$mathbb{R}cong Xtimes Y$ (homeomorphic) implies $X$ or $Y$ is a point.












4












$begingroup$


The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:




Suppose $mathbb{R}cong Xtimes Y$ (homeomorphic). Then $X$ or $Y$ is a point.




FWIW, the section 2.1 is about (path) connected components (i.e., $pi_0$) and the notion of homotopy.



I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $X,Y$ are arbitrary, and so I don't know how to handle this.



Any hints will be appreciated!






general-topology algebraic-topology







share|cite|improve this question









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  • 2




    $begingroup$
    math.stackexchange.com/questions/57375/… related
    $endgroup$
    – Zachary Selk
    Dec 22 '18 at 5:29
















4












$begingroup$


The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:




Suppose $mathbb{R}cong Xtimes Y$ (homeomorphic). Then $X$ or $Y$ is a point.




FWIW, the section 2.1 is about (path) connected components (i.e., $pi_0$) and the notion of homotopy.



I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $X,Y$ are arbitrary, and so I don't know how to handle this.



Any hints will be appreciated!






general-topology algebraic-topology







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    math.stackexchange.com/questions/57375/… related
    $endgroup$
    – Zachary Selk
    Dec 22 '18 at 5:29














4












4








4


1



$begingroup$


The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:




Suppose $mathbb{R}cong Xtimes Y$ (homeomorphic). Then $X$ or $Y$ is a point.




FWIW, the section 2.1 is about (path) connected components (i.e., $pi_0$) and the notion of homotopy.



I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $X,Y$ are arbitrary, and so I don't know how to handle this.



Any hints will be appreciated!






general-topology algebraic-topology







share|cite|improve this question









$endgroup$




The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:




Suppose $mathbb{R}cong Xtimes Y$ (homeomorphic). Then $X$ or $Y$ is a point.




FWIW, the section 2.1 is about (path) connected components (i.e., $pi_0$) and the notion of homotopy.



I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $X,Y$ are arbitrary, and so I don't know how to handle this.



Any hints will be appreciated!






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 '18 at 3:30









ColescuColescu

3,20511136




3,20511136








  • 2




    $begingroup$
    math.stackexchange.com/questions/57375/… related
    $endgroup$
    – Zachary Selk
    Dec 22 '18 at 5:29














  • 2




    $begingroup$
    math.stackexchange.com/questions/57375/… related
    $endgroup$
    – Zachary Selk
    Dec 22 '18 at 5:29








2




2




$begingroup$
math.stackexchange.com/questions/57375/… related
$endgroup$
– Zachary Selk
Dec 22 '18 at 5:29




$begingroup$
math.stackexchange.com/questions/57375/… related
$endgroup$
– Zachary Selk
Dec 22 '18 at 5:29










2 Answers
2






active

oldest

votes


















6












$begingroup$

This comes from the following theorem:



Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) in X times Y$, the space $X times Y -{(x_0,y_0)}$ is still path connected.



In your case, $X times Y$ must be path connected since $mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $mathbb{R}-pt$ is no longer path connected, but $X times Y -{(x_0,y_0)}$ is. So all you have to do is believe in the theorem.



Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X times Y -{(x_0,y_0)}$. We must show there exists a path between these two points, the path living in $X times Y -{(x_0,y_0)}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.



First, we could have $a=cneq x_0$. Use the fact that ${a} times Ycong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=dneq y_0$.



Next, the case $a=c=x_0$. It follows that $b neq y_0$ and $d neq y_0$. Here's where you need the assumptions that $|X|, |Y| geq 2$. We can choose a point $x'in X$ which is different from $x_0$. Now create the needed path in three segments:
$$
(a,b) to (x',b) to (x',d) to (c,d)
$$
working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.



Finally, there is the case $a neq c$ and $b neq d$. Construct two different three-legged paths
$$
(a,b) to (c,b) to (c,d)
$$
and
$$
(a,b) to (a,d) to (c,d).
$$
It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
    $endgroup$
    – Cameron Buie
    Dec 23 '18 at 22:59






  • 1




    $begingroup$
    @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
    $endgroup$
    – Moishe Cohen
    Dec 27 '18 at 12:48












  • $begingroup$
    @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
    $endgroup$
    – Cameron Buie
    Dec 27 '18 at 13:12










  • $begingroup$
    I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
    $endgroup$
    – Randall
    Dec 27 '18 at 13:22



















2












$begingroup$

Hint: By assumption, $Xtimes Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $Xtimes Y$ minus a point must still be path-connected.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    This comes from the following theorem:



    Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) in X times Y$, the space $X times Y -{(x_0,y_0)}$ is still path connected.



