Why is this imbedding continuous?












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Define the imbedding $i_{x_{1}}: Y rightarrow X times Y$ by $i_{x_{1}}(y) = (x_1,y)$, where $X$ and $Y$ are topological spaces. Then supposedly $i_{x_{1}}$ is continuous, but I fail to see why; we need to show that $i_{x_{1}}^{-1}(O)$ is open, where $O$ is any open subset of $X times Y$. I don't see why this is the case; can anyone provide me a proof of why the imbedding is continuous? Thanks in advance for your time and efforts.






general-topology continuity







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    you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 16:21


















0












$begingroup$


Define the imbedding $i_{x_{1}}: Y rightarrow X times Y$ by $i_{x_{1}}(y) = (x_1,y)$, where $X$ and $Y$ are topological spaces. Then supposedly $i_{x_{1}}$ is continuous, but I fail to see why; we need to show that $i_{x_{1}}^{-1}(O)$ is open, where $O$ is any open subset of $X times Y$. I don't see why this is the case; can anyone provide me a proof of why the imbedding is continuous? Thanks in advance for your time and efforts.






general-topology continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 16:21
















0












0








0





$begingroup$


Define the imbedding $i_{x_{1}}: Y rightarrow X times Y$ by $i_{x_{1}}(y) = (x_1,y)$, where $X$ and $Y$ are topological spaces. Then supposedly $i_{x_{1}}$ is continuous, but I fail to see why; we need to show that $i_{x_{1}}^{-1}(O)$ is open, where $O$ is any open subset of $X times Y$. I don't see why this is the case; can anyone provide me a proof of why the imbedding is continuous? Thanks in advance for your time and efforts.






general-topology continuity







share|cite|improve this question









$endgroup$




Define the imbedding $i_{x_{1}}: Y rightarrow X times Y$ by $i_{x_{1}}(y) = (x_1,y)$, where $X$ and $Y$ are topological spaces. Then supposedly $i_{x_{1}}$ is continuous, but I fail to see why; we need to show that $i_{x_{1}}^{-1}(O)$ is open, where $O$ is any open subset of $X times Y$. I don't see why this is the case; can anyone provide me a proof of why the imbedding is continuous? Thanks in advance for your time and efforts.






general-topology continuity




general-topology continuity






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asked Dec 2 '18 at 16:18









Steven WagterSteven Wagter

1569




1569








  • 1




    $begingroup$
    you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 16:21
















  • 1




    $begingroup$
    you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 16:21










1




1




$begingroup$
you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
$endgroup$
– mathworker21
Dec 2 '18 at 16:21






$begingroup$
you just need to check $i_{x_i}^{-1}(O)$ is open when $O$ is of the form $Utimes V$ for $U$ open in $X$ and $V$ open in $Y$ (since such sets form a basis for the product topology). And $i_{x_1}^{-1}(Utimes V) = emptyset$ if $x_1 not in U$ and $V$ if $x_1 in U$.
$endgroup$
– mathworker21
Dec 2 '18 at 16:21












2 Answers
2






active

oldest

votes


















2












$begingroup$

A function $f : X to Y times Z$ is continuous iff $pi_Y circ f$ and $pi_Z circ f$ are both continuous. (univeral property of continuity of maps into products).



In our case:
$pi_Ycirc i_{x_1} = mathrm{id}_Y$ (the identity) and $pi_X circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:53










  • $begingroup$
    @StevenWagter what book are you using then?
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:55










  • $begingroup$
    Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:56










  • $begingroup$
    @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:58



















1












$begingroup$

One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    2












    $begingroup$

    A function $f : X to Y times Z$ is continuous iff $pi_Y circ f$ and $pi_Z circ f$ are both continuous. (univeral property of continuity of maps into products).



    In our case:
    $pi_Ycirc i_{x_1} = mathrm{id}_Y$ (the identity) and $pi_X circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:53










    • $begingroup$
      @StevenWagter what book are you using then?
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:55










    • $begingroup$
      Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:56










    • $begingroup$
      @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:58
















    2












    $begingroup$

    A function $f : X to Y times Z$ is continuous iff $pi_Y circ f$ and $pi_Z circ f$ are both continuous. (univeral property of continuity of maps into products).



    In our case:
    $pi_Ycirc i_{x_1} = mathrm{id}_Y$ (the identity) and $pi_X circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:53










    • $begingroup$
      @StevenWagter what book are you using then?
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:55










    • $begingroup$
      Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:56










    • $begingroup$
      @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:58














    2












    2








    2





    $begingroup$

    A function $f : X to Y times Z$ is continuous iff $pi_Y circ f$ and $pi_Z circ f$ are both continuous. (univeral property of continuity of maps into products).



    In our case:
    $pi_Ycirc i_{x_1} = mathrm{id}_Y$ (the identity) and $pi_X circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.






    share|cite|improve this answer









    $endgroup$



    A function $f : X to Y times Z$ is continuous iff $pi_Y circ f$ and $pi_Z circ f$ are both continuous. (univeral property of continuity of maps into products).



    In our case:
    $pi_Ycirc i_{x_1} = mathrm{id}_Y$ (the identity) and $pi_X circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 16:29









    Henno BrandsmaHenno Brandsma

    107k347114




    107k347114












    • $begingroup$
      Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:53










    • $begingroup$
      @StevenWagter what book are you using then?
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:55










    • $begingroup$
      Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:56










    • $begingroup$
      @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:58


















    • $begingroup$
      Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:53










    • $begingroup$
      @StevenWagter what book are you using then?
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:55










    • $begingroup$
      Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
      $endgroup$
      – Steven Wagter
      Dec 4 '18 at 8:56










    • $begingroup$
      @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
      $endgroup$
      – Henno Brandsma
      Dec 4 '18 at 8:58
















    $begingroup$
    Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:53




    $begingroup$
    Thank you; this wasn't an explicit theorem in my book, so I'm glad I asked.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:53












    $begingroup$
    @StevenWagter what book are you using then?
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:55




    $begingroup$
    @StevenWagter what book are you using then?
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:55












    $begingroup$
    Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:56




    $begingroup$
    Bert Mendelson's introduction to topology; I just saw it was one of the exercises which I haven't done yet though.
    $endgroup$
    – Steven Wagter
    Dec 4 '18 at 8:56












    $begingroup$
    @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:58




    $begingroup$
    @StevenWagter it’s a very handy fact to know about maps into products so do that exercise too. There are solutions on this site already and it’s shown in other standard texts like Munkres.
    $endgroup$
    – Henno Brandsma
    Dec 4 '18 at 8:58











    1












    $begingroup$

    One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.






        share|cite|improve this answer









        $endgroup$



        One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 16:40









        Chris CusterChris Custer

        11.6k3824




        11.6k3824






























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