Prove that $S^2 vee S^2$ is path connected?












1












$begingroup$


Prove that $S^2 vee S^2$ is path connected?



Let $a$ and $b$ be two points in a topological space $X$. A path in $X$ from $a$ to $b$ is a continuous map $f$ from [0,1] to $X$ s.t $f$(0) = $a$ and $f$(1)=$b$.



A topological space X is said to be path connected iff given any two points $a$ and $b$ , there exists a path in $X$ from a to b



I can draw but i can’t give a function.






general-topology path-connected







share|cite|improve this question











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  • 4




    $begingroup$
    What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
    $endgroup$
    – freakish
    Dec 20 '18 at 20:32












  • $begingroup$
    @freakisb i mean Spherical coordinates
    $endgroup$
    – Angel
    Dec 20 '18 at 20:34










  • $begingroup$
    I don't even know how that answers my questions...
    $endgroup$
    – freakish
    Dec 20 '18 at 20:34






  • 1




    $begingroup$
    $X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 20:50








  • 1




    $begingroup$
    The symbol you're looking for is times, I think.
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 20:51
















1












$begingroup$


Prove that $S^2 vee S^2$ is path connected?



Let $a$ and $b$ be two points in a topological space $X$. A path in $X$ from $a$ to $b$ is a continuous map $f$ from [0,1] to $X$ s.t $f$(0) = $a$ and $f$(1)=$b$.



A topological space X is said to be path connected iff given any two points $a$ and $b$ , there exists a path in $X$ from a to b



I can draw but i can’t give a function.






general-topology path-connected







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
    $endgroup$
    – freakish
    Dec 20 '18 at 20:32












  • $begingroup$
    @freakisb i mean Spherical coordinates
    $endgroup$
    – Angel
    Dec 20 '18 at 20:34










  • $begingroup$
    I don't even know how that answers my questions...
    $endgroup$
    – freakish
    Dec 20 '18 at 20:34






  • 1




    $begingroup$
    $X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 20:50








  • 1




    $begingroup$
    The symbol you're looking for is times, I think.
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 20:51














1












1








1





$begingroup$


Prove that $S^2 vee S^2$ is path connected?



Let $a$ and $b$ be two points in a topological space $X$. A path in $X$ from $a$ to $b$ is a continuous map $f$ from [0,1] to $X$ s.t $f$(0) = $a$ and $f$(1)=$b$.



A topological space X is said to be path connected iff given any two points $a$ and $b$ , there exists a path in $X$ from a to b



I can draw but i can’t give a function.






general-topology path-connected







share|cite|improve this question











$endgroup$




Prove that $S^2 vee S^2$ is path connected?



Let $a$ and $b$ be two points in a topological space $X$. A path in $X$ from $a$ to $b$ is a continuous map $f$ from [0,1] to $X$ s.t $f$(0) = $a$ and $f$(1)=$b$.



A topological space X is said to be path connected iff given any two points $a$ and $b$ , there exists a path in $X$ from a to b



I can draw but i can’t give a function.






general-topology path-connected




general-topology path-connected






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 0:31









Math1000

19.4k31746




19.4k31746










asked Dec 20 '18 at 20:27









Angel Angel

73




73








  • 4




    $begingroup$
    What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
    $endgroup$
    – freakish
    Dec 20 '18 at 20:32












  • $begingroup$
    @freakisb i mean Spherical coordinates
    $endgroup$
    – Angel
    Dec 20 '18 at 20:34










  • $begingroup$
    I don't even know how that answers my questions...
    $endgroup$
    – freakish
    Dec 20 '18 at 20:34






  • 1




    $begingroup$
    $X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 20:50








  • 1




    $begingroup$
    The symbol you're looking for is times, I think.
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 20:51














  • 4




    $begingroup$
    What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
    $endgroup$
    – freakish
    Dec 20 '18 at 20:32












  • $begingroup$
    @freakisb i mean Spherical coordinates
    $endgroup$
    – Angel
    Dec 20 '18 at 20:34










  • $begingroup$
    I don't even know how that answers my questions...
    $endgroup$
    – freakish
    Dec 20 '18 at 20:34






  • 1




    $begingroup$
    $X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 20:50








  • 1




    $begingroup$
    The symbol you're looking for is times, I think.
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 20:51








4




4




$begingroup$
What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
$endgroup$
– freakish
Dec 20 '18 at 20:32






$begingroup$
What is $s^2$? What is $*$? Did you mean the $2$-dimensional sphere $S^2$ and the wedge sum? I don't think its the Cartesian product cause otherwise how can you draw it? That would require some serious skill. :)
$endgroup$
– freakish
Dec 20 '18 at 20:32














