While the fundamental group $\pi\_1$ preserves products, it is not true in general that an inverse limit of simply connected topological spaces is simply connected. I would like to know if similar things can happen with topological groups.


Let $G=\varprojlim\_{n}(G\_n,p\_{n+1,n}:G\_{n+1}\to G\_n)$ be the inverse limit of an inverse sequence of 1-connected topological groups $G\_n$ and continuous homomorphisms. If $e$ is the identity element of $G$, must $\pi\_1(G,e)$ be trivial? What if each $G\_n$ is a Lie group?



Must an inverse limit of simply connected groups be simply connected?
It seems that the inverse limit of simply connected Lie groups is simply connected. The argument uses the well-known fact that the second homotopy group of any Lie group is trivial (see e.g. [Homotopy groups of Lie groups](https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups)).


Applying the long exact sequence of homotopy groups to the bonding homomorphism $p\_{n+1,n}:G\_{n+1}\to G\_n$ between the simply connected topological groups, we conclude that its kernel $K\_{n+1}=p\_{n+1,n}^{-1}(e)$ has trivial homotopy group $\pi\_1(K\_{n+1})=\pi\_2(G\_n)=0$. Using this fact, and also the fact that a locally trivial bundle over a contractible space is trivial, for any loop $\gamma:\partial\mathbb D\to G$ defined on the boundary of the unit disk $\mathbb D$ we can inductively construct a sequence of maps $\bar\gamma\_n:\mathbb D\to G\_n$ such that $p\_{n+1,n}\circ \bar\gamma\_{n+1}=\bar\gamma\_n$ and $\bar\gamma\_n|\partial\mathbb D=p\_{\infty,n}\circ\gamma$ for all $n$. Here $p\_{\infty,n}:G\to G\_n$ denotes the limit projection. Then the maps $\bar \gamma\_n$ determine a map $\bar \gamma:\mathbb D\to G$ extending the map $\gamma$ and witnessing that the topological group $G$ is simply connected.


I hope that this argument is correct (this is a question to specialists in Algebraic Topology).


