Lifting Free Group Actions on $S^1$ to $\mathbb{R}$ The free group $F_2$ acts faithfully on the circle $S^1$ by homeomorphisms.  Someone told me that given any such action on $S^1$, we can lift it to a faithful action $F_2\curvearrowright \mathbb{R}$ by homeomorphisms.  Why isn't there an obstruction?  It seems to me like we should have to worry about lifts composing correctly.
 A: We don't have to worry about lifts composing correctly because the group is free, so there aren't any relations to worry about!  Letting $a$ and $b$ be the generators of $F_2$ acting on $S^1$, we can lift them to homeomorphisms $\bar{a},\bar{b}:\mathbb{R}\to\mathbb{R}$.  Then $\bar{a}$ and $\bar{b}$ generate an action of $F_2$ on $\mathbb{R}$ which is a lift of our action on $S^1$.  This action is faithful since the action in $S^1$ is faithful.
A bit more abstractly, let $G$ be the group of homeomorphisms of $S^1$ and let $H$ be the group of homeomorphisms of $\mathbb{R}$ that descend to a homeomorphism of $S^1$.  There is a homomorphism $p:H\to G$, taking an element of $H$ to the element of $G$ it descends to.  Since every element of $G$ can be lifted to a homeomorphism of $\mathbb{R}$, $p$ is surjective.
Now, our faithful action on $S^1$ is just an injective homomorphism $f:F_2\to G$.  If $a$ and $b$ are the generators of $F_2$, pick $x,y\in H$ such that $p(x)=f(a)$ and $p(y)=f(b)$ and let $g:F_2\to H$ be the unique homomorphism such that $g(a)=x$ and $g(b)=y$.  Then $pg=f$ (since they agree on $a$ and $b$), so $g$ is a lift of our action to $\mathbb{R}$.  Since $f$ is injective and $pg=f$, $g$ is injective too, so the lifted action is faithful.