    In your case, $X times Y$ must be path connected since $mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $mathbb{R}-pt$ is no longer path connected, but $X times Y -{(x_0,y_0)}$ is. So all you have to do is believe in the theorem.



    Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X times Y -{(x_0,y_0)}$. We must show there exists a path between these two points, the path living in $X times Y -{(x_0,y_0)}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.



    First, we could have $a=cneq x_0$. Use the fact that ${a} times Ycong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=dneq y_0$.



    Next, the case $a=c=x_0$. It follows that $b neq y_0$ and $d neq y_0$. Here's where you need the assumptions that $|X|, |Y| geq 2$. We can choose a point $x'in X$ which is different from $x_0$. Now create the needed path in three segments:
    $$
    (a,b) to (x',b) to (x',d) to (c,d)
    $$
    working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.



    Finally, there is the case $a neq c$ and $b neq d$. Construct two different three-legged paths
    $$
    (a,b) to (c,b) to (c,d)
    $$
    and
    $$
    (a,b) to (a,d) to (c,d).
    $$
    It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
      $endgroup$
      – Cameron Buie
      Dec 23 '18 at 22:59






    • 1




      $begingroup$
      @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
      $endgroup$
      – Moishe Cohen
      Dec 27 '18 at 12:48












    • $begingroup$
      @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
      $endgroup$
      – Cameron Buie
      Dec 27 '18 at 13:12










    • $begingroup$
      I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
      $endgroup$
      – Randall
      Dec 27 '18 at 13:22
















    6












    $begingroup$

    This comes from the following theorem:



    Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) in X times Y$, the space $X times Y -{(x_0,y_0)}$ is still path connected.



    In your case, $X times Y$ must be path connected since $mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $mathbb{R}-pt$ is no longer path connected, but $X times Y -{(x_0,y_0)}$ is. So all you have to do is believe in the theorem.



    Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X times Y -{(x_0,y_0)}$. We must show there exists a path between these two points, the path living in $X times Y -{(x_0,y_0)}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.



    First, we could have $a=cneq x_0$. Use the fact that ${a} times Ycong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=dneq y_0$.



    Next, the case $a=c=x_0$. It follows that $b neq y_0$ and $d neq y_0$. Here's where you need the assumptions that $|X|, |Y| geq 2$. We can choose a point $x'in X$ which is different from $x_0$. Now create the needed path in three segments:
    $$
    (a,b) to (x',b) to (x',d) to (c,d)
    $$
    working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.



    Finally, there is the case $a neq c$ and $b neq d$. Construct two different three-legged paths
    $$
    (a,b) to (c,b) to (c,d)
    $$
    and
    $$
    (a,b) to (a,d) to (c,d).
    $$
    It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
      $endgroup$
      – Cameron Buie
      Dec 23 '18 at 22:59






    • 1




      $begingroup$
      @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
      $endgroup$
      – Moishe Cohen
      Dec 27 '18 at 12:48












    • $begingroup$
      @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
      $endgroup$
      – Cameron Buie
      Dec 27 '18 at 13:12










    • $begingroup$
      I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
      $endgroup$
      – Randall
      Dec 27 '18 at 13:22














    6












    6








    6





    $begingroup$

    This comes from the following theorem:



    Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) in X times Y$, the space $X times Y -{(x_0,y_0)}$ is still path connected.



    In your case, $X times Y$ must be path connected since $mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $mathbb{R}-pt$ is no longer path connected, but $X times Y -{(x_0,y_0)}$ is. So all you have to do is believe in the theorem.



    Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X times Y -{(x_0,y_0)}$. We must show there exists a path between these two points, the path living in $X times Y -{(x_0,y_0)}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.



    First, we could have $a=cneq x_0$. Use the fact that ${a} times Ycong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=dneq y_0$.



