# Must a space that is locally injective image of $\mathbb{R}^n$ be a manifold?

Suppose $$X\subseteq\mathbb{R}^m$$ s.t. for any $$x\in X$$ and any open $$U\subseteq\mathbb{R}^m$$ that contains $$x$$, there exists a smaller open set $$V\subseteq U$$ also containing $$x$$, so that $$V\cap X$$ is the image of some injective continuous map $$f:\mathbb{R}^n\rightarrow\mathbb{R}^m$$.

Any $$n$$-manifold embedded in $$\mathbb{R}^m$$ clearly satisfies this property; conversely must such an $$X$$ be an $$n$$-manifold? Is it true at least for $$n=1$$? The usual counterexamples such as the topologist's sine curve and the figure 8 are ruled out by this definition.

I am asking in the continuous category, but feel free to consider the smooth variants. Clarification: I mean must $$X$$ be a topological manifold if we further assume each $$f$$ is smooth, and must it be a smooth manifold if $$f$$ is an immersion.

• The only possible way this fails is if the (local) inverse $f^{-1}:f(\omega)\to\omega$ is not continuous/$C^k$ (here $\omega\subset \mathbb R^n$ is a neighborhood of the origin). Or am I missing something? Apr 17, 2022 at 8:28
• Following on the previous comment, unless I'm misunderstanding the hypothesis, I think the accepted answer to this question provides a negative answer to this question: math.stackexchange.com/questions/1720851/… Apr 17, 2022 at 8:49
• @GregFriedman But this is the figuere 8 which, as the OP says, isn't a counterexample. Apr 17, 2022 at 10:27
• For what it is worth: in 1D the answer is yes. It is not hard to check that if $f:I\to J$ is a continuou bijection between two intervals then $f^{-1}$ is also continuous. I suspect this is not true anymore in higher dimensions. Apr 18, 2022 at 5:20
• @TobiasDiez By shrinking $U$ (and then $V$), you can assume that $U$ is an open ball. But then you can assume $U=\mathbb R^n$ as well, because an open ball is homeomorphic to $\mathbb R^n$. Apr 21, 2022 at 17:49

Here an outline of an argument for an affirmative answer in the case n=1: By replacing $$X$$ by $$X\cap V$$, we can assume $$X=g(\mathbb R)$$ for a continuous injective map $$g:\mathbb R\rightarrow \mathbb R^m$$.

We prove (under the given assumptions on X) that $$g$$ is a proper map to its image (and thus a homeomorphism to its image). Since $$g$$ is locally on $$\mathbb R$$ a homeomorphism to its image, it suffices to show that no $$x\in X$$ is the limit of a sequence $$g(t_k)$$ with $$|t_k|\rightarrow \infty$$.

Assume by contradiction, that there is $$x\in X$$ with $$g(t_k)\rightarrow x$$ for some $$t_k\rightarrow \infty$$.

(*) Assume further for simplicity that $$g(t)$$ converges to $$z\neq x$$ at the other end of $$\mathbb R$$, as $$t\rightarrow -\infty$$.

1. We will show that in this situation $$\lim_{t\rightarrow \infty}g(t)=x$$.

Indeed, if there is another accumulation point $$y\in \mathbb{R}^m$$ of $$g(t)$$ as $$t\rightarrow \infty$$, pick an open neighborhood $$V$$ of $$x$$ such that $$y,z\notin \overline{V}$$.

Then $$g^{-1}(V)$$ consists of a countably infinite disjoint union of nonempty open intervals $$I_k\subset \mathbb R$$ with compact closures (these intervals tend to $$\infty$$, since $$g$$ only accumulates to $$x$$ as $$t\rightarrow \infty$$). The images of $$J_k:=g(I_k)\subset V$$ are homeomorphic to open intervals. The closures of the $$J_k$$'s in $$\mathbb R^m$$ are pairwise disjoint and their boundary $$\overline{J_k}\setminus J_k\subset \partial V$$, consists of two points for each $$k$$. Thus each $$J_k$$ is closed in $$V$$.

Assume now $$V$$ is also such that $$X\cap V$$ is the image of a continuous injective map $$f:\mathbb R\rightarrow \mathbb R^m$$. Since we can write $$X\cap V= \bigsqcup_k J_k$$ as a disjoint union of countably many nonempty closed subsets, the properties of $$f$$ imply that $$\mathbb R$$ is the disjoint union of countably many nonempty closed subsets. But this is not possible (e.g. by an application of the Baire category theorem), thus we find a contradiction.

