I'm trying to prove that an immersion $f:U\subset\mathbb R^n\rightarrow \mathbb R^n\times\mathbb R^m$ is a local homeomorphism. I was able to show that if $f$ is injective, then $f$ is homeomorphism, so if I can prove that there is an open set $V_a$, for each $a\in U$, where $f$ is injective, and so $f|_{V_a}$ is a homeomorphism.
I've tryied to suppose that for every open set $a\in V$, there is a $y\in V$ such $f(y)=f(a)$ and then get an absurd, but I was not able.
I don't even know if its true that $f'(a)$ injective implies locally injective, but it feels right.
Any hint, or counter example?
Thanks in advance.
Suppose $f$ is $C^1$and $f'(a)$ is injective. Then $f'(a)(\mathbb R^n)$ is an $n$-dimensional linear subspace of $\mathbb R^n \times \mathbb R^m .$ We can choose a linear $T: \mathbb R^n \times \mathbb R^m \to \mathbb R^{n}$ such that $T$ is injective on $(f'(a)(\mathbb R^n)).$ Then $T\circ f$ is a $C^1$ mapping of $\mathbb R^n$ into $\mathbb R^{n}.$ Note that $(T \circ f)'(a) = T\circ f'(a),$ which is nonsingular. By the inverse function theorem, $T\circ f$ is injective in a neighborhood $V_a.$ This implies $f$ is injective in $V_a$ as desired.
Previous answer: A differentiable (but not $C^1$) example: On the real line, let $f(x) = x + 2x^2\sin (1/x), a = 0.$ Then $f'(a)$ is injective, but $f$ is not injective in any neighborhood of $a.$