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Could any one give me hint to solve this one?

$f:G\rightarrow H$ is continuous homomorphism into a locally compact Hausdorff group $H$. Then we need to show $f$ is necessarily open. all spaces 2nd countable, hausdorff and locally compact

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You still have into, not onto. It would have been nice to edit in such a way that my answer doesn't seem completely off. – Martin Jan 26 '13 at 8:54
up vote 2 down vote accepted

You must be missing some hypotheses.

If $f$ is into, not onto, then this is very wrong: take the inclusion $x \mapsto (x,0)$ from $\mathbb{R}$ to $\mathbb{R}^2$.

But even if $f$ is onto this is still wrong. Let $G$ be the additive group of real numbers with the discrete topology and $H$ the same group with the usual topology. Both groups are locally compact and the identity mapping $f \colon G \to H$ is continuous and bijective, but it is not open.

Answer to the modified question which probably is supposed to read:

Let $G$ and $H$ be locally compact, Hausdorff and second countable topological groups. Let $f \colon G \to H$ be a continuous surjective homomorphism. Then $f$ is open.

Recall Pettis's theorem: if $U$ is a non-meager subset of a Baire topological group then $U^{-1}U$ is a neighborhood of the identity.

Let $K$ be the kernel of $f$ and let $p \colon G \to G/K$ be the canonical projection. There is a unique $g \colon G/K \to H$ such that $f = gp$, and $g$ is a continuous and bijective homomorphism. Since $p$ is clearly open, it suffices to prove that $g^{-1} \colon H \to G/K$ is continuous. This follows from the fact that the graph of $g$ is closed, hence the graph of $g^{-1}$ is closed and thus $g^{-1}$ is Borel measurable. Now it is a standard consequence of Pettis's theorem that a Borel measurable homomorphism from a Baire group to a second countable group is continuous, hence $g^{-1}$ is continuous, since $H$ is Baire and $G/K$ is second countable.

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One can prove that the statement is true if both $G$ and $H$ are assumed to be second countable and $f$ is surjective, but I'm not sure if this is what you really want. – Martin Jan 26 '13 at 6:23
thank you, I will check it once more! – La Belle Noiseuse Jan 26 '13 at 6:32
my spaces are second countable, hausdorff, locally compact. – La Belle Noiseuse Jan 26 '13 at 8:35
I sketched a proof. I don't have the time to look for references right now. – Martin Jan 26 '13 at 9:19
Pettis's theorem?Why is pettis's thm? – David Chan May 9 '15 at 3:30

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