb{R}^2$



          Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Well not really explicit at all, but Naturally we can consider $S^2$ to be $text{bd}(B_{mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $mathbb{R}^2$ that is ${(x,y,0)|x,yinmathbb{R}}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $mathbb{R}^2$. Complements of closed disc are neighborhoods of $infty$ as we define an open set containing $infty$ to be the complement of a compact set in $mathbb{R}^2$



            Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.






            share|cite|improve this answer









            $endgroup$



            Well not really explicit at all, but Naturally we can consider $S^2$ to be $text{bd}(B_{mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $mathbb{R}^2$ that is ${(x,y,0)|x,yinmathbb{R}}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $mathbb{R}^2$. Complements of closed disc are neighborhoods of $infty$ as we define an open set containing $infty$ to be the complement of a compact set in $mathbb{R}^2$



            Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 14 '18 at 3:09









            MelodyMelody

            78812




            78812























                0












                $begingroup$

                I will use the following fact: if $equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X equiv Y$ imples $X^* equiv Y^*$. The idea is to take a homeomorphism $f : X to Y$ and define $f^* :X^* to Y^*$ via $f^*(x) = f(x)$ when $x in X$ and $f^*(infty_X) = infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.



                Now, for any $k geq 1$, the map



                $$
                g(x_1, dots, x_{k+1}) = frac{1}{1-x_{k+1}}(x_1, dots, x_k)
                $$



                from $X_k := S^k setminus {(0,dots,0,1)}$ to $mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,dots,1)$ with $x in S^k$, and $g(x)$ is the intersection of the line with ${x_{k+1} = 0} subset mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* to (mathbb{R}^k)^*$. You can check that $S^k equiv (X_k)^*$ via $f(x) = x$ when $x neq N_k$, and $f(N_k) := infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $mathbb{R}^k$,



                $$
                begin{align}
                Phi : & S^k rightarrow (mathbb{R}^k)^* \
                &xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
                end{align}
                $$



                It is clear from here that $Phi(N_k) = infty$ and $Phi(0,0,dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 to S^2$ such that $a mapsto N_k$ and $b mapsto (0,0,dots,-1)$ and post-compose it with $Phi$. Can you take it from here?






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  I will use the following fact: if $equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X equiv Y$ imples $X^* equiv Y^*$. The idea is to take a homeomorphism $f : X to Y$ and define $f^* :X^* to Y^*$ via $f^*(x) = f(x)$ when $x in X$ and $f^*(infty_X) = infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.



                  Now, for any $k geq 1$, the map



                  $$
                  g(x_1, dots, x_{k+1}) = frac{1}{1-x_{k+1}}(x_1, dots, x_k)
                  $$



                  from $X_k := S^k setminus {(0,dots,0,1)}$ to $mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,dots,1)$ with $x in S^k$, and $g(x)$ is the intersection of the line with ${x_{k+1} = 0} subset mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* to (mathbb{R}^k)^*$. You can check that $S^k equiv (X_k)^*$ via $f(x) = x$ when $x neq N_k$, and $f(N_k) := infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $mathbb{R}^k$,



                  $$
                  begin{align}
                  Phi : & S^k rightarrow (mathbb{R}^k)^* \
                  &xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
                  end{align}
                  $$



                  It is clear from here that $Phi(N_k) = infty$ and $Phi(0,0,dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 to S^2$ such that $a mapsto N_k$ and $b mapsto (0,0,dots,-1)$ and post-compose it with $Phi$. Can you take it from here?






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    I will use the following fact: if $equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X equiv Y$ imples $X^* equiv Y^*$. The idea is to take a homeomorphism $f : X to Y$ and define $f^* :X^* to Y^*$ via $f^*(x) = f(x)$ when $x in X$ and $f^*(infty_X) = infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.



                    Now, for any $k geq 1$, the map



                    $$
                    g(x_1, dots, x_{k+1}) = frac{1}{1-x_{k+1}}(x_1, dots, x_k)
                    $$



                    from $X_k := S^k setminus {(0,dots,0,1)}$ to $mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,dots,1)$ with $x in S^k$, and $g(x)$ is the intersection of the line with ${x_{k+1} = 0} subset mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* to (mathbb{R}^k)^*$. You can check that $S^k equiv (X_k)^*$ via $f(x) = x$ when $x neq N_k$, and $f(N_k) := infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $mathbb{R}^k$,



                    $$
                    begin{align}
                    Phi : & S^k rightarrow (mathbb{R}^k)^* \
                    &xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
                    end{align}
                    $$



                    It is clear from here that $Phi(N_k) = infty$ and $Phi(0,0,dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 to S^2$ such that $a mapsto N_k$ and $b mapsto (0,0,dots,-1)$ and post-compose it with $Phi$. Can you take it from here?






                    share|cite|improve this answer











                    $endgroup$



                    I will use the following fact: if $equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X equiv Y$ imples $X^* equiv Y^*$. The idea is to take a homeomorphism $f : X to Y$ and define $f^* :X^* to Y^*$ via $f^*(x) = f(x)$ when $x in X$ and $f^*(infty_X) = infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.



                    Now, for any $k geq 1$, the map



                    $$
                    g(x_1, dots, x_{k+1}) = frac{1}{1-x_{k+1}}(x_1, dots, x_k)
                    $$



                    from $X_k := S^k setminus {(0,dots,0,1)}$ to $mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,dots,1)$ with $x in S^k$, and $g(x)$ is the intersection of the line with ${x_{k+1} = 0} subset mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* to (mathbb{R}^k)^*$. You can check that $S^k equiv (X_k)^*$ via $f(x) = x$ when $x neq N_k$, and $f(N_k) := infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $mathbb{R}^k$,



                    $$
                    begin{align}
                    Phi : & S^k rightarrow (mathbb{R}^k)^* \
                    &xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
                    end{align}
                    $$



                    It is clear from here that $Phi(N_k) = infty$ and $Phi(0,0,dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 to S^2$ such that $a mapsto N_k$ and $b mapsto (0,0,dots,-1)$ and post-compose it with $Phi$. Can you take it from here?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 14 '18 at 3:18

























                    answered Dec 14 '18 at 3:11









                    Guido A.Guido A.

                    7,3901730




                    7,3901730























                        0












                        $begingroup$

                        If you know complex function theory, working on the Riemann sphere $mathbb{C} cup {infty}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $infty$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          If you know complex function theory, working on the Riemann sphere $mathbb{C} cup {infty}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $infty$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            If you know complex function theory, working on the Riemann sphere $mathbb{C} cup {infty}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $infty$.






                            share|cite|improve this answer









                            $endgroup$



                            If you know complex function theory, working on the Riemann sphere $mathbb{C} cup {infty}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $infty$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 14 '18 at 5:02









                            Henno BrandsmaHenno Brandsma

                            108k347114




                            108k347114






























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