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Let $X$ be a topological manifold of dimension $d$, and let $F$ be a collection of continuous maps from $X$ into $\mathbf{R}^d$ such that:

  • $F$ separates points of $X$, i.e. for any two distinct points $x,y\in X$ there is $\varphi\in F$ such that $\varphi(x)\ne\varphi(y)$;

  • for every $x\in X$ there is an open neighbourhood $U$ of $x$ and $\varphi\in F$ such that $\varphi|_U$ is a homeomorphism of $U$ onto an open set $W$ in $\mathbf{R}^d$.

Does it follow that $F$ generates the topology of $X$, i.e. the original topology of $X$ is the minimal topology in which all elements of $F$ are continuous?

Since $F$ consists of continuous maps, it generates the topology weaker than the original one, and since $F$ separates points, the generated topology is Hausdorff.

It seems that this topology is also locally Euclidean, but I cannot articulate the right reasoning. Then we get that the identity is a continuous bijection between the original and generated topologies, and since they both are locally Euclidean it follows that this map is open, and therefore a homeomorphism. However, I feel concerned about the very last argument, since it does not use Hausdorff property, without which the statement is wrong.

PS You may assume that all the objects considered are actually smooth or holomorphic, but I don't think it is relevant here.

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  • $\begingroup$ There is no argument that guarantees second countability. Do you want that as part of your definition of manifold? $\endgroup$ – Sebastian Goette Jul 12 '17 at 7:13
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    $\begingroup$ Consider the set $F$ of all functions on $\mathbb R$ which vanish at infinity and at zero. In the induced topology there is no bounded neighborhood of zero. $\endgroup$ – user1688 Jul 12 '17 at 7:59
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This is just a variation of Corbennick's comment.

The statement in the question is wrong for any non-compact manifold: take $x\in X$ and consider all maps into $\mathbf{R}^{d}$, such that their limit at infinity equals their value at $x$. This collection generates the topology of the one point compactification of $X$ factorized by gluing $\infty$ with $x$, which is not the original one.

One has to tweak this argument for the holomorphic case, however, because holomorphic functions cannot have limit at infinity unless they are constants.

Let $X=D$ be the unit disk and let $F'$ be the set of all functions $f$ on $\overline{D}$ which are continuous on $\overline{D}$, holomorphic on $D$ and such that $f(0)=f(1)$. Let $F$ be the set of restrictions of the members of $F'$ on $D$. Since $F$ contains functions $z^2-z$ and $z^3-z$, it follows that at every point there is a local biholomorphism. However, $F$ generates the topology of the "folded" disk, not equivalent to the original one.

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