$begingroup$
@freakisb i mean Spherical coordinates
$endgroup$
– Angel
Dec 20 '18 at 20:34




$begingroup$
@freakisb i mean Spherical coordinates
$endgroup$
– Angel
Dec 20 '18 at 20:34












$begingroup$
I don't even know how that answers my questions...
$endgroup$
– freakish
Dec 20 '18 at 20:34




$begingroup$
I don't even know how that answers my questions...
$endgroup$
– freakish
Dec 20 '18 at 20:34




1




1




$begingroup$
$X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:50






$begingroup$
$X * Y$ is not usual notation for the wedge of pointed spaces $X$ and $Y$: the wedge is usually written $X vee Y$. $X * Y$ is the usual notation for the topological join of $X$ and $Y$. As you now write $X mathop{x} Y$, you almost certainly mean the product $X times Y$ (as Asaf suggests) and you definitely need to start using MathJax to typeset your questions.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:50






1




1




$begingroup$
The symbol you're looking for is times, I think.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:51




$begingroup$
The symbol you're looking for is times, I think.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:51










1 Answer
1






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oldest

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$begingroup$

Let $p$ be the joining point, which belongs to both copies of $S^2$.



Let $x_1,x_2in S^2vee S^2$. There exist paths $f_1, f_2$ such that $f_i(0)=x_i$ and $f_i(1)=p$. Then consider the path



$$F(t) = begin{cases} f_1(2t) , tin[0,.5) \ f_2(2-2t) , tin[0.5,1]end{cases}$$



Note, $F(t)$ is sometimes called the path composition of $f_1, f_2^{-1}$ and might be notated $f_1*f_2^{-1}$ (the "inverse" because here I traverse $f_2$ backwards).



There was also nothing special about $S^2$ here.






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    2












    $begingroup$

    Let $p$ be the joining point, which belongs to both copies of $S^2$.



    Let $x_1,x_2in S^2vee S^2$. There exist paths $f_1, f_2$ such that $f_i(0)=x_i$ and $f_i(1)=p$. Then consider the path



    $$F(t) = begin{cases} f_1(2t) , tin[0,.5) \ f_2(2-2t) , tin[0.5,1]end{cases}$$



    Note, $F(t)$ is sometimes called the path composition of $f_1, f_2^{-1}$ and might be notated $f_1*f_2^{-1}$ (the "inverse" because here I traverse $f_2$ backwards).



    There was also nothing special about $S^2$ here.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Let $p$ be the joining point, which belongs to both copies of $S^2$.



      Let $x_1,x_2in S^2vee S^2$. There exist paths $f_1, f_2$ such that $f_i(0)=x_i$ and $f_i(1)=p$. Then consider the path



      $$F(t) = begin{cases} f_1(2t) , tin[0,.5) \ f_2(2-2t) , tin[0.5,1]end{cases}$$



      Note, $F(t)$ is sometimes called the path composition of $f_1, f_2^{-1}$ and might be notated $f_1*f_2^{-1}$ (the "inverse" because here I traverse $f_2$ backwards).



      There was also nothing special about $S^2$ here.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $p$ be the joining point, which belongs to both copies of $S^2$.



        Let $x_1,x_2in S^2vee S^2$. There exist paths $f_1, f_2$ such that $f_i(0)=x_i$ and $f_i(1)=p$. Then consider the path



        $$F(t) = begin{cases} f_1(2t) , tin[0,.5) \ f_2(2-2t) , tin[0.5,1]end{cases}$$



        Note, $F(t)$ is sometimes called the path composition of $f_1, f_2^{-1}$ and might be notated $f_1*f_2^{-1}$ (the "inverse" because here I traverse $f_2$ backwards).



        There was also nothing special about $S^2$ here.






        share|cite|improve this answer











        $endgroup$



        Let $p$ be the joining point, which belongs to both copies of $S^2$.



        Let $x_1,x_2in S^2vee S^2$. There exist paths $f_1, f_2$ such that $f_i(0)=x_i$ and $f_i(1)=p$. Then consider the path



        $$F(t) = begin{cases} f_1(2t) , tin[0,.5) \ f_2(2-2t) , tin[0.5,1]end{cases}$$



        Note, $F(t)$ is sometimes called the path composition of $f_1, f_2^{-1}$ and might be notated $f_1*f_2^{-1}$ (the "inverse" because here I traverse $f_2$ backwards).



        There was also nothing special about $S^2$ here.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 21 '18 at 4:23

























        answered Dec 20 '18 at 21:33









        David PetersonDavid Peterson

        8,86821935




        8,86821935






























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