geometric-topology
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.
The following is essentially a fleshed-out version of the proof of Proposition 4.2 in Ghys's beautiful article Groups acting on the circle.
It is convenient to translate the question into a question about continuous functions on $\mathbb{R}$. Viewing the circle $S^1 = \mathbb{R}/\mathbb{Z}$ as a quotient of $\mathbb{R}$, a continuous function $f\colon S^1 \to S^1$ can be lifted over the covering projection $\pi \colon \mathbb{R} \to S^1$ to a continuous function $F \colon \mathbb{R} \to \mathbb{R}$ such that $f \circ \pi = \pi \circ F$: on the fundamental domain $[0,1)$ the function $F$ is determined by $f$ up to addition of an integer, and once that integer is chosen, there is only one way to extend $F$ continuously to all of $\mathbb{R}$.
A continuous function $F \colon \mathbb{R} \to \mathbb{R}$ is the lift of a continuous function $f \colon S^1 \to S^1$ if and only if there is $d \in \mathbb{Z}$ such that for all $x$ the equation $F(x+1) = F(x) + d$ holds. It is not hard to see that $d = \deg{f}$. Using this [or by elementary considerations with the intermediate value theorem] we see that $f$ can only be a homeomorphism if $d = \pm 1$: the lifts of a homeomorphism must satisfy $F(x+1) = F(x) \pm 1$. The lift $F$ of a homeomorphism $f$ is a homeomorphism because continuous bijections $\mathbb{R} \to \mathbb{R}$ have continuous inverses.
Let $\tilde{H} = \operatorname{Homeo}_\mathbb{Z}(\mathbb{R})$ be the group of homeomorphisms $F \colon \mathbb{R} \to \mathbb{R}$ such that $F(x+1) = F(x) +1$. The compact-open topology on $\tilde{H}$ coincides with the uniform topology and is metrized by $d(F,G) = \sup_{x \in [0,1]} \lvert F(x) - G(x)\rvert$ since $F-G$ is $1$-periodic. If $F,G \in \tilde{H}$ then their convex combinations $(1-t)F + tG$ belong to $\tilde{H}$ for all $t \in [0,1]$: they are strictly increasing and surjective. The map $$h\colon [0,1] \times \tilde{H} \times \tilde{H} \longrightarrow \tilde{H}, \quad(t,F,G) \longmapsto (1-t)F + tG$$ is clearly continuous. This shows that $\tilde{H}$ is contractible, as we can take $G$ to be the identity $G(x) = x$ and for all $F \in \tilde{H}$ we have $h(0,F,G) = F$ and $h(1,F,G) = G$.
The map $F \mapsto f$ sending $F \in \tilde{H}$ to the homeomorphism of $S^1$ it covers is a continuous homomorphism $p\colon \tilde{H} \to H = \operatorname{Homeo}_+(S^1)$ onto the group of orientation-preserving homeomorphisms (i.e., those $f \colon S^1 \to S^1$ preserving the cyclic order of triples of pairwise distinct points on the circle). The kernel of $p$ can be identified with $\mathbb{Z} = \langle \tau \rangle$ generated by the unit translation $\tau(x) = x + 1$. In fact, $p$ identifies $\tilde{H}$ with the universal covering group of $H$.
Observe that $f = p(F)$ is a rotation with angle $[\alpha] \in \mathbb{R/Z}$ if and only if $F(x) = x + \alpha$ is a translation.
Now notice that each $F \in \tilde{H}$ can be uniquely written as $F(x) = x + \alpha_F + \varphi_F(x)$ with $\alpha_F = \int_{0}^1 [F(x)-x]\,dx \in \mathbb{R}$ and $\varphi(x) = F(x) - x -\alpha$ is a $1$-periodic function with mean zero: $\int_{0}^1 \varphi(t)\,dt = 0$. Since $F_n \to F$ in $\tilde{H}$ means uniform convergence, we see that $\alpha_F$ and $\varphi_F$ depend continuously on $F$. Thus, the map $$ \tilde{k} \colon [0,1] \times \tilde{H} \longrightarrow \tilde{H} \quad \tilde{k}(t,F) = x + \alpha_F + (1-t)\varphi_F $$ is continuous and $\tilde{k}(0,F) = F$ while $\tilde{k}(1,F) = x+\alpha_F$ covers a rotation, i.e., an element of $SO(2)$. If $F(x) = x + \alpha$ is a translation then $\tilde k(t,F) = F$ for all $t \in [0,1]$, so $\tilde k$ is constant on the group of translations.