1. Obtain from 1. and (*) a contradiction to our assumptions on $$X$$:

$$g(t)$$ converges to $$x$$ as $$t\rightarrow \infty$$, but does not accumulate at $$x$$ for $$t\rightarrow -\infty$$. Thus $$X\setminus {x}$$ has locally near $$x$$ 3 connected components, which shows that $$X\setminus{x}$$ locally near x cannot be the continuous image of $$\mathbb{R}\setminus p$$. This contradicts the assumptions on $$X$$.

1. Removing the assumption $$(*)$$. There are several cases to consider:

a) If $$x$$ is not an accumulation point of $$g(t)$$ as $$t\rightarrow -\infty$$ then the argument from 1. goes through by choosing $$V$$ so small that it contains no such accumulation point, e.g. such that $$g((-\infty,0])\cap V=\emptyset$$ (and by ignoring $$z$$). Step 2. applies now as before to yield a contradiction to the assumptions on $$X$$.

b) If $$x$$ is an accumulation point of $$g(t)$$ as $$t\rightarrow -\infty$$ but not the limit of $$g(t)$$ as $$t\rightarrow -\infty$$, then the argument from 1. still goes through to derive a contradiction, if we choose $$z$$ as another accumulation point of $$g$$ as $$t\rightarrow -\infty$$ (there are now intervals $$I_k$$ tending to both $$\pm \infty$$). This contradiction obtained as in 1. shows (without any further assumptions) that no $$x\in X$$ can be "accumulation point but not limit" at both ends of $$\mathbb R$$. Up to switching $$\pm\infty$$, we are thus in the situation c) if $$(*)$$ does not hold:

c) $$x$$ is the limit of $$g(t)$$ as $$t\rightarrow -\infty$$ (and an accumulation point, but not the limit, of $$g(t)$$ as $$t\rightarrow \infty$$). In this situation, we still follow the basic strategy of 1., picking V and z as before, but the argument changes as slightly: The connected components of $$g^{-1}(V)$$ can be enumerated as follows. One component has the form $$I_0:=(-\infty,a)$$, one (potentially different) component is an open interval $$I_0'\subset \mathbb R$$ containing $$g^{-1}(x)$$, and the other components form a countable sequence of pairwise disjoint open intervals $$I_k, k\geq 1$$ tending to $$\infty$$ (and not containing $$x$$). The images $$g(I_k), k\geq 1$$ are closed in V as before, but also $$g(I_0\cup I_0')$$ is closed in V (since the only accumulation point of $$g(I_0)$$ in $$V\setminus g(I_0)$$ is $$x\in g(I_0')$$) and all of these sets are pairwise disjoint. Thus one obtains a contradiction as in 1., and it follows that $$x$$ is the limit of $$g$$ at both $$\pm \infty$$. Then $$X\setminus x$$ has (similar as in 2.) locally near $$x$$ 4 connected components, and thus it cannot be the continuous image of $$\mathbb R\setminus p$$.

I feel like this is true for $$n=1$$. Here is an outline. Suppose $$X$$ is Hausdorff and

$$\{U\subseteq X\mid U\text{ is open and there exists a continuous bijection }f:\mathbb{R}\rightarrow U\}$$

form a basis of $$X$$. It suffices to show that each of these $$f$$ is actually an embedding. It is certainly an embedding restricted to any finite interval, and the only way it can fail to be an embedding is that there exists a sequence $$t_n\rightarrow\infty$$ s.t. $$f(t_n)\rightarrow a$$ where $$a\in\mathrm{Im}f$$ (or a sequence $$t_n\rightarrow-\infty$$, which is similar).

1. If there is only one such point $$a$$, then $$X$$ locally looks like a "cross road" at $$a$$, and cannot satisfy the definition there, for reason similar to the figure 8.

2. Now suppose there exist two such points $$a,b$$. Take disjoint neighborhoods $$U\ni a,V\ni b$$ in $$X$$, such that there is a continuous bijection $$g:\mathbb{R}\rightarrow U$$. $$f^{-1}(U)$$ is an open subset of $$\mathbb{R}$$ that contains a sequence $$(J_i)_{i\geq 1}$$ of open intervals that "go to infinity", and we can assume $$\overline{f(J_i)}\cap\overline{f(J_j)}=\emptyset$$ for $$i\neq j$$ since $$f^{-1}(U)$$ is "interleaved" with $$f^{-1}(V)$$. Then the desired contradiction should follow from the claim below, which I believe is true.

Claim: if $$f$$, a continuous bijection to a Hausdorff space, is defined on countably many disjoint open intervals $$(J_i)_{i\geq 1}$$, and $$\overline{f(J_i)}\cap\overline{f(J_j)}=\emptyset$$ for $$i\neq j$$, then $$\mathrm{Im}f$$ cannot be path-connected.