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Let $T^2$ be the torus and let $\mathcal{C}^{\infty}(T^2, \mathbb{R}^3)$ be the space of smooth functions from $T^2$ to $\mathbb{R}^3$ endowed with the norm $\|f\| = \sup_x |f(x)| + \sup_x \|df_x\|$.

Is a generic function $f \in \mathcal{C}^{\infty}(T^2, \mathbb{R}^3)$ an immersion?

That is, is the set $$ \{f \in \mathcal{C}^{\infty}(T^2, \mathbb{R}^3) \,|\, \forall x,\, \text{rank}\,df_x = 2 \} $$ open and dense in $\mathcal{C}^\infty(T^2, \mathbb{R}^3)$?


Openess is clear. What I'm not sure about is if any smooth function can be well approximated by an immersion.

This seems to be true for embeddings in $\mathcal{C}^\infty(M, R^N)$ where $N > 2 \dim M$, as a corollary of Whitney's embedding theorem proof.

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  • $\begingroup$ I think "generic" would be better understood than "general" for this meaning. $\endgroup$ Aug 2, 2017 at 13:46
  • $\begingroup$ @AnthonyCarapetis Thank you, I corrected it. $\endgroup$
    – Olivier
    Aug 2, 2017 at 13:59
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    $\begingroup$ The word "generic" doesn't mean the same as "holds on an open and dense subset". A property of elements of a topological space is "generic" if it holds on a countable intersection of dense open sets. For example, irrational numbers are generic among numbers. $\endgroup$
    – Ben McKay
    Aug 2, 2017 at 16:45
  • $\begingroup$ @BenMcKay Thanks for the clarification. That definition makes more sense. $\endgroup$
    – Olivier
    Aug 8, 2017 at 12:35

1 Answer 1

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Consider the following function $$ f: S^1 \times [-1,1] \to \mathbb R^3\,,\quad f(\theta,z) = (z \cos \theta, z \sin \theta, z)\,,$$ that maps a cylinder into $\mathbb R^3$. If needed, we can extend it to a map from the torus into $\mathbb R^3$.

I would claim that it is not possible to approximate $f$ by an immersion in the $C^1$-norm, because $f$ is orientation-reversing for $z < 0$ and orientation-preserving for $z > 0$, while an immersion could be only one of those.

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  • $\begingroup$ I'm not sure I completely understand the example. How do you orient $\text{Im}(f)$ in the first place? $\endgroup$
    – Olivier
    Aug 2, 2017 at 15:35
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    $\begingroup$ @Olivier: You don't orient all of the image; only orient the part where $z \ne 0$, by projecting to the horizontal plane. For any choice of a point of the surface on which $z > 0$ and another on which $z<0$, any sufficiently small perturbation of $f$ will compose with projection to the horizontal plane to give the same orientations. $\endgroup$
    – Ben McKay
    Aug 2, 2017 at 16:49

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