Defining continuity for functions between surfaces












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In general, considering function $f: M_{1} rightarrow M_{2}$ between 2 manifolds, how does one formalize the idea of that function being continuous? Specifically, I am asking this in the context of needing to prove that if a neighbourhood of a point $x$ in the first manifold is in its' interior, then $f(x)$ is in the interior of $M_{2}$






general-topology differential-geometry







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    1












    $begingroup$


    In general, considering function $f: M_{1} rightarrow M_{2}$ between 2 manifolds, how does one formalize the idea of that function being continuous? Specifically, I am asking this in the context of needing to prove that if a neighbourhood of a point $x$ in the first manifold is in its' interior, then $f(x)$ is in the interior of $M_{2}$






    general-topology differential-geometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      In general, considering function $f: M_{1} rightarrow M_{2}$ between 2 manifolds, how does one formalize the idea of that function being continuous? Specifically, I am asking this in the context of needing to prove that if a neighbourhood of a point $x$ in the first manifold is in its' interior, then $f(x)$ is in the interior of $M_{2}$






      general-topology differential-geometry







      share|cite|improve this question











      $endgroup$




      In general, considering function $f: M_{1} rightarrow M_{2}$ between 2 manifolds, how does one formalize the idea of that function being continuous? Specifically, I am asking this in the context of needing to prove that if a neighbourhood of a point $x$ in the first manifold is in its' interior, then $f(x)$ is in the interior of $M_{2}$






      general-topology differential-geometry




      general-topology differential-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 6:35









      Perturbative

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      asked Jan 4 at 6:24









      Aryaman GuptaAryaman Gupta

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          $begingroup$

          Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.



          Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, phi)$ where $U$ is an open set of $M_1$ and $phi : U to phi[U] subseteq mathbb{R}^{2}$ is a homeomorphism and where $phi[U]$ is an open subset of $mathbb{R}^2$.



          To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, psi)$ where $V$ is an open set of $M_2$ and $psi : V to psi[V] subseteq mathbb{R}^{2}$ is a homeomorphism for which $psi[V]$ is an open subset of $mathbb{R}^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh yes, forgot to add that in, thanks @Arthur
            $endgroup$
            – Perturbative
            Jan 4 at 7:02










          • $begingroup$
            Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
            $endgroup$
            – Aryaman Gupta
            Jan 4 at 7:40










          • $begingroup$
            @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
            $endgroup$
            – Perturbative
            Jan 4 at 8:24














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          $begingroup$

          Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.



          Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, phi)$ where $U$ is an open set of $M_1$ and $phi : U to phi[U] subseteq mathbb{R}^{2}$ is a homeomorphism and where $phi[U]$ is an open subset of $mathbb{R}^2$.



          To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, psi)$ where $V$ is an open set of $M_2$ and $psi : V to psi[V] subseteq mathbb{R}^{2}$ is a homeomorphism for which $psi[V]$ is an open subset of $mathbb{R}^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh yes, forgot to add that in, thanks @Arthur
            $endgroup$
            – Perturbative
            Jan 4 at 7:02










          • $begingroup$
            Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
            $endgroup$
            – Aryaman Gupta
            Jan 4 at 7:40










          • $begingroup$
            @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
            $endgroup$
            – Perturbative
            Jan 4 at 8:24


















          1












          $begingroup$

          Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.



          Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, phi)$ where $U$ is an open set of $M_1$ and $phi : U to phi[U] subseteq mathbb{R}^{2}$ is a homeomorphism and where $phi[U]$ is an open subset of $mathbb{R}^2$.



          To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, psi)$ where $V$ is an open set of $M_2$ and $psi : V to psi[V] subseteq mathbb{R}^{2}$ is a homeomorphism for which $psi[V]$ is an open subset of $mathbb{R}^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh yes, forgot to add that in, thanks @Arthur
            $endgroup$
            – Perturbative
            Jan 4 at 7:02










          • $begingroup$
            Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
            $endgroup$
            – Aryaman Gupta
            Jan 4 at 7:40










          • $begingroup$
            @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
            $endgroup$
            – Perturbative
            Jan 4 at 8:24
















          1












          1








          1





          $begingroup$

          Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.



          Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, phi)$ where $U$ is an open set of $M_1$ and $phi : U to phi[U] subseteq mathbb{R}^{2}$ is a homeomorphism and where $phi[U]$ is an open subset of $mathbb{R}^2$.



          To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, psi)$ where $V$ is an open set of $M_2$ and $psi : V to psi[V] subseteq mathbb{R}^{2}$ is a homeomorphism for which $psi[V]$ is an open subset of $mathbb{R}^2$.






          share|cite|improve this answer











          $endgroup$



          Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.



          Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, phi)$ where $U$ is an open set of $M_1$ and $phi : U to phi[U] subseteq mathbb{R}^{2}$ is a homeomorphism and where $phi[U]$ is an open subset of $mathbb{R}^2$.



          To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, psi)$ where $V$ is an open set of $M_2$ and $psi : V to psi[V] subseteq mathbb{R}^{2}$ is a homeomorphism for which $psi[V]$ is an open subset of $mathbb{R}^2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 7:01

























          answered Jan 4 at 6:31









          PerturbativePerturbative

          4,46621554




          4,46621554












          • $begingroup$
            Ahh yes, forgot to add that in, thanks @Arthur
            $endgroup$
            – Perturbative
            Jan 4 at 7:02










          • $begingroup$
            Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
            $endgroup$
            – Aryaman Gupta
            Jan 4 at 7:40










          • $begingroup$
            @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
            $endgroup$
            – Perturbative
            Jan 4 at 8:24




















          • $begingroup$
            Ahh yes, forgot to add that in, thanks @Arthur
            $endgroup$
            – Perturbative
            Jan 4 at 7:02










          • $begingroup$
            Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
            $endgroup$
            – Aryaman Gupta
            Jan 4 at 7:40










          • $begingroup$
            @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
            $endgroup$
            – Perturbative
            Jan 4 at 8:24


















          $begingroup$
          Ahh yes, forgot to add that in, thanks @Arthur
          $endgroup$
          – Perturbative
          Jan 4 at 7:02




          $begingroup$
          Ahh yes, forgot to add that in, thanks @Arthur
          $endgroup$
          – Perturbative
          Jan 4 at 7:02












          $begingroup$
          Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
          $endgroup$
          – Aryaman Gupta
          Jan 4 at 7:40




          $begingroup$
          Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it.
          $endgroup$
          – Aryaman Gupta
          Jan 4 at 7:40












          $begingroup$
          @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
          $endgroup$
          – Perturbative
          Jan 4 at 8:24






          $begingroup$
          @AryamanGupta No problem, I'm glad to help. Since $mathbb{R}^n$ is a topological space this answer also applies to $mathbb{R}^n$, but since $mathbb{R}^n$ is a metric space to show continuity of a function $f : mathbb{R}^n to mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $epsilon-delta$ formulation of continuity, in this case we say $f$ is continuous at $x in mathbb{R}^n$ if $forall epsilon > 0$ there exists a $delta > 0$ such that $d(x, y) < delta implies d(f(x), f(y)) < epsilon$
          $endgroup$
          – Perturbative
          Jan 4 at 8:24




















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