Prove that it is an homeomorphism I'm studying for an exam and I wanna prove that a function is a homemorphism, I've made some translation of that function. Now my problem is to prove that $f: \mathbb{R}^n \to \mathbb{R}^n$, given by $f(x)= x - a \vert \vert x \vert \vert$ is an homeomorphism (where $a$ is s.t $\vert \vert a \vert \vert<1 $).
I'm really sure that $f$ is a bijection, so I have to ways to solve this problem:

*

*find $f^{-1}$ and hope that it is clearly continuous (I'm trying that)

*Using sequences and the continuity of $f$ to prove that $f^{-1}$ is continuous without find the inverse.

I guess the second way is harder than the first one, so can you help me how to find the inverse??
 A: Index $f$ by $a$ - that is, write $f = f_a(x) = x - a ||x||$.
Given arbitrary $x$, $y$, we see that if $f_a(x) - f_a(y) = (x - y) - a (||x|| - ||y||)$. Then $|f(x) - f(y)| \geq ||x - y|| - ||a (||x|| - ||y||)|| \geq ||x - y|| - ||a|| |(||x|| - ||y||)| \geq ||x - y|| - ||a|| (||x - y||) = ||x - y||(1 - ||a||)$. This holds for any $a$.
Consider some point $p$. Then consider the map $g : \mathbb{S}^{n - 1} \to \mathbb{R}^n$ given by $g(s) = \frac{||p|| + 1}{1 - ||a||} s$.
And consider the map $g' : \mathbb{S}^{n - 1} \to \mathbb{R}^n$ given by $g' = f_a \circ g$.
We see that $g$ and $g'$ are homotopy equivalent in $\mathbb{R}^n \setminus \{p\}$. This is because the map $h_t(s) = f_{ta} \circ g$ is continuous and $p$ is not in its range. This is because $||h_t(s)|| = ||f_{ta}(s)|| = ||f_{ta}(s) - f_{ta}(0)|| \geq ||s - 0||(1 - ||a||) = ||s|| (1 - ||a||) = \frac{||p|| + 1}{1 - ||a||} (1 - ||a||) = ||p|| + 1 > ||p||$.
Now we see that $p \in Im(f_a)$. For suppose that $p \notin Im(f_a)$. Then we could contract $g'$ to a single point in $\mathbb{R} \setminus \{p\}$ via the homotopy $h_t(s) = t g'(s)$, since $t g'(s) = t f_a(g(s)) = f_a(t g(s))$. But since $g'$ and $g$ are homotopy equivalent, we could thus contract $g$ to a single point in $\mathbb{R} \setminus \{p\}$. But since $p$ is within the boundary of the sphere $g$, this is not possible (one can prove this using homology / higher homotopy groups).
Therefore, $f$ is a bijection.
Now consider its inverse $g$. Note that for any $x, y$, $||x - y|| = ||f(g(x)) - f(g(y))|| \geq (1 - ||a||) ||g(x) - g(y)||$.
Therefore, for any $\epsilon > 0$, if $||x - y|| < \epsilon(1 - ||a||)$ then $||g(x) - g(y)|| \leq \frac{||x - y||}{1 - ||a||} < \epsilon$. So $g$ is continuous.
