Suppose I have a manifold $M$ (without boundary) and a subset $S$ such that it is locally an image of an immersion. Namely, for any $s\in S$ there exists an open set $U_s \subset M$ with $s \in U_s$, a manifold $N_s$ and an immersion $i_s: N_s \to M$ such that $S\cap U_s $ is the image of $i_s$. Does there exist a manifold $N$ and an immersion $i$ such that $S$ is the image of $i$?
What if I assume $S$ is compact? Does there exist a closed manifold $N$ with this property in this case?
Sure. Just let $N$ be the disjoint union of the manifolds $N_s$, and let $i$ be the map whose restriction to each $N_s$ is $i_s$. (If you require manifolds to be second-countable, you should first take a countable set of $U_s$'s that cover $S$.)
On the other hand, you cannot require $N$ to be closed if $S$ is compact. For instance, let $\Theta\subset\mathbb{R}^2$ be the union of the unit circle and $(-1,1)\times\{0\}$. Then $\Theta$ is compact and is the image of an immersion: namely, it is the image of an immersion $S^1\sqcup(-1,1)\to\mathbb{R}^2$ which is the identity on $S^1$ and sends and $(-1,1)$ to $(-1,1)\times\{0\}$. But $\Theta$ is not the image of any immersion from a closed manifold. (Proving this rigorously is a bit messy, but the idea is that such an immersion would need to include a parametrization of the segment $(-1,1)\times\{0\}$, but then this parametrization cannot be extended smoothly past $(-1,0)$ or $(1,0)$.)
