Commutator subgroup of connected group. Let $G$ be a compact, connected (Lie) group. I read in some paper that the commutator subgroup, $[G,G]$, which is the subgroup of $G$ generated by all its commutators, is also connected.
Elements of $[G,G]$ have the form $[g_1,g'_1] \ldots [g_k,g'_k],$ where $k\in \mathbb{N}$ and $g_j,g'_j \in G$, for $j=1,\ldots,k$, and my only thought is that, since a continuous image of a connected set is also connected, I would like to find a continuous function $f:G \to G$, such that $f(G)=[G,G]$. Is this possible?

The argument is a bit more complicated than what you suggest.  For each $k$, there is a continuous map $f:G^{2k}\to G$ sending $(g_1,g_1',\dots,g_k,g_k')$ to $[g_1,g_1']\dots[g_k,g_k']$.  So for each $k$, the set of products of $k$ commutators is connected.  These connected sets each contain the identity element, so the union of all of them is connected as well.  That is, $[G,G]$ is connected.
(A similar argument shows that if $A\subseteq G$ is a connected subset of a topological group which contains the identity element, then the subgroup generated by $A$ is connected.  So you could also apply this with $A$ being the set of commutators.)
