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In their book Differential Topology, the authors Guillemin and Pollack give the following counterexample to the claim that the image of an injective immersion $f: X \to Y$ must necessarily be a submanifold of $Y$:

Let $g : \Bbb R \to S^1$ be the local diffeomorphism $g(t) = (\cos 2\pi t, \sin 2\pi t)$. Define $G : \Bbb R^2 \to S^1 \times S^1$ by $G(x, y) = (g(x), g(y))$. $G$ is a local diffeomorphism of the plane onto the torus. In fact, looking at $G$ on the fundamental unit square, one may consider it to be a construction of the torus by gluing opposite sides of the square together. Now define a map of $\Bbb R$ into the torus by restricting $G$ to a straight line through the origin in $\Bbb R^2$ with irrational slope. Since G is a local diffeomorphism, this is an immersion that wraps $ \Bbb R$ around the torus. Moreover, the irrationality of the slope implies that the immersion is one-to-one and that its image is a dense subset of the torus!

On explaining what goes wrong here they continue:

Notice that the injective immersion behaves strangely because it maps too many points "near infinity" in $\Bbb R$ into small regions of the image. Perhaps prohibiting this behavior will sufficiently tame immersions. The general topological analog of points "near infinity" is the exterior of a compact subset in a given space, the compact set being thought of as very large.

Now, my interpretation of "mapping too many points "near infinity" into small regions of the image" is if we take any $\epsilon$ ball $B_{\epsilon}$ such that $B_{\epsilon}\cap S^1 \times S^1 \neq \phi$ and any $R>0$, we can find infinitely many points $x>R$ such that $f(x)\in B_{\epsilon}\cap S^1 \times S^1$. Is my interpretation correct? And if so, how does that cause a problem here?

Why does the topological analog of points "near infinity" require the set to be compact? I don't see why it cannot be thought of as any large set. What's so special about compactness here?

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Your interpretation is correct. What goes wrong with this situation is that you can then find a sequence of points $(x_n)$ in $\mathbb{R}$ which go to infinity (and in particular, do not converge in $\mathbb{R}$), but such that the points $f(x_n)$ are contained in smaller and smaller balls which force them to converge to a point of $S^1\times S^1$. Moreover, we can choose such a sequence to converge to any point of $S^1\times S^1$ at all. So in some sense the map $f$ badly fails to preserve the topology--there are sequences which diverge in the domain but which become sequences converging to any point you want in the codomain.

As for the role of compactness, the text explains this somewhat poorly. The role of compactness is actually to ensure that you have a small set, not a large set. The idea is that a compact set is small enough that you can't go out "to infinity" within the compact set, and so all the ways to go to infinity are outside it. That is, the complement of your compact set will be an entire "neighborhood" of infinity. On the other hand, the fact that you're allowed to take an arbitrary compact set means that you can have a "large" set in some sense. That is, even though within your compact set you can't go out to infinity, you can choose the compact set to get as close as you want to infinity (e.g., in the case of $\mathbb{R}$, you can choose the compact set to be $[-R,R]$ for some very large $R$). As you take larger and larger compact sets, the complements form smaller and smaller "neighborhoods" of infinity which tell you what it means to "approach infinity". This is the same way that taking smaller and smaller neighborhoods $(-\epsilon,\epsilon)$ of $0\in\mathbb{R}$ tells you what it means to converge to $0$.

In the background of all of this discussion is the theorem that a continuous injection from a compact space to a Hausdorff space is automatically an embedding with closed image. In particular, if you take the badly behaved map $f:\mathbb{R}\to S^1\times S^1$ and restrict it to any compact subset of $\mathbb{R}$, then it is perfectly well-behaved (in particular, its image is closed and it is a homeomorphism to its image). So the sort of bad behavior we're worried about here cannot occur when restricted to a compact set. This motivates why it is natural to look only at what happens outside of arbitrarily large compact sets if you want to impose restrictions to prevent this sort of pathology.

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  • $\begingroup$ Wow....just wonderfully written. $\endgroup$ Jul 28, 2017 at 19:32
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I remember this problem. .. though I must admit that as an undergraduate I didn't catch the part about it being a counterexample. .. I just remember $R$, the line with irrational slope, being dense in the torus, $(S^1×S^1) \cong {R^2/Z^2}$... Well, i guess having the domain space compact might very well rule out the "pathology" that is detected here... as I recall everything in guilleman and pollack is done in $R^n$. Hence we have heine-borel and compact means closed and bounded. The key appears to be boundedness. Thus too many points "near" $\infty $ couldn't be mapped into small balls $B_\epsilon$...

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