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Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be a continuous map such that $\forall \ x : |x-f(x)|<5777$.

Show that $f$ is surjective.

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    This is not true: let f map each (x,y) to the point (a,b) with a the integer part of x and b that of y. – 2017-01-21
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    sorry sorry, oops a continuous one – 2017-01-21
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    By the way: **please** write complete sentences. You are not being charged by the word here... – 2017-01-21
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    @peter, and obviously the point of my comment is that assumptions should be made explicit. – 2017-01-21
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    Yes, of course. – 2017-01-21

2 Answers 2

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Consider a homeomorphism $h: \Bbb R^2\to D^2$ where $D^2$ is the open unit disc, $h(v):=v/(|v|+1)$. Let $\tilde{f}: D^2\to D^2$ be $hfh^{-1}$. The constraints $|f(x)-x|<5777$ implies that whenever $v\in D^2$ is close enough to $w\in \partial D^2$, then $\tilde{f}(v)$ is close enough to $w$. In other words, $\tilde{f}$ extends to a continuous map $F: \overline{D^2}\to \overline{D^2}$ such that $F(v)=v$ for $v\in\partial D^2$.

Any continuous map from a closed unit disc into itself that restricts to the identity on the boundary, is surjective by standard arguments (otherwise, we could construct a retraction of the disc into its boundary sphere).

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Let $a$ be an arbitrary point in $\mathbb R^2$; we wish to show that it is hit by $f$.

Consider the image under $f$ of a closed curve $\gamma$ that goes once around a circle centered on $a$ with radius $5778$. By the dog-walking lemma, $f(\gamma)$ winds around $a$ exactly once.

Now let the radius of the circle contract towards $0$; the image of the circle will deform homotopically into a point. However if $a$ is not in the range of $f$, a homotopic deformation of the curve cannot change its winding number around $a$, and the winding number of a stationary point is $0$ -- a contradiction!