This was something I learnt from Prof. Umut Varoglunes during the graduate differential topology class (Math 215B) at Stanford.

Suppose you had a differentiable map $(a,b) \to \mathbb R$. If it is locally injective, must it be (globally) injective?

The answer, is of course, yes. The map $f$ being locally injective at $x$ means that $f'(x) \neq 0$. If the derivative vanishes nowhere, that means it never changes the sign (by the Intermediate Value Theorem), and so any integral of $f'$ is strictly positive or negative. (This can also be argued in the contrapositive: suppose $f(x_1)=f(x_2)$, then $f$ attains some maxima on $[x_1,x_2]$)).

Suppose instead if you have a differentiable map from a connected open subset $U\subset \mathbb R^n$ to $\mathbb R^n$. Does it still hold that local injectivity implies (global) injectivity?

It is much less clear in this case. In fact, our argument for the $n=1$ case is hinting towards there being “algebraic topology” reasons (since the Intermediate Value Theorem is the 1-D case of theorems from AlgTop).

One possible piece of intuition (from Prof. Varolgunes) is as follows: suppose therwise that our map was not injective, then we may select $x_1,x_2\in U$ satisfying $f(x_1) = f(x_2)$. Then consider two paths $\gamma_1,\gamma_2$ from $x_1\to x_2$. The composition $\gamma_1 \cdot \gamma_2^{-1}$ forms a loop in $U$, and a “double loop” in the image. If the two components of the “double loop” do not exactly overlap, then we can show that local injectivity is violated.

But what if they overlap perfectly? It’s hard to see where this leads.

The punchline (from algebraic topology) is as follows: $f$ being locally an isomorphism means that $f$ is a covering space map, and there are no multi-covers of $\mathbb R^n$ since it is simply connected.