The curve given below (from John Lee, Introduction to Smooth Manifolds, 2.ed., p.111) is frequently used as an example of an immersed submanifold of $\mathbb R^2$ which is not an embedded submanifold. The arrow means that the curve approaches itself up to an arbitrarily small distance without meeting it. As a parametrization $\gamma:\mathbb R \to \mathbb R^2$, the curve is clearly an injective immersion, since there is no crossing point and the velocity vector never vanishes.
As a set of points, i.e. as a subset of $\mathbb R^2$, the curve $\gamma(\mathbb R)$ shall be endowed with the final topology with respect to $\gamma$, which is the topology induced by the natural topology of $\mathbb R$, and with respect to this topology every point of the curve, including p, has an open neighborhood homeomorphic to an open subset of $\mathbb R$. Thus, the curve is a topological 1-manifold. Furthermore, $(\gamma(\mathbb R),\gamma^{-1},1)$ is a chart giving $\gamma(\mathbb R)$ a smooth structure.
Question 1: What I don't see clearly is why this smooth structure makes the inclusion map $\iota:\gamma(\mathbb R) \to \mathbb R^2 $ an immersion, this being the requirement for $\gamma(\mathbb R)$ to be an immersed submanifold of $\mathbb R^2$.
The requirement for the curve to be an embedded submanifold of $\mathbb R^2$ is that, $\gamma(\mathbb R)$ this time being endowed with the subspace topology of $\mathbb R^2$, the inclusion map $\iota$ is a topological and therefore smooth embedding, which means that $\iota$ induces a homeomorphism $\varphi: \gamma(\mathbb R) \to \iota(\gamma(\mathbb R))$.
Question 2: It is unclear to me, what precisely is the argument showing that such a homeomorphism is not induced in the case of this curve. I think it has to do with open neighborhoods of the point p within the respective topologies, but I cannot find out what exactly it is.
Thanks in advance for help.
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Best Answer
Question 1:
If $G = \gamma(\mathbb R) \subset \mathbb R^2$ as usual receives the subspace topology, then the function $\bar \gamma : \mathbb R \stackrel{\gamma}{\to} G$ is a continuous bijection, but not a homeomorphism (see Question 2). But we can endow $G$ with a unique new topology $\tau$ making $\bar \gamma : \mathbb R \to G_\tau = (G,\tau)$ a homeomorphism (this can be done for each bijection without any further assumptions). Next $G_\tau$ can be endowed with a unique smooth structure making $\bar \gamma$ a diffeomorphism. Obviously $G_\tau$ is a copy of $\mathbb R$ which lies in $\mathbb R^2$ on the level sets, but is not a subspace (or a submanifold). of $\mathbb R^2$.
We know that $\gamma : \mathbb R \to \mathbb R^2$ is an immersion and $\bar \gamma$ is a diffeomorphism, thus $$\iota = \gamma \circ \bar \gamma^{-1} : G_\tau \to \mathbb R^2$$ is an immersion.
Question 2:
You ask why $\bar \gamma : \mathbb R \to G$ is not a homeomorphism. Consider a point $x \in G$ lying right of $p$. Then $G \setminus \{x\}$ is connected (as a subspace of $\mathbb R^2$), but $\mathbb R \setminus \{ \bar \gamma^{-1}(x)\}$ is not connected.