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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.

stb2084
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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.

  • I'm not familiar with homotopy groups, but is it easy to verify that $f|{S_D}$ represents a nontrivial element of $\pi{d-1}\left(B_{\frac{3D}{2}}\setminus{z}\right)$ when there exist a point $z\in B_{\frac{D}{2}}\setminus f(B_D)$ ? – stb2084 Nov 20 '15 at 13:10
  • Um.. I do not think that's easy. But there are some observations: $B_{\frac 32 D} \setminus{z}$ deformation retract onto $S_D$, and $f : S_D\to S_D$ can be thought as the identity map. In $d = 2$, it is clear that the identity map $f: \mathbb S^1 \to \mathbb S^1$ represents a nontrivial element in $\pi_1(\mathbb S^1)$. @stb2084 –  Nov 20 '15 at 13:18
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    I have overlooked your explanation that "the line joining $x$ to $f(x)$ are completely in $I_D$". I was able to understand your answer now! Thank you! – stb2084 Nov 20 '15 at 13:53
  • For what $x$ "the line joining $x$ to $f(x)$ are completely in $I_D$"? For "line", do you mean straight line? –  Nov 30 '15 at 21:48
  • Yes, I do @Jack. –  Nov 30 '15 at 22:17
  • For what $x$ "the line joining $x$ to $f(x)$ are completely in $I_D$"? –  Dec 01 '15 at 01:12
  • @Jack For all $x\in S_D$. –  Dec 01 '15 at 02:21
  • @JohnMa did you mean $f:S_D\to I_D$ can be thought of as an identity map in your comment above? – user2154420 Dec 01 '15 at 04:52
  • Sort of, I am saying that it is homotopic to the identity map. @user2154420 –  Dec 01 '15 at 05:49
  • I don't see why for all $x\in S_D$, the line joining $x$ to $f(x)$ are completely in $I_D$. Would you elaborate? –  Dec 01 '15 at 15:41