For two lifts $F_1(x) = x + \alpha_{F_1} + \varphi_{F_1}(x)$ and $F_{2}(x) = x + \alpha_{F_2} + \varphi_{F_2}(x)$ of the homeomorphism $f \colon S^1 \to S^1$ we have $\varphi_{F_1} = \varphi_{F_2}$ and $\alpha_{F_1} - \alpha_{F_2} \in \mathbb{Z}$. This implies that $\tilde{k}(t,F_1)$ and $\tilde{k}(t,F_2)$ cover the same homeomorphism of $S^1$ and hence $\tilde{k}$ yields a continuous map $k\colon [0,1] \times H \to H$ which is a deformation retraction from $\operatorname{Homeo}_+(S^1)$ to $SO(2)$.
The argument for the group of all homeomorphisms of $S^1$ is essentially the same: If $f$ reverses orientation then its lifts are of the form $F(x+1) = F(x) - 1$ and $F$ is strictly decreasing. Now define $\alpha_F = \int_{0}^1 [F(x) + x]\,dx$ and $\varphi_F = F + x - \alpha_F$ to obtain a homotopy from $F$ to $\alpha_F - x$, covering the reflection at the line with angle $\alpha_F/2$.
Since asking the question i have learned that the canonical reference for this is the very detailed book "The convenient setting of global analysis" by Kriegl and Michor. In this book, they introduce the theory of infinite dimensional smooth manifolds. Let me sketch the argument.
They work with convinient vector spaces which are locally convex vector space satisfying a condition called $c^\infty$-completeness and a notion of smooth mappings between them. It turns out that one has to introduce a new topology called the $c^\infty$-topology on the locally convex vector spaces, because in the locally convex topology, smooth mappings are not necessarily continuous. Then infinite dimensional manifolds are defined analogously to ordinary smooth manifolds using convenient vector spaces equipped with the $c^\infty$-topology instead of Euclidian space. The set $C^\infty(M,N)$ equipped with the Whitney $C^\infty$-topology becomes an infinite dimensional manifold, and $\text{Emb}(M,N)\subset C^\infty(M,N)$ and $\text{Diff}(F)\subset C^\infty(F,F)$ are submanifolds, and $\text{Diff}(F)$ is a Lie group.
Let $F$ be a closed smooth finite dimensional manifold, and let $\ell^2(\mathbb{N})$ denote the Hilbert space of sequences bounded in the norm $\lvert \lvert \cdot \rvert \rvert_2$. The Lie group $\text{Diff}(F)$ acts on the embedding space $\text{Emb}(F,\ell^2(\mathbb{N}))$ by composition and yields a principal $\text{Diff}(F)$-bundle $$ \text{Emb}(F,\ell^2(\mathbb{N})) \rightarrow \text{Emb}(F,\ell^2(\mathbb{N}))/\text{Diff}(F). $$ The total space is contractible which implies that the base space is homotopy equivalent to $\text{BDiff}(F)$. Denote the base space with $B_\infty(F)$. As they mention in the book, the base space can be thought of as the nonlinear version of the infinite Grassmannian. Consider the associated bundle $$ \text{Emb}(F,\ell^2(\mathbb{N})) \times_{\text{Diff}(F)}F \rightarrow B_\infty(F), $$ and denote its total space by $E_\infty(F)$. This associated bunde has fiber $F$ and is the universal $F$-bundle (Theorem 44.24 in the book). Since $B_\infty(F)\simeq B\text{Diff}(F)$, we get the wanted bijection in the question.
Best Answer
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.