Every image of an immersion is an image of an injective immersion?

Question

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 $N$).

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*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.

Answer 1

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.

Answer 2

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.