Showing quotient topology is the same as direct image topology

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Let $(Y, tau_Y)$ be a pseudo metric space, and let $tau_Y$ be the associated topology. Let $sim$ be the equivalence relation defined by $$x sim y text if d_Y(x,y)=0,$$
for $x, y in Y$. Set $X = Y/sim$, and let $(X, d_X)$ be the quotient metric space. Consider the mapping $P: Y rightarrow X$ defined by $x mapsto [x]$. Let $tau_D$ be the direct image of topology $tau_Y$ on $X$ via the mapping $P$, and let $tau_X$ be the topology of the metric space $(X, d_X)$. Show that $tau_D = tau_X$.



I'm not sure how to start on this problem. I think I need to prove that open sets of $tau_D$ are contained in open sets of $tau_X$ and vice versa. Open balls generate $tau_X$ because it's a metric topology, but the problem is that I don't know what the open sets of $tau_D$ are. Any assistance is appreciated.










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    Let $(Y, tau_Y)$ be a pseudo metric space, and let $tau_Y$ be the associated topology. Let $sim$ be the equivalence relation defined by $$x sim y text if d_Y(x,y)=0,$$
    for $x, y in Y$. Set $X = Y/sim$, and let $(X, d_X)$ be the quotient metric space. Consider the mapping $P: Y rightarrow X$ defined by $x mapsto [x]$. Let $tau_D$ be the direct image of topology $tau_Y$ on $X$ via the mapping $P$, and let $tau_X$ be the topology of the metric space $(X, d_X)$. Show that $tau_D = tau_X$.



    I'm not sure how to start on this problem. I think I need to prove that open sets of $tau_D$ are contained in open sets of $tau_X$ and vice versa. Open balls generate $tau_X$ because it's a metric topology, but the problem is that I don't know what the open sets of $tau_D$ are. Any assistance is appreciated.










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      Let $(Y, tau_Y)$ be a pseudo metric space, and let $tau_Y$ be the associated topology. Let $sim$ be the equivalence relation defined by $$x sim y text if d_Y(x,y)=0,$$
      for $x, y in Y$. Set $X = Y/sim$, and let $(X, d_X)$ be the quotient metric space. Consider the mapping $P: Y rightarrow X$ defined by $x mapsto [x]$. Let $tau_D$ be the direct image of topology $tau_Y$ on $X$ via the mapping $P$, and let $tau_X$ be the topology of the metric space $(X, d_X)$. Show that $tau_D = tau_X$.



      I'm not sure how to start on this problem. I think I need to prove that open sets of $tau_D$ are contained in open sets of $tau_X$ and vice versa. Open balls generate $tau_X$ because it's a metric topology, but the problem is that I don't know what the open sets of $tau_D$ are. Any assistance is appreciated.










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      Let $(Y, tau_Y)$ be a pseudo metric space, and let $tau_Y$ be the associated topology. Let $sim$ be the equivalence relation defined by $$x sim y text if d_Y(x,y)=0,$$
      for $x, y in Y$. Set $X = Y/sim$, and let $(X, d_X)$ be the quotient metric space. Consider the mapping $P: Y rightarrow X$ defined by $x mapsto [x]$. Let $tau_D$ be the direct image of topology $tau_Y$ on $X$ via the mapping $P$, and let $tau_X$ be the topology of the metric space $(X, d_X)$. Show that $tau_D = tau_X$.



      I'm not sure how to start on this problem. I think I need to prove that open sets of $tau_D$ are contained in open sets of $tau_X$ and vice versa. Open balls generate $tau_X$ because it's a metric topology, but the problem is that I don't know what the open sets of $tau_D$ are. Any assistance is appreciated.







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      edited Sep 7 at 3:02

























      asked Sep 7 at 2:40









      Pawnee

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          I think you can prove it by showing $tau_D$ and $tau_X$ contain each other.



          • $tau_Dsubset tau_X$. This is because each open ball in $tau_Y$ is mapped onto an open ball in $tau_ X$.


          • $tau_Xsubset tau_D$. This is because $tau_D$ is the quotient topology, and $Pcolon (Y,tau_Y)to(X,tau_X)$ is continuous, since $P^-1(B_X(x,varepsilon))=bigcup_P(y)=xB_Y(y,varepsilon)$ for each $xin X$ and $varepsilon> 0$, where $B_X(x,varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $varepsilon$.






