Alexander contraction of the boundary-fixed disk homeomorphism group

The Alexander deformation is a family of boundary-fixed homeomorphisms. For $s\in (0,1]$ and $h\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ define $H_s(h)(x):=\{\begin{array}{ll}sh(x\mathrm{/}s),& \parallel x{\parallel }_2\le s,\\ x,& \parallel x{\parallel }_2\ge s,\end{array}{H}_0(h):={\mathrm{id}}_{D^2}\mathrm{.}$ The two formulas agree when $\parallel x{\parallel }_2=s$, because then $x\mathrm{/}s\in \mathrm{\partial }{D}^2$ and $h$ fixes $\mathrm{\partial }{D}^2$ pointwise, so $sh(x\mathrm{/}s)=s(x\mathrm{/}s)=x$; hence $H_s(h)$ is well defined and continuous. It maps $D^2$ into $D^2$ (both branches land in the closed unit ball) and it fixes $\mathrm{\partial }{D}^2$ pointwise. Its inverse is $H_s(h^{-1})$: for $\parallel y{\parallel }_2\le s$ one has $\parallel s{h}^{-1}(y\mathrm{/}s){\parallel }_2\le s$ and $H_s(h)(s{h}^{-1}(y\mathrm{/}s))=sh(h^{-1}(y\mathrm{/}s))=y$, while for $\parallel y{\parallel }_2\ge s$ both maps fix $y$. Thus $H_s(h)$ is a continuous bijection of the compact disc with the inverse just displayed, hence a homeomorphism by [L1] that fixes the boundary pointwise, and $H_1(h)=h$ while $H_0(h)=\mathrm{id}$.

Continuity in the homeomorphism for fixed time. For $s\in [0,1]$ and $h,h^{\mathrm{\prime }}\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ every $x$ satisfies $\parallel H_s(h)(x)-H_s(h^{\mathrm{\prime }})(x){\parallel }_2\le d(h,h^{\mathrm{\prime }})$, where $d$ is the uniform distance: for $\parallel x{\parallel }_2\ge s$ both values equal $x$, and for $\parallel x{\parallel }_2\le s$ the difference is $s\parallel h(x\mathrm{/}s)-h^{\mathrm{\prime }}(x\mathrm{/}s){\parallel }_2\le d(h,h^{\mathrm{\prime }})$ (the case $s=0$ is the constant map). By [L2] the uniform distance metrizes the compact-open topology on the group, so $h\mapsto H_s(h)$ is continuous for each fixed $s$.

Joint continuity of the deformation. Fix $h_0$. The evaluation map $(s,x)\mapsto H_s(h_0)(x)$ is continuous on $[0,1]\times D^2$: for $s>0$ the two formulas are continuous and agree at $\parallel x{\parallel }_2=s$, while at $s=0$ the estimate $\parallel H_s(h_0)(x)-x{\parallel }_2\le 2s$ gives continuity. Since $[0,1]\times D^2$ is compact, this map is uniformly continuous, so $d(H_s(h_0),H_{s_0}(h_0))\to 0$ as $s\to s_0$. By step 2.1, $d(H_s(h),H_{s_0}(h_0))\le d(H_s(h),H_s(h_0))+d(H_s(h_0),H_{s_0}(h_0))\le d(h,h_0)+d(H_s(h_0),H_{s_0}(h_0)),$ which tends to zero as $(s,h)\to (s_0,h_0)$. Thus $(s,h)\mapsto H_s(h)$ is continuous for the uniform topology, hence for the compact-open topology by [L2].

The contraction. Define $F(t,h):=H_{1-t}(h)$ for $(t,h)\in [0,1]\times {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$. By step 3.1 the map $F$ is continuous into the compact-open topology, and by step 1.1 each $F(t,-)$ has values in the group; moreover $F(0,h)=H_1(h)=h$ and $F(1,h)=H_0(h)=\mathrm{id}$, so by [L4] the map $F$ is a homotopy from the identity map of ${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ to the constant map at the identity element, that is, the group is contractible.