A bounded Euclidean homeomorphism has a compactly supported local model

Choose $R>1$ so large that $K\cup f(K)$ lies in the open radius-$R$ ball; this is possible because the continuous image of a compact set is compact, and compact Euclidean sets are bounded. Set $L=R+1$. Define an increasing homeomorphism $\rho :[0,\mathrm{\infty })\to [0,L)$ by $\rho (r)=r$ for $r\le R$ and $\rho (r)=R+(r-R)\mathrm{/}(1+r-R)$ for $r\ge R$. The radial map $y(0)=0$, $y(x)=\rho (\parallel x\parallel )x\mathrm{/}\parallel x\parallel$ for $x\mathrm{\ne }0$, is a homeomorphism from ${\mathbb{R}}^n$ onto the open radius-$L$ ball, and it is the identity on the closed radius-$R$ ball. On the open radius-$L$ ball put $g=y\circ f\circ y^{-1}$. This is a homeomorphism there.

Extend $g$ by the identity on the radius-$L$ sphere. To verify continuity there, put $r=\parallel x\parallel$ and $s=\parallel f(x)\parallel$. The displacement bound gives $s\ge r-C$. For $r>C$, insert $f(x)\mathrm{/}r$ between the two unit vectors to obtain $\parallel \frac{f(x)}{s}-\frac{x}{r}\parallel \le \frac{\mathrm{\mid }s-r\mathrm{\mid }}{r}+\frac{C}{r}\le \frac{2C}{r}\mathrm{.}$ Consequently $\parallel y(f(x))-y(x)\parallel \le \mathrm{\mid }\rho (s)-\rho (r)\mathrm{\mid }+\rho (r)\frac{2C}{r}\le (L-\rho (s))+(L-\rho (r))+\frac{2LC}{r}\mathrm{.}$ The right side tends to zero as $r\to \mathrm{\infty }$, uniformly over directions, since $s\ge r-C$ and $\rho$ is increasing with limit $L$. For an interior point $z=y(x)$ approaching a boundary point $b$, the equality $\parallel z\parallel =\rho (r)$ forces $r\to \mathrm{\infty }$, and $\parallel g(z)-b\parallel \le \parallel g(z)-z\parallel +\parallel z-b\parallel$ tends to zero by this bound. This proves continuity directly. The inverse $f^{-1}$ has the same displacement bound: if $z=f(x)$ then $\parallel f^{-1}(z)-z\parallel =\parallel x-f(x)\parallel \le C$. Applying the same calculation to $f^{-1}$ gives the continuous inverse extension. Extend once more by the identity outside the closed radius-$L$ ball to obtain a homeomorphism $G$ of ${\mathbb{R}}^n$. The formulas and their inverse formulas agree on the joining sphere.

The set $U=\{x:\parallel x\parallel <R,\parallel f(x)\parallel <R\}$ is an open neighborhood of $K$. On $U$ both $y^{-1}(x)=x$ and $y(f(x))=f(x)$, so $G(x)=f(x)$. Thus any piecewise linearity of $f$ on an open neighborhood of $K$ is retained after intersecting that neighborhood with $U$. This is an equality of actual local maps, rather than an approximation statement about their images.

For $0<t\le 1$ define $G_t(x)=tG(x\mathrm{/}t)$ when $\parallel x\parallel \le tL$, and $G_t(x)=x$ when $\parallel x\parallel \ge tL$; put $G_0(x)=x$. The branches agree at $\parallel x\parallel =tL$ because $G$ fixes the radius-$L$ sphere. Each $G_t$ is a homeomorphism, with inverse given by the same formula using $G^{-1}$; it fixes every point outside the radius-$tL$ ball. The formulas give joint continuity for $t>0$. At $t=0$, both $x$ and its image lie in the radius-$tL$ ball whenever a point moves, so $\parallel G_t(x)-x\parallel \le 2tL$. This proves joint continuity at zero, uniformly in $x$. The identical estimate applies to the inverse family, hence $(x,t)\mapsto (G_t(x),t)$ is a homeomorphism with continuous inverse $(x,t)\mapsto (G_t^{-1}(x),t)$. Therefore $G_t$ is an ambient isotopy, supported in the radius-$L$ ball, from the identity to $G$. Together with step 2.1 this proves all conclusions.