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Does there exists an example of an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or continuous in a neighbourhood of identity) and satisfy the propery that $s(x^{-1}) =s(x)^{-1}$. One can see that the extension $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{S}^1 \to 1$ does not admit such a section.

For discrete groups there are many extensions admitting such a section. For example, consider the extension $0 \to \mathbb{Z} \stackrel{i}{\to} \mathbb{Z} \times \mathbb{Z}_2 \stackrel{p}{\to} \mathbb{Z}_4 \to 0$ with $i(x)= (2x, [x])$ and $p(x, [y])=[x + 2y]$. Clearly, the section $s$ defined by $s([0])=(0,[0])$, $s([1])=(-1,[1])$, $s([2])=(0,[1])$ and $s([3])=(1,[1])$ satisfy the property that $s(x^{-1})= s(x)^{-1}$. I am looking for an example in the setting of compact connected topological groups.

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Yes: a semi-direct product. Or do you want some sort of non-triviality condition? –  David Roberts Dec 24 '10 at 8:36
    
Thanks. But I want a non-trivial extension. –  gurs Dec 24 '10 at 9:04
    
Let me write my question again. Does there exist an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or only continous in a neighbourhood of identity) and satisfy the property that $s(x^{-1})=s(x)^{-1}$ for all $x \in G$ but is not a homomorphism. –  gurs Dec 24 '10 at 9:13
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Take the extension $1 \to \mathbb{R} \to \mathbb{R}^2 \to \mathbb{R} \to 0$ and the section $x \mapsto (x,x^3)$. –  Johannes Ebert Dec 24 '10 at 10:06
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gurs says (moved from an answer): Can someone give an example in the setting of compact topological groups or in the setting of finite dimensional Banach spaces. –  Anton Geraschenko Dec 31 '10 at 18:16
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up vote 10 down vote accepted

A very easy example (inspired by an answer by Laurent Moret-Bailly that he removed immediately after posting): Take $0 \to 3\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \to 0$ and define a section $s: \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}$ by $s(0) = 0$, $s(1) = 1$ and $s(2) = -1$.

Here's an example from functional analysis that I happen to like, but it certainly is rather involved:

Take a non-complemented subspace $F$ of a Banach space $E$. The quotient map $p:E \to E/F$ admits a continuous and homogeneous right inverse $\sigma$ by a theorem of Bartle-Graves and Michael, see E. Michael "Continuous selections, I", Ann. Math. Vol. 63, No. 2 (Mar., 1956), pp. 361-382, Proposition 7.2. Homogeneity yields in particular $\sigma(-x) = -\sigma(x)$, but $\sigma$ cannot be linear because $F$ is not complemented in $E$.

You can take for instance $F = c_{0}$, the space of sequences converging to zero, inside the Banach space $E = \ell^{\infty}$ of bounded sequences.

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gurs says (moved from an answer): Dear Theo Buehler, many thanks for the example. Can someone give an example in the setting of compact topological groups or in the setting of finite dimensional Banach spaces. –  Anton Geraschenko Dec 31 '10 at 18:16
    
Thanks, Anton. @gurs: no, there is no example involving finite-dimensional Banach spaces because for such spaces all linear maps are continuous. If you want compact groups, try finite groups first, that shouldn't be too hard. –  Theo Buehler Dec 31 '10 at 18:21
    
@gurs: You can tweak the discrete example to be finite (and thus compact), 0 → 3ℤ/9ℤ → ℤ/9ℤ → ℤ/3ℤ → 0. Perhaps you want and example where $E$ is connected (and compact?). @Theo: Is there some reason to use 3 instead of 2, as Laurent did? –  Anton Geraschenko Dec 31 '10 at 18:40
    
@Anton: yes, gurs wanted in addition $s(x^{-1}) = s(x)^{-1}$ and this doesn't work with $2$. I agree that it would be nice to have a finite-dimensional connected (and compact) example, but I cannot come up with one right now. –  Theo Buehler Dec 31 '10 at 18:54
    
@Theo: Of course. Thanks. –  Anton Geraschenko Dec 31 '10 at 19:08
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