# Finding the boundary of the continuous image of a compact simply-connected Lie group

Statement of the problem

Given a continuous map $f:G \rightarrow D^2$ where $G$ is a compact simply connected Lie group and $D^2$ is the unit disk in the plane, I have shown that:

• There exists a simple (non-self-intersecting, except at a single point) loop $\gamma \subset G$ that maps injectively to a simple loop $f(\gamma) \subset D^2$.

• Furthermore, $f(\gamma)$ intersects $\partial D^2$ at several isolated points. Call $P \subset G$ the isolated set of points in $\gamma$ that map to $\partial D^2$.

• Any small neighborhood of $\gamma$ maps to the interior of the region bounded by $f(\gamma)$.

Is this sufficient info to prove that $f(\gamma)$ is the boundary of the image $f(G)$?

The image is this

Attempt at a solution

$f$ continuous, $G$ compact implies $f(G)$ compact and $f(G)\subset D^2$ has a boundary $B$. Let $\gamma\subset G$ be as described above and call $S$ the open set that has boundary $f(\gamma)$. Assume that $B$ $\neq$ $\gamma$ along some path $\phi \subset D^2$. Then $\phi \subset D^2 \setminus S$.

Then $\exists x \in G$ s.t. $f(x) \in \phi$ and $f(x) \not\in S\cup f(\gamma)$. Then $x$ is not in a neighborhood of $\gamma$, by definition of $\gamma$.

Claim: Because $G$ is simply connected, we can define a continuous map

\begin{align} & h:[0,1] \times G\rightarrow G \\ &h(0,x)= x,\hspace{5mm} h(1,x)=y\in \gamma \\ \end{align}

Furthermore, we can choose this $h$ so that $f h(t,x) \not\in f(\gamma), \forall t \neq 1$.

(Is it possible to prove that such an $h$ exists, speaking rougly: that I can always continuously deform a point not in a neighborhood of $\gamma$ to a point in $\gamma$, while avoiding other subsets of $G$ that might map to $f(\gamma)$?)

Then $\forall \epsilon > 0$, $\exists t < 1$, such that $h(t,x)$ is in the open ball $B(y,\epsilon)$. But then for some $\tilde{t}<1, h(\tilde{t},x) \in S$ (by definition of $\gamma$: small neighborhoods of $\gamma$ map to $S$).

But this is impossible, because $fh$ is continuous and $fh(t,x)\not\in f(\gamma)\,,\forall t\neq 1$. Then $\phi$ cannot exist and $f(\gamma)=\partial f(G)$.

I'm looking for either an answer using some other machinery, a resolution of my attempt, or some references (really my knowledge of topology/homotopy/differential geometry is quite limited).

• No, this is not true. You've given us nothing about what the map looks like away from this small neigjborhood of $\gamma$; you could engineer the map to be surjective away from this small neighborhood. – user98602 Jul 1 '16 at 15:45
• What if I restrict the domain of $f$ to some open ball $H$ centered on a point in $\gamma$ and use the canonical diffeomorphisms of an atlas on $G$ to define locally $\tilde{f}: H\subset \mathbb{C}^n \rightarrow \mathbb{C}$, then show that $\tilde{f}$ is holomorphic? Then by the open mapping theorem for several complex variables, $\tilde{f}(H)$ is open, but I claim I've shown that $f(H)$ maps to some subset of $f(\gamma)$ and points "nearby" inside of the region bounded by $f(\gamma)$ and so $\tilde{f}(H)$ can only be open if $f(\gamma)$ is the boundary of $f(G)$? Fair? – Jonathan Rayner Jul 5 '16 at 23:54