Surjectivity of a continuous map between $\mathbb{R}^d$s Let $f:\mathbb{R}^{d}\to\mathbb{R}^d$ be a continuous map. 
Show that if $\displaystyle\sup_{x\in\mathbb{R}^d}|f(x)-x|<\infty$, then $f$ is surjective. 
I encountered this problem more than 3 years ago in my class of exercises in calculus when I was an undergraduate student of my college. I have no idea since then. Can anyone solve this? Thank you. 

Let 
$$c > \sup_{x\in \mathbb R^d} \|f(x) - x\|$$
be fixed. For each $y\in \mathbb R^d$ (the image), let $D$ be large so that $D \ge 2c, 2\|y\|$. Restrict $f$ to $S_D = \{x\in \mathbb R^d: \|x\| = D\}$. Then if $x\in S_D$, 
$$\|f(x)\| = \|f(x) - x +x\| \le c+D \le \frac 32 D.$$ 
On the other hand, 
$$D=\|x\| = \|x - f(x) + f(x)\|\le c +\|f(x)\| \Rightarrow \|f(x)\| \ge D-c \ge \frac 12 D.$$
So 
$$f : S_D \to \{ x\in \mathbb R^d : \frac 12 D \le \|x\| \le \frac 32 D\} = I_D$$
Note that the line joining $x$ to $f(x)$ are completely in $I_D$, so $ f : S_D \to I_D$ is homotopic to the identity and so there is $x_0 \in \mathbb R^d$, $\|x_0\| < D$ so that $f(x_0) = y$. As $y$ is arbitrary, $f$ is surjective. 
Remark To explain more, note that if $\|x\| \le D$, then 
$$\|f(x)\| \le c+D \le \frac 32 D.$$
So we can restrict $f$ to $B_D = \{ \|x \| \le D\}$ to get
$$f : B_D \to B_{\frac 32 D}.$$
We have that $f|_{S_D} \subset I_D$, so $f(B_D)$ must contain $B_{\frac 12D}$: if not, then $f(B_D)$ misses a point $z$. But $f|_{S_D}$ represent an nontrivial element in $\pi_{d-1} (B_{\frac 32 D} \setminus \{z\}) \cong \mathbb Z$, which is not possible as $\pi_{d-1} (B_D)$ is trivial. 