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          • Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
            – Pawnee
            Sep 7 at 11:24







          • 1




            Actually never mind, I see why that is. Thanks again!
            – Pawnee
            Sep 7 at 11:27










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          I think you can prove it by showing $tau_D$ and $tau_X$ contain each other.



          • $tau_Dsubset tau_X$. This is because each open ball in $tau_Y$ is mapped onto an open ball in $tau_ X$.


          • $tau_Xsubset tau_D$. This is because $tau_D$ is the quotient topology, and $Pcolon (Y,tau_Y)to(X,tau_X)$ is continuous, since $P^-1(B_X(x,varepsilon))=bigcup_P(y)=xB_Y(y,varepsilon)$ for each $xin X$ and $varepsilon> 0$, where $B_X(x,varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $varepsilon$.






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          • Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
            – Pawnee
            Sep 7 at 11:24







          • 1




            Actually never mind, I see why that is. Thanks again!
            – Pawnee
            Sep 7 at 11:27














          up vote
          1
          down vote













          I think you can prove it by showing $tau_D$ and $tau_X$ contain each other.



          • $tau_Dsubset tau_X$. This is because each open ball in $tau_Y$ is mapped onto an open ball in $tau_ X$.


          • $tau_Xsubset tau_D$. This is because $tau_D$ is the quotient topology, and $Pcolon (Y,tau_Y)to(X,tau_X)$ is continuous, since $P^-1(B_X(x,varepsilon))=bigcup_P(y)=xB_Y(y,varepsilon)$ for each $xin X$ and $varepsilon> 0$, where $B_X(x,varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $varepsilon$.






          share|cite|improve this answer




















          • Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
            – Pawnee
            Sep 7 at 11:24







          • 1




            Actually never mind, I see why that is. Thanks again!
            – Pawnee
            Sep 7 at 11:27












          up vote
          1
          down vote










          up vote
          1
          down vote









          I think you can prove it by showing $tau_D$ and $tau_X$ contain each other.



          • $tau_Dsubset tau_X$. This is because each open ball in $tau_Y$ is mapped onto an open ball in $tau_ X$.


          • $tau_Xsubset tau_D$. This is because $tau_D$ is the quotient topology, and $Pcolon (Y,tau_Y)to(X,tau_X)$ is continuous, since $P^-1(B_X(x,varepsilon))=bigcup_P(y)=xB_Y(y,varepsilon)$ for each $xin X$ and $varepsilon> 0$, where $B_X(x,varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $varepsilon$.






          share|cite|improve this answer












          I think you can prove it by showing $tau_D$ and $tau_X$ contain each other.



          • $tau_Dsubset tau_X$. This is because each open ball in $tau_Y$ is mapped onto an open ball in $tau_ X$.


          • $tau_Xsubset tau_D$. This is because $tau_D$ is the quotient topology, and $Pcolon (Y,tau_Y)to(X,tau_X)$ is continuous, since $P^-1(B_X(x,varepsilon))=bigcup_P(y)=xB_Y(y,varepsilon)$ for each $xin X$ and $varepsilon> 0$, where $B_X(x,varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $varepsilon$.







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          answered Sep 7 at 3:32









          josephz

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          1,4002419











          • Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
            – Pawnee
            Sep 7 at 11:24







          • 1




            Actually never mind, I see why that is. Thanks again!
            – Pawnee
            Sep 7 at 11:27
















          • Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
            – Pawnee
            Sep 7 at 11:24







          • 1




            Actually never mind, I see why that is. Thanks again!
            – Pawnee
            Sep 7 at 11:27















          Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
          – Pawnee
          Sep 7 at 11:24





          Thanks for your help. So $tau_D subset tau_X$ is basically given to me by definition of the direct image topology. Got it. But since $Y$ is a pseudo metric space, why are the open sets of $Y$ generated by open balls? That doesn't make intuitive sense to me.
          – Pawnee
          Sep 7 at 11:24





          1




          1




          Actually never mind, I see why that is. Thanks again!
          – Pawnee
          Sep 7 at 11:27




          Actually never mind, I see why that is. Thanks again!
          – Pawnee
          Sep 7 at 11:27

















           

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