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Uniqueness of the dimension of a manifold



The 2019 Stack Overflow Developer Survey Results Are InDimension of submanifold is lees or equal to the dimension of the manifold.Any neighbourhood of a point $x$ in a manifold $X$ ($dim X geq 2$) has a subneighbourhood $V$ of $x$ such that $V setminus {x}$ is connectedExtending a smooth vector field on a manifoldTangent space of manifold has two unit vectors orthogonal to tangent space of its boundaryIs any orientable smooth manifold of dimension $3$ with two independent vector fields parallelizable?Proving a subset is not a submanifoldwhy there is smooth curve in any direction on manifold?On the definition of derivative between smooth manifolds in euclidean spaceWhen we construct a regular sub manifold, can we set any coordinate of the coordinate system equal to 0?Show the disjoint union of 2 n-manifolds is an n-manifoldI'm having trouble understanding manifolds.












2












$begingroup$


Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).



I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.



However, I got stuck.
Any help would be appreciated.










share|cite|improve this question









$endgroup$












  • $begingroup$
    If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
    $endgroup$
    – reuns
    Mar 22 at 1:56








  • 1




    $begingroup$
    The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
    $endgroup$
    – o.h.
    Mar 22 at 7:11
















2












$begingroup$


Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).



I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.



However, I got stuck.
Any help would be appreciated.










share|cite|improve this question









$endgroup$












  • $begingroup$
    If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
    $endgroup$
    – reuns
    Mar 22 at 1:56








  • 1




    $begingroup$
    The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
    $endgroup$
    – o.h.
    Mar 22 at 7:11














2












2








2





$begingroup$


Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).



I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.



However, I got stuck.
Any help would be appreciated.










share|cite|improve this question









$endgroup$




Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).



I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.



However, I got stuck.
Any help would be appreciated.







calculus multivariable-calculus manifolds






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 21 at 23:48









Gabi GGabi G

553210




553210












  • $begingroup$
    If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
    $endgroup$
    – reuns
    Mar 22 at 1:56








  • 1




    $begingroup$
    The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
    $endgroup$
    – o.h.
    Mar 22 at 7:11


















  • $begingroup$
    If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
    $endgroup$
    – reuns
    Mar 22 at 1:56








  • 1




    $begingroup$
    The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
    $endgroup$
    – o.h.
    Mar 22 at 7:11
















$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56






$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56






1




1




$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11




$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11










3 Answers
3






active

oldest

votes


















2












$begingroup$

If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
    $$
    H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
    $$

    where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
    $$
    H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
    $$

    This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)



    Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
      $endgroup$
      – Gabi G
      Mar 22 at 9:21










    • $begingroup$
      @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
      $endgroup$
      – o.h.
      Mar 22 at 15:01



















    0












    $begingroup$

    I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?



    It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.



    It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
      $endgroup$
      – Gabi G
      Mar 22 at 9:19










    • $begingroup$
      @GabiG a topological manifold does not need to have a differentiable structure.
      $endgroup$
      – Zest
      Mar 23 at 1:49












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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.






        share|cite|improve this answer











        $endgroup$



        If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 24 at 3:16

























        answered Mar 22 at 3:37









        triitrii

        85317




        85317























            1












            $begingroup$

            For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
            $$
            H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
            $$

            where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
            $$
            H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
            $$

            This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)



            Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
              $endgroup$
              – Gabi G
              Mar 22 at 9:21










            • $begingroup$
              @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
              $endgroup$
              – o.h.
              Mar 22 at 15:01
















            1












            $begingroup$

            For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
            $$
            H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
            $$

            where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
            $$
            H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
            $$

            This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)



            Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
              $endgroup$
              – Gabi G
              Mar 22 at 9:21










            • $begingroup$
              @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
              $endgroup$
              – o.h.
              Mar 22 at 15:01














            1












            1








            1





            $begingroup$

            For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
            $$
            H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
            $$

            where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
            $$
            H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
            $$

            This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)



            Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.






            share|cite|improve this answer









            $endgroup$



            For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
            $$
            H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
            $$

            where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
            $$
            H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
            $$

            This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)



            Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 22 at 7:03









            o.h.o.h.

            6917




            6917








            • 1




              $begingroup$
              Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
              $endgroup$
              – Gabi G
              Mar 22 at 9:21










            • $begingroup$
              @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
              $endgroup$
              – o.h.
              Mar 22 at 15:01














            • 1




              $begingroup$
              Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
              $endgroup$
              – Gabi G
              Mar 22 at 9:21










            • $begingroup$
              @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
              $endgroup$
              – o.h.
              Mar 22 at 15:01








            1




            1




            $begingroup$
            Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
            $endgroup$
            – Gabi G
            Mar 22 at 9:21




            $begingroup$
            Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
            $endgroup$
            – Gabi G
            Mar 22 at 9:21












            $begingroup$
            @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
            $endgroup$
            – o.h.
            Mar 22 at 15:01




            $begingroup$
            @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
            $endgroup$
            – o.h.
            Mar 22 at 15:01











            0












            $begingroup$

            I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?



            It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.



            It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
              $endgroup$
              – Gabi G
              Mar 22 at 9:19










            • $begingroup$
              @GabiG a topological manifold does not need to have a differentiable structure.
              $endgroup$
              – Zest
              Mar 23 at 1:49
















            0












            $begingroup$

            I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?



            It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.



            It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
              $endgroup$
              – Gabi G
              Mar 22 at 9:19










            • $begingroup$
              @GabiG a topological manifold does not need to have a differentiable structure.
              $endgroup$
              – Zest
              Mar 23 at 1:49














            0












            0








            0





            $begingroup$

            I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?



            It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.



            It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.






            share|cite|improve this answer









            $endgroup$



            I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?



            It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.



            It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 22 at 4:34









            jmerryjmerry

            17k11633




            17k11633












            • $begingroup$
              Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
              $endgroup$
              – Gabi G
              Mar 22 at 9:19










            • $begingroup$
              @GabiG a topological manifold does not need to have a differentiable structure.
              $endgroup$
              – Zest
              Mar 23 at 1:49


















            • $begingroup$
              Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
              $endgroup$
              – Gabi G
              Mar 22 at 9:19










            • $begingroup$
              @GabiG a topological manifold does not need to have a differentiable structure.
              $endgroup$
              – Zest
              Mar 23 at 1:49
















            $begingroup$
            Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
            $endgroup$
            – Gabi G
            Mar 22 at 9:19




            $begingroup$
            Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
            $endgroup$
            – Gabi G
            Mar 22 at 9:19












            $begingroup$
            @GabiG a topological manifold does not need to have a differentiable structure.
            $endgroup$
            – Zest
            Mar 23 at 1:49




            $begingroup$
            @GabiG a topological manifold does not need to have a differentiable structure.
            $endgroup$
            – Zest
            Mar 23 at 1:49


















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