Immersion of $M^n$ into $\mathbb{R}^n$, is $M^n$ orientable? Compact?

Question

Say we have an immersion of $M^n$ into $\mathbb{R}^n$ (same dimension). I have two questions.

1. Is $M^n$ orientable?
2. Is $M^n$ compact?

Thanks in advance!

Answer 1

1) Immersions are local embeddings, so you can use a sufficiently fine open covering of $M^n$ to construct an atlas from your immersion. All the resulting transition maps are positively oriented, so this atlas is automatically an oriented atlas.

2) No, $M^n$ is not compact unless it is the empty manifold.

Theorem: Let $M^d$, $N^d$ be closed, smooth, connected, oriented $d$-dim. manifolds. Let $f: M^d \to N^d$ be smooth. Let $y \in N^d$ be a regular value of $f$. Then $$\text{deg}(f) = \sum_{x\in f^{-1}(y)} \text{sign det}(T_xf).$$

Assume $M^n$ is compact, nonempty and wlog. connected. In that case, you could extend your immersion into an immersion $f: M^n \to S^n$ using any embedding $\mathbb{R}^n \to S^n$.

Now apply the theorem to $f$ using any orientation of $S^n$ and the orientation of $M^n$ induced by the immersion. We see that all values $y \in S^n$ are regular values, and $\text{sign det}(T_xf)$ is always 1 by choice of orientation. This entails that all $y \in S^n$ have the same number of preimages, namely $|f^{-1}(y)| = \text{deg}(f)$. This is not possible, however, since $f$ is not surjective.