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Let $ G $ be a metrizable group. If (i) $ K $ is a closed normal subgroup of $ G $ and (ii) both $ K $ and $ G/K $ are complete, then $ G $ is complete.

Here is how I am proceeding:

It can be assumed w.l.o.g. that the topology of $ G $ is induced by a right-invariant metric $ d $. Let $ (x_{n})_{n \in \mathbb{N}} $ be a right Cauchy sequence in $ G $. It suffices to show that some neighborhood $ V $ of $ e $ in $ G $ is complete. I don’t know how to proceed after this. Please help me.

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The process of helping you will be more expeditious if you use MathJax to write all of the mathematical expressions. :) –  Haskell Curry Jan 18 '13 at 3:34
    
I'm a little confused a metric on $G$ doesn't necessarily descend to a metric on $G/K.$ For example consider the case of $\mathbb{R}$ with the metric $d(x,y) := |arctan(x-y)|$ and $K := \mathbb{Z}.$ On the other hand, if it did this case would be a counterexample to your claim. –  jspecter Jan 18 '13 at 4:54
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@jspecter: You take the infimum of the metric on the cosets. –  user641 Jan 18 '13 at 15:43
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@K.Ghosh: Consider a Cauchy sequence $x_i$ in $G$; mapping it to $G/K$ gives a Cauchy sequence in $G/K$, which converges to some coset $Kx$. Now think about the Cauchy sequence $x_ix^{-1}$. –  user641 Jan 18 '13 at 15:45
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@jspecter: The quotient of any topological group by a closed normal subgroup is Hausdorff. –  Emil Jeřábek Jan 18 '13 at 19:04

1 Answer 1

It seems that there is no need to prove anything, you can simply use the references.

Lemma 1. [Eng, Th. 4.3.26] A topological space $X$ is metrizable by a complete metric iff $X$ is \v Cech-complete and metrizable.

Lemma 2. ([Bro] or [Cho]) A first countable $T_1$ topological group $G$ is Raikov complete iff $G$ is \v Cech-complete.

Tkachenko in [Tka] wrote that Vilenkin shows that the class of Weil-complete topological groups is closed under extensions: if $N$ is a closed normal of subgroup of a $T_1$ topological group $G$ and both groups $N$ and $G/N$ are Weil-complete, then $G$ is Weil-complete, and following Vilenkin’s reasoning, Graev [Gra] proves a similar result for Raikov-complete topological groups.

Referenses

[Bro] L.G. Brown, {\it Topologically complete groups}, Proc. Amer. Math. Soc. {\bf 35} (1972), 593--600.

[Cho] M.M. Choban, {\it On completions of topological groups}, Vestnik Moskov. Univ. Ser. Mat. Mekh. (1) (1970), 33--38 (in Russian).

[Eng] R. Engelking. {\it General Topology}. -- M.: Mir, 1986 (in Russian).

[Gra] M.I. Graev, {\it Theory of topological groups I}, Uspekhy Mat. Nauk {\bf 5} (1950), 3—56 (in Russian).

[Tka] M.G. Tkachenko, {\it Topological groups for topologists: part I}, Bol. Soc. Mat. Mexicana (3), Vol. 5, 1999.

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Welcome to the site and thanks for your contribution! I think the statement Lemma 1 should be: $X$ is metrizable by a complete metric iff $X$ is Cech-complete and metrizable. // We only have restricted LaTeX functionality here. To get italic and boldface text you can use markdown: *italics* and **bold** are formatted italics and bold. –  Martin Apr 12 '13 at 6:06
    
@Martin Thanks, I have corrected the formulation of Lemma 1. Also thanks for the information about markdown. But I think that sometimes pure LaTeX parts of answers may be good too, because they can be pasted directly into the article. :-) –  Alex Ravsky Apr 12 '13 at 6:24

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