For which topological rings $A$ does there exist a continuous section (as a set map at least) of the quotient morphism $GL_n(A) \to GL_n(A/I)$, where $I$ denotes a nilpotent ideal in $A$?
It should work for Frechet spaces.
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If $x'$ is invertible in $M_n(A/I)$ with inverse $y'$, then lifting $x'$ and $y'$ to $M_n(A)$ yields a pair of elements $x,y$ such that $xy = 1 + z$ with $z \in M_n(I)$. Now, $M_n(I)$ is nilpotent and the Neumann series yields an inverse for $1+z$ in $M_n(A)$: $$(1+z)^{-1} = \sum_{n \geq 0} (-z)^n.$$
One can now see that $y(1+z)^{-1}$ is an inverse of $x$.
So your question boils down to the question whether one can find a continuous section of $A \to A/I$. The Bartle-Graves Theorem states that every surjection of Banach spaces has continuous section. I think this also holds for Frechet spaces by work of Ernest Michael (The Annals of Mathematics, 2nd Ser., Vol. 63, No. 2. (Mar., 1956), pp. 361-382.)
answered Oct 22, 2010 at 12:47
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$\begingroup$ When $n=1$, then the topology on A* = {GL}_{1}(A) is not necessarily the subspace topology (e.g. adeles/ideles). This was intentionally left ambiguous in the question. What would happen then? Your answer works when ${GL}_{n}(A)$ has the subspace topology from $M_n(A)$. $\endgroup$ Commented Oct 23, 2010 at 15:05
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$\begingroup$ I do not know. I think that there are many results about continuous section of surjective homomorphisms between polish groups. Maybe you will find what you need there. $\endgroup$ Commented Oct 23, 2010 at 16:12