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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.

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    $\begingroup$ The image is not a submanifold of $\mathbb{R}^2$. $\endgroup$
    – quasi
    Commented May 25, 2017 at 21:11
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    $\begingroup$ Because as a subset of $\mathbb{R}^2$, the intersection point is present (in the image). $\endgroup$
    – quasi
    Commented May 25, 2017 at 21:19
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    $\begingroup$ 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. $\endgroup$
    – Brian Tung
    Commented May 25, 2017 at 21:24
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    $\begingroup$ Yes. To $\mathbb{R}^2$, the image is just a figure $8$, intersection point included. $\endgroup$
    – quasi
    Commented May 25, 2017 at 21:36
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    $\begingroup$ I love the infectivity typo! $\endgroup$
    – lesath82
    Commented May 25, 2017 at 21:47

2 Answers 2

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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)

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  • $\begingroup$ 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 $f: Z \to X$? $\endgroup$ Commented May 26, 2017 at 15:42
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    $\begingroup$ 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. $\endgroup$
    – Ngicco
    Commented May 28, 2017 at 8:37
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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.

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