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This example comes from Victor Guillemin and Alan Pollack's Differential Topology. They are trying to emphasize that just because an immersion mapping $f: X \to Y$ is injective (we're talking about the infectivity of $f$, not $df_{x}$), the image of $f$ may not necessarily be a submanifold of $Y$. They supply the following figure as an example: figure eight with the intersection reomved (mapping of $\mathbb{R} \to \mathbb{R}^2$

I understand why this is an injective mapping but do not understand why the image would not be a manifold. The intersection is removed so I do not see where any other trouble points may lie.

Anmol Bhullar
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    The image is not a submanifold of $\mathbb{R}^2$. – quasi May 25 '17 at 21:11
  • I don't see why it is not a submanifold of $\mathbb{R}^2$. Can you give me some insight into why :p – Anmol Bhullar May 25 '17 at 21:16
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    Because as a subset of $\mathbb{R}^2$, the intersection point is present (in the image). – quasi May 25 '17 at 21:19
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    Just because the mapping is injective doesn't mean that the image isn't itself a cross at that point. Presumably the would-be intersection point is the image of zero, and it would be the image of "the point at infinity", but that point doesn't exist in $\mathbb{R}^1$. Nevertheless, in the vicinity of that point, no matter how close you "look", it's still a cross. – Brian Tung May 25 '17 at 21:24
  • Let me see if I understand you and @quasi correctly: No matter what how small the neighbourhood around 0 we are looking at, it still looks like a cross because the lines (the ones that don't touch 0) are infinitesimally close to 0 and would be present in the neighbourhood. Thus it is impossible a find a neighbourhood around 0 which can be parameterized which implies the image fails to be a manifold. – Anmol Bhullar May 25 '17 at 21:31
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    Yes. To $\mathbb{R}^2$, the image is just a figure $8$, intersection point included. – quasi May 25 '17 at 21:36
  • Sounds good, I've never thought about it that way before. If you wanted to write it as an answer, I'd accept it. – Anmol Bhullar May 25 '17 at 21:44
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    Note that from a formal point of view the figure $8$ is a manifold but not a submanifold of $\mathbb{R}^2$ – JJR May 25 '17 at 21:44
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    @Anmol Bhullar: Thanks, I appreciate the offer, but I just gave the idea. You finished it. – quasi May 25 '17 at 21:47
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    I love the infectivity typo! – lesath82 May 25 '17 at 21:47
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    @Anmol Bhullar: Also, as JJR notes, the figure 8 can be a manifold, simply by coordinatizing it from the domain of $f$, and of course, it would then be a submanifold of itself. Still, as you explained, the image is not a submanifold of $\mathbb{R}^2$. – quasi May 25 '17 at 21:52

2 Answers2

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The smooth map $$ F: \mathbb{R} \rightarrow \mathbb{R}^2, \quad t \mapsto F(t)= (2 \cos (t-\frac{\pi}{2}), \sin 2(t-\frac{\pi}{2}))$$ is an non-injective immersion such that $F(\mathbb{R})$ is an "eight" figure (or better, "infinity").

If you consider $F_{|_{]0, 2\pi[}}= F(\mathbb{R})$ it holds that this map is injective (removing the intersection).

Consider the smooth function $$g: \mathbb{R} \rightarrow \mathbb{R}^2, \quad t \mapsto g(t)= \pi + 2 \arctan t.$$ It holds that $G=F \circ g$ is an immersion and $$ G(\mathbb{R})= F(]0, 2\pi[) $$ (the infinity figure).

However, if you consider the topology on $ G(\mathbb{R})$ induced by $\mathbb{R}^2$, it holds that $ G(\mathbb{R})$ is compact while trivially $\mathbb{R}$ is not.

As a consequence, $G$ is an injective immersion but it is not an embedding.

(I assumed that submanifold means an embedded manifold for you. Probably we are using diffent notation. Let me know if you want further information and I hope I got your question)

Ngicco
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  • My definition of a submanifold comes straight from the book I mentioned: If $X$ and $Z$ are both manifolds in $\mathbb{R}^n$ and $Z\subset X$, then $Z$ is a manifold of $X$. I'm not quite sure what you mean by an embedded manifold, is that just the image of an embedding $f: Z \to X$? – Anmol Bhullar May 26 '17 at 15:42
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    Yes, it is. And it is related to induced topology. Unfortunately I don't know the book you cited. Maybe my answer is not the one you needed, but it seems to me that the "eight" figure is a manifold by that definition. Surely, it is not an embedded manifold. – Ngicco May 28 '17 at 08:37
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As for the answer by Ngicco, I think $G$ is not an embedding doesn't necessarily mean the image of $G$, the figure eight, cannot be an embedded submanifold. See Problem 5.4 in "Introduction to smooth manifolds" by Lee.

My intuition is that at the cross point, there are "two tangent directions", so it cannot be a 1-manifold. I cannot formalize this intuition into a proof.

My tentative proof would be like this (though without much intuition). Let the image of the figure eight curve be $\gamma$ and let the cross point be $p$. If $\gamma$ is an embedded submanifold of $\mathbb{R}^2$ of dimension 1, then there is a chart $(U,\phi)$ of $\mathbb{R}^2$ containing $p$, such that $\phi(U\cap \gamma)=\phi(U)\cap \mathbb{R}$, that is, $U\cap \gamma$ is a 1-slice of $U$. Choose a ball $B_\epsilon(\phi(p))$ small enough that is contained in $\phi(U)$, for the purpose that $B_\epsilon(\phi(p)) \cap \mathbb{R}$ is an interval which we call $I$. Now restrict $\phi$ to be a homeomorphism between the interval and its counterpart $\phi^{-1}(I)$ in $\gamma$. Take away the point $\phi(p)$ in the interval and the corresponding point $p$ in $\phi^{-1}(I)$, the remaining part of the former has two connected components while the remaining part of the latter has at least four connected components (at the beginning we should choose $U$ small enough such that $U\cap \gamma -\{p\}$ has four connected components). Therefore $\phi$ cannot be a homeomorphism and we prove by contradiction that $\gamma$ cannot be an embedded submanifold of dimension 1.

liyiontheway
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