# Is this an alternative definition of manifolds?

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 in $$\mathbb{R}^m$$ clearly satisfies this property; conversely must such an $$X$$ be an $$n$$-manifold? At least I cannot come up with a $$1$$-dimensional counterexample; the usual examples such as the topologist's sine curve are ruled out by this definition, since it implies locally connectedness. It also rules out the figure 8 since the X shape cannot be injective image of $$\mathbb{R}$$.

I am asking in the continuous category, but feel free to assume smoothness for simplification.

Clarification: at first I just let $$X$$ be any topological space, so people thought I was looking for examples of non-standard manifolds; also someone pointed out that $$\mathbb{R}^n$$ with discrete topology is a counterexample. Therefore I decided to require $$X$$ to be a subset of some Euclidean space, but this is really the case which I think is more interesting.

Update: I feel like at least this is true for $$n=1$$. Here is an outline, although I'm not confident about any of these steps.

(i) Show that if $$Y\subseteq X$$ is the injective image of $$\mathbb{R}$$, then it cannot be the injective image of disjoint copies of $$\mathbb{R}$$.

(ii) If $$f,g$$ are both injective map from $$\mathbb{R}$$ to $$X$$ and $$f(\mathbb{R})\cap g(\mathbb{R})\neq\emptyset$$, then $$f(\mathbb{R})\cup g(\mathbb{R})$$ is either also such an image or a circle.

(iii) Each point of $$X$$ is contained in either a circle or the image of some injective $$f:\mathbb{R}\rightarrow X$$ that is maximal (no larger set is such an image).

(iv) Conclude that $$X$$ is a $$1$$-manifold.

• Both the long line and the line with two origins satisfy your definition, but are not manifolds (because they fail to be second-countable and Hausdorff respectively). Dec 23, 2021 at 19:11
• You ask whether any $X$ satisfying your definition must be a manifold; I (and Vercassivelaunos below) gave examples of spaces satisfying your definition which are not manifolds. EDIT: or do you not include second-countable and Hausdorff in the definition of manifolds in the first place? Dec 23, 2021 at 19:27
• Ah, then I think you should mention that in your question - in my experience they are included by default. Dec 23, 2021 at 19:34
• It does get annoying when you keep modifying the question after we've thought about the question and provided answers. At least make your edits clear additional modifications, rather than changing the ground-rules every time. A number of us have given this serious thought, but when you yank the rug out from under us every time, it may result in our ignoring your future questions. :) Dec 23, 2021 at 20:20
• Using Invariance of Domain, your open neighborhood $V$ is homeomorphic to $\mathbb R^n$, so $X$ is an $n$-manifold. Dec 31, 2021 at 3:54