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Let $M,N$ be smooth connected* manifolds, with or without boundary. Let $f:M \to N$ be a smooth immersion. Can we realize $f(M)$ as an image of some injective immersion into $N$?

That is, does there exist a manifold $\tilde M$ and an injective smooth immersion $j:\tilde M \to N$, such that $j(\tilde M)=f(M)$?

I am particularly interested in the case where $\dim M=\dim N$, and $\partial M \neq \emptyset$. (If $\partial M = \emptyset$, the image is open, hence an embedded submanifold of $N$).


*As commented below, if we assume $M$ is not connected, there probably are examples where the image cannot be realized as an image of an injective immersion.

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    $\begingroup$ Consider two dense geodesics (with different slopes) on the flat 2-torus. It is unlikely that the image of the corresponding map ${\mathbb R}\sqcup {\mathbb R}\to T^2$ is the image of an injective immersion. I am also unsure about the motivation for the question. $\endgroup$ – Moishe Kohan Nov 11 '16 at 20:43
  • $\begingroup$ @MoisheCohen Thanks. Your example shows I want to restrict to the case where I have an image of an immersion whose domain is a connected manifold. Regarding the motivation, I have a situation where I have an image of an immersion, and I would like to know whether or not it is an immersed submanifold in the target manifold. The point is that being an immersed subamnifold is equivalent to being the image of some injective immersion (according to the definition I am working with, there are different conventions about this). $\endgroup$ – Asaf Shachar Nov 11 '16 at 20:48
  • $\begingroup$ $f \colon t \mapsto (t^3-t, t^2)$. With a connected domain, you can only get three of the four branches of the self-intersection with an injective immersion, I think. $\endgroup$ – Daniel Fischer Nov 11 '16 at 20:58
  • $\begingroup$ If $\dim M = \dim N$, then $U := f(M)$ is a connected open subset of $N$, and thus a connected manifold. Which has an obvious embedding into $N$. $\endgroup$ – Daniel Fischer Nov 11 '16 at 21:13
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    $\begingroup$ The first example of Cohen can be modoified so that it start from $\mathbb R$ (by making the image worst). $\endgroup$ – user99914 Nov 11 '16 at 21:25

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