A theorem of a continuous map $f: S^1 to S^1$












2












$begingroup$


This is a theorem from my lecture notes:



If the continuous map $f: S^1 to S^1$ extends to a continuous map $F: B(0,1) to S^1$ the $f$ is homotopic to a constant map.



The proof just defines a homotopy $G$ by $G(z,u) = F(uz)$ so $G(z,1) = F(z) = f(z)$ for $Zin S^1$ and $G(z,0) = F(0)$.



I don't understand why the criterion that $f$ ends to a continuous map $F: B(0,1) to S^1$ is necessary? Surely you can still just define $G(z,u) = f(uz)$ to get the same result?






general-topology homotopy-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
    $endgroup$
    – Mario Carneiro
    Dec 17 '12 at 18:26








  • 4




    $begingroup$
    What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
    $endgroup$
    – WimC
    Dec 17 '12 at 18:30
















2












$begingroup$


This is a theorem from my lecture notes:



If the continuous map $f: S^1 to S^1$ extends to a continuous map $F: B(0,1) to S^1$ the $f$ is homotopic to a constant map.



The proof just defines a homotopy $G$ by $G(z,u) = F(uz)$ so $G(z,1) = F(z) = f(z)$ for $Zin S^1$ and $G(z,0) = F(0)$.



I don't understand why the criterion that $f$ ends to a continuous map $F: B(0,1) to S^1$ is necessary? Surely you can still just define $G(z,u) = f(uz)$ to get the same result?






general-topology homotopy-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
    $endgroup$
    – Mario Carneiro
    Dec 17 '12 at 18:26








  • 4




    $begingroup$
    What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
    $endgroup$
    – WimC
    Dec 17 '12 at 18:30














2












2








2


0



$begingroup$


This is a theorem from my lecture notes:



If the continuous map $f: S^1 to S^1$ extends to a continuous map $F: B(0,1) to S^1$ the $f$ is homotopic to a constant map.



The proof just defines a homotopy $G$ by $G(z,u) = F(uz)$ so $G(z,1) = F(z) = f(z)$ for $Zin S^1$ and $G(z,0) = F(0)$.



I don't understand why the criterion that $f$ ends to a continuous map $F: B(0,1) to S^1$ is necessary? Surely you can still just define $G(z,u) = f(uz)$ to get the same result?






general-topology homotopy-theory







share|cite|improve this question









$endgroup$




This is a theorem from my lecture notes:



If the continuous map $f: S^1 to S^1$ extends to a continuous map $F: B(0,1) to S^1$ the $f$ is homotopic to a constant map.



The proof just defines a homotopy $G$ by $G(z,u) = F(uz)$ so $G(z,1) = F(z) = f(z)$ for $Zin S^1$ and $G(z,0) = F(0)$.



I don't understand why the criterion that $f$ ends to a continuous map $F: B(0,1) to S^1$ is necessary? Surely you can still just define $G(z,u) = f(uz)$ to get the same result?






general-topology homotopy-theory




general-topology homotopy-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '12 at 18:22









user53076user53076

686723




686723












  • $begingroup$
    If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
    $endgroup$
    – Mario Carneiro
    Dec 17 '12 at 18:26








  • 4




    $begingroup$
    What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
    $endgroup$
    – WimC
    Dec 17 '12 at 18:30


















  • $begingroup$
    If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
    $endgroup$
    – Mario Carneiro
    Dec 17 '12 at 18:26








  • 4




    $begingroup$
    What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
    $endgroup$
    – WimC
    Dec 17 '12 at 18:30
















$begingroup$
If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
$endgroup$
– Mario Carneiro
Dec 17 '12 at 18:26






$begingroup$
If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim.
$endgroup$
– Mario Carneiro
Dec 17 '12 at 18:26






4




4




$begingroup$
What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
$endgroup$
– WimC
Dec 17 '12 at 18:30




$begingroup$
What is $f(uz)$ when $u < 1$? Note that $uz not in S^1$ in this case.
$endgroup$
– WimC
Dec 17 '12 at 18:30










1 Answer
1






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1












$begingroup$

In this case, looking at the picture clarifies the situation.



You start with a map $f : S^1 to S^1$ that extends to a map $F : D^2 to S^1$.



You want a homotopy $H : S^1 times I to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = operatorname{const}_e$ for some $e in S^1$.



Essentially what we do is collapse $S^1 times {1}$ to a point, so that $(S^1 times I) / (S^1 times {1}) cong D^2$ (the cone construction).



Collapsing Subspace to a Point.



In this case we define $H := F circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$



In particular, $H(x, t) = Fbig(x(1 -t) big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = operatorname{const}_e$.



The key part is that for $t in (0, 1]$, $x(1 - t) in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.






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    active

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    $begingroup$

    In this case, looking at the picture clarifies the situation.



    You start with a map $f : S^1 to S^1$ that extends to a map $F : D^2 to S^1$.



    You want a homotopy $H : S^1 times I to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = operatorname{const}_e$ for some $e in S^1$.



    Essentially what we do is collapse $S^1 times {1}$ to a point, so that $(S^1 times I) / (S^1 times {1}) cong D^2$ (the cone construction).



    Collapsing Subspace to a Point.



    In this case we define $H := F circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$



    In particular, $H(x, t) = Fbig(x(1 -t) big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = operatorname{const}_e$.



    The key part is that for $t in (0, 1]$, $x(1 - t) in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      In this case, looking at the picture clarifies the situation.



      You start with a map $f : S^1 to S^1$ that extends to a map $F : D^2 to S^1$.



      You want a homotopy $H : S^1 times I to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = operatorname{const}_e$ for some $e in S^1$.



      Essentially what we do is collapse $S^1 times {1}$ to a point, so that $(S^1 times I) / (S^1 times {1}) cong D^2$ (the cone construction).



      Collapsing Subspace to a Point.



      In this case we define $H := F circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$



      In particular, $H(x, t) = Fbig(x(1 -t) big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = operatorname{const}_e$.



      The key part is that for $t in (0, 1]$, $x(1 - t) in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        In this case, looking at the picture clarifies the situation.



        You start with a map $f : S^1 to S^1$ that extends to a map $F : D^2 to S^1$.



        You want a homotopy $H : S^1 times I to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = operatorname{const}_e$ for some $e in S^1$.



        Essentially what we do is collapse $S^1 times {1}$ to a point, so that $(S^1 times I) / (S^1 times {1}) cong D^2$ (the cone construction).



        Collapsing Subspace to a Point.



        In this case we define $H := F circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$



        In particular, $H(x, t) = Fbig(x(1 -t) big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = operatorname{const}_e$.



        The key part is that for $t in (0, 1]$, $x(1 - t) in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.






        share|cite|improve this answer









        $endgroup$



        In this case, looking at the picture clarifies the situation.



        You start with a map $f : S^1 to S^1$ that extends to a map $F : D^2 to S^1$.



        You want a homotopy $H : S^1 times I to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = operatorname{const}_e$ for some $e in S^1$.



        Essentially what we do is collapse $S^1 times {1}$ to a point, so that $(S^1 times I) / (S^1 times {1}) cong D^2$ (the cone construction).



        Collapsing Subspace to a Point.



        In this case we define $H := F circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$



        In particular, $H(x, t) = Fbig(x(1 -t) big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = operatorname{const}_e$.



        The key part is that for $t in (0, 1]$, $x(1 - t) in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 21:02









        Robert CardonaRobert Cardona

        5,28023499




        5,28023499






























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