What (some) people call the Alexander trick is a proof of the statement that every homeomorphism of $S^1$ can be extended to a homeomorphism of $D^2$. Explicitly, the map $f: S^1 \rightarrow S^1$ is extended to $D^2 = \{(r, x) : r \in [0, 1], x \in S^1\}$ by taking $rx$ to $rf(x)$.
Furthermore, I'm inclined to say that this in fact defines a continuous map $\text{Homeo}(S^1) \rightarrow \text{Homeo}(D^2)$ (with the compact-open topology). I'm wondering if anyone has a simple proof of this.
Call the map for $\phi$. If $X$ is compact the compact-open topology on $C(X,Y)$ is the same as the topology induced by uniform convergence. So to check that $\phi$ is continuous w.r.t. this topology, suppose $\{f_n: S^1 \rightarrow S^1\}_n$ converges uniformly to $f: S^1 \rightarrow S^1$.Then $|\phi(f_n)(rx)-\phi(f)(rx)|=|rf_n(x)-rf(x)|=|r||f_n(x)-f(x)|\leq |f_n(x)-f(x)|$ for $r\in [0,1]$, $x\in S^1$, showing that $\phi(f_n) \rightarrow \phi(f)$ uniformly. So $\phi$ is continuous.
