If $f : M \to \mathbb{R}^N$ is an immersion then $\exists \delta > 0$ such that $f$ is injective on all balls of radius $\delta$?

Question

Theorem:

Let $M$ is a smooth compact manifold with complete metric $\rho$. Let $f : M \to \mathbb{R}^N$ be an immersion. Then there exists a $\delta > 0$ such that for any $m_1,m_2 \in M$ that satisfy $\rho(m_1,m_2) < \delta$, it follows $f(m_1) = f(m_2) \iff m_1 = m_2$.

Proof Idea:

For each $m \in M$ there exists an open neighbourhood $U_m \subset M$ such that $m \in U$ and $f\rvert_{U_m}$ is an embedding. In particular $f\rvert_{U_m}$ is injective. For each point $m \in M$ we define a ball $B_m \subset U_m$ centred at $m \in M$. The collection of balls $\{ B_m \ | \ m \in M \}$ forms an open cover of $M$.

Can I somehow ensure that there the infimum over the radius of the balls is strictly positive? Then I can define the infimum as $\delta > 0$ and this will (I hope) complete the proof?

Is there a different strategy for proving this result?

If this theorem is false, are there any extra conditions I can impose on the map $f$ or the manifold $M$ so that the Theorem becomes true?

Answer 1

No this is exactly the right idea. You can use that you manifold is already covered by finitely many of the balls $B_m$, which is imposed by the compactness.

And, starting from there, you should replace your $\delta$ by something smaller to ensure that each two points may be regarded in one of the balls $B_m$.

$\textbf{Edit:}$ My first answer was kind of a quick shot only answering your first part of the question. Sorry for that. The answer to you question in the comments below is no, its not that easy.

$\textbf{Another edit:}$ In your second proof I can't see why the radius is continuous, I mean vividly I would expect it, but proving that may be difficult. However, I remember some theorem which states something like that the compactness implies that you can find a $\delta>0$ such that any ball of radius $\delta$ is already contained in at least one of the covering sets. With this you can finish your proof by your last paragraph. However, I can't remember the name and I can't find it at the moment.

Answer 2

Is this a convincing proof?

For each $m \in M$ there exists an open neighbourhood $U_m \subset M$ such that $m \in U$ and $f\rvert_{U_m}$ is an embedding. In particular $f\rvert_{U_m}$ is injective. The collection of sets $\{ U_m \ | \ m \in M \}$ forms an open cover of $M$ from which we can extract a finite subcover (using the compactness of $M$) $\{ U_i \ | \ i \in I \}$ for some finite set $I$.
For each $m \in M$ there exists a $J \subset I$ such that $$ m \in \bigcap_{j \in J} U_j. $$ Let $B_m$ be the largest ball centred at $m \in M$ such that there exists a $j \in J$ such that $B_m \subset U_j$. The radius of $B_m$ is a continuous function of $m$. A continuous function on a compact manifold attains its minimum $\delta > 0$.
If we let $m_1, m_2 \in M$ satisfy $\rho(m_1,m_2) < \delta$ then $m_2 \in B_{m_1}$ so both $m_1,m_2 \in B_{m_1}$. Now there exists a $j \in I$ such that $m_1,m_2 \in B_{m_1} \subset U_j$. Since $f\rvert_{U_j}$ is injective it follows $f(m_1) = f(m_2) \iff m_1 = m_2$.

EDIT: I think it's better to use Lebesgue's number lemma directly:

$f$ is an immersion hence for each $m \in M$ there exists an open neighbourhood $U_m \subset M$ such that $m \in U$ and $f\rvert_{U_m}$ is an embedding. The collection of sets $\{ U_m \ | \ m \in M \}$ forms an open cover of $M$. Then by Lebesgue's number lemma, there exists a $\delta > 0$ such that every set of diameter $\delta$ is contained in some set in $\{ U_m \ | \ m \in M \}$. Then for any pair $m_1,m_2 \in M$ such that $\rho(m_1,m_2) < \delta$ we can choose a ball $B$ of diameter $\delta$ to cover $m_1$ and $m_2$. Now $B$ is contained in some set in $\{ U_m \ | \ m \in M \}$ on which $f$ is injective so $m_1 = m_2 \iff f(m_1) = f(m_2)$.