Surjective vector field on $\mathbb{R}^n$ Let $V: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous with the property
$$\frac{\langle V(x), \, x\rangle}{|x|} \, \to \infty \quad \text{as} \quad |x| \to \infty  \qquad \qquad (1)$$
where $\langle \cdot \, , \cdot \rangle$ denote the standard inner product and $| \cdot | = \sqrt{\langle \cdot \, , \cdot \rangle}$ is the Euclidian norm on $\mathbb{R}^n$.
I have to show that $V$ is surjective.
My attempt:
 Take $z \in \mathbb{R}^n$ and define $\varphi: \mathbb{R}^n \rightarrow \mathbb{R}^n, \, \varphi(x): = V(x) - z$. The aim is to show that $\varphi$ has a zero. So assume by contradiction that $\varphi(x)\neq 0 \, \, \, \forall \, x \in \mathbb{R}^n$. Let $R > 0$. We define an auxiliary function $\psi : \mathbb{R}^n \rightarrow \mathbb{R}^n, \, \psi(x): = R \cdot \frac{\varphi(x)}{|\varphi(x)|}$. Then, $\psi$ is a continuous self-mapping $\psi: \overline{B}_R(0) \rightarrow \overline{B}_R(0)$ with $\text{im}(\psi) \subset \partial B_R(0)$. Using Schauder's fixed point theorem, there exists $x_0 \in \overline{B}_R(0)$ s.t. $\psi(x_0) = x_0$. In particular, $x_0 \in \partial B_R(0)$.
Now, I tried to get a contradiction with the assumption $(1)$ taking $R \to \infty$ without success. (Maybe it's not the right thing to do)
Any suggestions? Thanks in advance!
 A: Here's an idea: $\varphi(x)$ cannot be tangent to spheres of big radius. This means that $\frac{\varphi(x)}{|\varphi(x)|}$ has non-zero degree on some sphere of big radius $R$. But at the same time this vector field is defined on the whole $R$-ball, so, it's degree has to be zero then. That's the contradiction.
A: Assume $\varphi(x)$ has no zeros.
It follows from the given property that there is $R$ such that $\forall x \in S_R(0)$ $\varphi(x)$ in not tangent to $S_R(0).$ Let's assume that $\varphi(x)$ is outward-pointing (consider $-\varphi(x)$ otherwise). Then it's outward-pointing on the whole sphere by continuity.
Having such a vector field, it's possible to construct a retraction $r$ of $B_R(0)$ onto $S_R(0):$
for $x \in B_R(0)$ define $r(x)$ to be the intersection of the ray $(x; \varphi(x))$ with the sphere. Note that $r(x) = x$ on the sphere and $r$ is continuous.
Using Schauder's fixed point theorem we can prove that there's no such retraction: consider $$F: B_R(0) \to B_R(0)$$
$$F(x) = - r(x). $$
This map cannot have a fixed point. Contradiction.