    Next, the case $a=c=x_0$. It follows that $b neq y_0$ and $d neq y_0$. Here's where you need the assumptions that $|X|, |Y| geq 2$. We can choose a point $x'in X$ which is different from $x_0$. Now create the needed path in three segments:
    $$
    (a,b) to (x',b) to (x',d) to (c,d)
    $$
    working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.



    Finally, there is the case $a neq c$ and $b neq d$. Construct two different three-legged paths
    $$
    (a,b) to (c,b) to (c,d)
    $$
    and
    $$
    (a,b) to (a,d) to (c,d).
    $$
    It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.






    share|cite|improve this answer









    $endgroup$



    This comes from the following theorem:



    Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) in X times Y$, the space $X times Y -{(x_0,y_0)}$ is still path connected.



    In your case, $X times Y$ must be path connected since $mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $mathbb{R}-pt$ is no longer path connected, but $X times Y -{(x_0,y_0)}$ is. So all you have to do is believe in the theorem.



    Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X times Y -{(x_0,y_0)}$. We must show there exists a path between these two points, the path living in $X times Y -{(x_0,y_0)}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.



    First, we could have $a=cneq x_0$. Use the fact that ${a} times Ycong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=dneq y_0$.



    Next, the case $a=c=x_0$. It follows that $b neq y_0$ and $d neq y_0$. Here's where you need the assumptions that $|X|, |Y| geq 2$. We can choose a point $x'in X$ which is different from $x_0$. Now create the needed path in three segments:
    $$
    (a,b) to (x',b) to (x',d) to (c,d)
    $$
    working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.



    Finally, there is the case $a neq c$ and $b neq d$. Construct two different three-legged paths
    $$
    (a,b) to (c,b) to (c,d)
    $$
    and
    $$
    (a,b) to (a,d) to (c,d).
    $$
    It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 22 '18 at 5:40









    RandallRandall

    10.2k11230




    10.2k11230












    • $begingroup$
      The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
      $endgroup$
      – Cameron Buie
      Dec 23 '18 at 22:59






    • 1




      $begingroup$
      @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
      $endgroup$
      – Moishe Cohen
      Dec 27 '18 at 12:48












    • $begingroup$
      @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
      $endgroup$
      – Cameron Buie
      Dec 27 '18 at 13:12










    • $begingroup$
      I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
      $endgroup$
      – Randall
      Dec 27 '18 at 13:22


















    • $begingroup$
      The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
      $endgroup$
      – Cameron Buie
      Dec 23 '18 at 22:59






    • 1




      $begingroup$
      @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
      $endgroup$
      – Moishe Cohen
      Dec 27 '18 at 12:48












    • $begingroup$
      @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
      $endgroup$
      – Cameron Buie
      Dec 27 '18 at 13:12










    • $begingroup$
      I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
      $endgroup$
      – Randall
      Dec 27 '18 at 13:22
















    $begingroup$
    The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
    $endgroup$
    – Cameron Buie
    Dec 23 '18 at 22:59




    $begingroup$
    The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.)
    $endgroup$
    – Cameron Buie
    Dec 23 '18 at 22:59




    1




    1




    $begingroup$
    @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
    $endgroup$
    – Moishe Cohen
    Dec 27 '18 at 12:48






    $begingroup$
    @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $Xtimes Y$ to its factors.
    $endgroup$
    – Moishe Cohen
    Dec 27 '18 at 12:48














    $begingroup$
    @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
    $endgroup$
    – Cameron Buie
    Dec 27 '18 at 13:12




    $begingroup$
    @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete.
    $endgroup$
    – Cameron Buie
    Dec 27 '18 at 13:12












    $begingroup$
    I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
    $endgroup$
    – Randall
    Dec 27 '18 at 13:22




    $begingroup$
    I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote.
    $endgroup$
    – Randall
    Dec 27 '18 at 13:22











    2












    $begingroup$

    Hint: By assumption, $Xtimes Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $Xtimes Y$ minus a point must still be path-connected.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint: By assumption, $Xtimes Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $Xtimes Y$ minus a point must still be path-connected.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: By assumption, $Xtimes Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $Xtimes Y$ minus a point must still be path-connected.






        share|cite|improve this answer









        $endgroup$



        Hint: By assumption, $Xtimes Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $Xtimes Y$ minus a point must still be path-connected.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 5:35









        asdqasdq

        1,8311518




        1,8311518






























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