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Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^4$ be defined by $F(u,v) = (u+v, uv, u-v, v^3)$ and let $M = F(\mathbb{R}^2)$. Prove that $M$ is a smooth manifold.

The proof that my TA posted online was this:

If the differential $DF|_{(u,v)}: T_{(u,v)} \mathbb{R}^2 \rightarrow T_{F(u,v)} \mathbb{R}^4$ is non degenerate then $F$ defines a smooth immersion of $\mathbb{R}^2$ into $\mathbb{R}^4$ and hence $F$ defines local diffeomorphisms for the image $M$. To verify that $F$ is a smooth immersion, note that $\mathrm{rank}(F) = 2 = \dim(\mathbb{R}^2)$. Because $F$ is globally injective, then it is actually a global diffeomorphism onto its image $M$. Hence $M$ is a smooth $2$-manifold.

I am struggling to understand this proof. I get why $F$ is a smooth immersion, but I don't get why it defines local diffeomorphisms. It seems as if he is applying the inverse function theorem to $F: \mathbb{R}^2 \rightarrow \mathbb{R}^4$, but he can't do this because the domain and co-domain do not have the same dimension. You could apply the inverse function theorem to $F: \mathbb{R}^2 \rightarrow M$, but this would require you to already know that $M$ is a smooth $2$-manifold, which is what we are trying to prove. What am I missing here? I would ask my TA but classes are done for the semester, so I won't be seeeing him again.

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  • $\begingroup$ think about implicit function theorem instead of inverse function theorem $\endgroup$ – user26977 Apr 15 '16 at 8:17
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The proof your TA gave is incomplete: contrary to a widespread misconception a smooth injective immersion $F$ need not have as image a smooth manifold.
For two counterexamples look at Lee's Intoduction to Smooth Manifolds, Second Edition, Examples 4.19 (the notorious plane figure eight aka lemnicate) and Example 4.20, both on page 86.
The image of $F$ will however be a submanifold if $F$ is a proper injective immersion : same book, Proposition 4.22, page 87 (and preceding pages for excellent background).
Fortunately, in your case $F$ is indeed proper, as already demonstrated by its first and third components $u+v, u-v$ .

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    $\begingroup$ The question doesn't actually require $M$ to be a submanifold. So the immersion argument should really be ok. $\endgroup$ – Maik Pickl Apr 15 '16 at 8:31
  • $\begingroup$ @Maik No, I don't think that the immersion argument is ok: that $M$ (a naked set!) "is" a manifold does not make sense if you don't explain what its topology and differentiable structure are. The natural answer is to consider that the topology and differential structure are inherited from the ambient $\mathbb R^4$, i.e. that $M$ is a submanifold of $\mathbb R^4$. Lee's Proposition 4.22 gives sufficient conditions for that to be true. $\endgroup$ – Georges Elencwajg Apr 15 '16 at 9:00
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    $\begingroup$ I think this is merely a matter of opinion. An immersed manifold will of course carry the structure of a manifold. The topology induced by the immersion will just not agree with the subset-topology in general. If you insist on it being a submanifold I would agree with your point. But the question actually didn't ask for that. $\endgroup$ – Maik Pickl Apr 15 '16 at 9:06
  • $\begingroup$ Dear @ Maik, I think that by default a subset of numerical space is called a submanifold if the inclusion is an embedding, while an open immersion is a morphism rather than a subset, and the OP asked about the set $M=F(\mathbb R^2)$. However I agree that these are sterile terminology questions and I strongly encourage you to answer the question in your interpretation: competition makes for a healthy intellectual debate :-) $\endgroup$ – Georges Elencwajg Apr 15 '16 at 9:26
  • $\begingroup$ Allright, if the question is still open in roughly 5 hours I will answer. :) Unfortunately I have no time now. $\endgroup$ – Maik Pickl Apr 15 '16 at 9:32

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