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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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Asaf Shachar
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    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. – Moishe Kohan Nov 11 '16 at 20:43
  • @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). – Asaf Shachar Nov 11 '16 at 20:48
  • $f \colon t \mapsto (t^3-t, t^2)$. With a connected domain, you can only get three of the four branches of the self-intersection with an injective immersion, I think. – Daniel Fischer Nov 11 '16 at 20:58
  • 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 $N$. – Daniel Fischer Nov 11 '16 at 21:13
  • @DanielFischer You are right in the case where there are no boundaries. An immersion (even injective immersion) from a manifold with boundary does not have to be an open map (consider $x \to x+1$, when $M=N=[0,\infty)$. I should really be more careful, and stress this is the non-trivial part. – Asaf Shachar Nov 11 '16 at 21:15
  • Yes, the bad point is $f(1) = f(-1) = (0,1)$. If we look at an injective immersion $g$ such that $g(t) \in f((-\infty,-1))$ for some $t$, then $g$ cannot do anything other than trace the path of $f$ (with different speed, perhaps, and maybe in the other direction, then flip). When it reaches the intersection point from the loop for the second time, it must stop, and can't ever get into $f((1,+\infty))$. – Daniel Fischer Nov 11 '16 at 21:18
  • Ah, I forgot about the case with boundaries. In that case, something like $(t + is) \mapsto f(t) + i\cdot s \cdot f'(t)$ (identifying $\mathbb{R}^2$ with $\mathbb{C}$ for more convenient notation) on ${ z : 0 \leqslant \operatorname{Re} z \leqslant \varepsilon}$ is probably not obtainable from an injective immersion. – Daniel Fischer Nov 11 '16 at 21:23
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    The first example of Cohen can be modoified so that it start from $\mathbb R$ (by making the image worst). –  Nov 11 '16 at 21:25

2 Answers2

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This is a counterexample for manifolds without boundary: consider the map $\varphi : M = \mathbb{D}_1^2 \sqcup \mathbb{D}_2^2 \to \mathbb{R}^3$ given by $(u,v) \mapsto (u, v, 0)$ in the open unit disk $\mathbb{D}_1^2$ and $(u, v) \mapsto (u + 1, 0, v)$ in another open unit disk $\mathbb{D}_2^2$. Clearly $\varphi$ is an immersion and I claim that $\varphi(M)$ is not the image of an injective immersion $\psi : L \to \mathbb{R}^3$. If so, for every point $p \in \varphi(M)$ there’s a unique $q \in L$ with $\psi(q) = p$, so around $q$, $\psi$ defines a small embedded disk contained in $\varphi(M)$, which therefore can be either parallel to the $z=0$ plane or to the $y=0$ plane. This dichotomy allows us to partition the segment $I = \{(x, 0, 0) \mid -\frac{1}{2} \leq x \leq \frac{3}{2}\}$ into two sets $V_1$ and $V_2$. Of course these two sets are nonempty, so if we prove they are open, this will contradict the fact that $I$ is connected. If for $p = \psi(q) \in V_i$ we have that $p \in \psi(U) \subseteq \varphi(M)$ is a small embedded disk around $p$, then for any $p' \in \psi(U) \cap I$, $\psi(U)$ also serves as an embedded disk around $p'$, so $\psi(U) \cap I \subseteq V_i$, and we are done because $\psi(U)$ is open in $\varphi(\mathbb{D}_i^2)$ (thanks to the inverse function theorem) and so $\psi(U) \cap I$ is open in $I$.

If you insist that $M$ is connected, you can just add a strip joining the two disks away from the part where they intersect.

Gerard Orriols
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About the immersion $f \sqcup g : \mathbb{R} \sqcup \mathbb{R} \to \mathbb{T}^2$ as two dense wrapped lines, it turns out that it can be made injective! This is done by progressively restricting the domain of $f$ and $g$ to avoid self-intersections.

By changing the speed of the curves, we may suppose that $f(\mathbb{Z}) \cap g(\mathbb{R}) = f(\mathbb{R}) \cap g(\mathbb{Z}) = \varnothing$ and define $A_n = [-n,-n+2] \cup [n-2,n]$ for every $n \in \mathbb{Z}^+$. Start by removing $A_2 \cap g^{-1}(f([-1,1]))$ from the domain of $g$. Next, erase from the domain of $f$ the set $A_3 \cap f^{-1}(g([-2,2]))$. Then erase from the domain of $g$ the set $A_4 \cap g^{-1}(f([-3,3]))$, and so on. The final domains of $f$ and $g$ will be open, because the removed set is a locally finite union of closed sets, hence closed, so $f \sqcup g$ will still be a smooth immersion of a manifold. To see injectivity, suppose that $f(x) = g(y)$, with $x, y \in \mathbb{R}\setminus\mathbb{Z}$ in the final domains of $f$ and $g$, respectively. Let $n$ be odd such that $x \in A_n$ and $m$ even such that $y \in A_m$. If $n < m$ then $A_m \cap g^{-1}(f([-m+1,m-1]))$ was erased from the domain of $g$, and this contains $y$ because $x \in A_n \subseteq [-n,n] \subseteq [-m+1,m-1]$. The other case is handled equally.

Gerard Orriols
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