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Question:

Is there always a continuous surjection $f: H \times G/H \rightarrow G$?

where $G$ is a topological group, $H$ is a subgroup of $G$ and $G/H$ is given the quotient topology.

We know $H \times G/H$ is not necessarily homeomorphic to $G$, even in the case $H$ is a normal subgroup of $G$. (for example, take $G=\Bbb R$ and $H=\Bbb Z$). But that is evidently a stronger statement.

To find a counterexample, we may look for topological properties that are preserved by continuous maps. However, compactness and connectedness are already ruled out. For example, see If $H$ and $G/H$ are compact, then $G$ is compact. and If $H$ and $\frac GH$ are connected so is $G$

As commented by @MoisheKohan, it's not a very interesting question. What if I require $H$ to be closed?

YuiTo Cheng
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2 Answers2

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An easy examples to the original question is: Take $G={\mathbb R}$ (the additive group) and $H={\mathbb Q}< G$. Then $H$ is dense in $G$, hence, every point is dense in $G/H$, hence, every continuous map $G/H\to G$ is constant. Thus, the image of every continuous map $F= H\times G/H\to G$ has (at most) countable image, hence, cannot be surjective.

As for the case when $H$ is closed, I do not know. It is not hard to see (using Hahn-Mazurkevich theorem) that for every Lie group $G$ and closed subgroup $H<G$, there is a continuous surjection $F\to G$. The same holds in the case when $G$ is metrizable, locally compact and totally disconnected. One should be able to prove the existence of a surjective continuous map when $G$ is a locally compact metrizable group (but I did not think about details).

One last thing: I find the question rather unnatural since it essentially ignores the group structure of $G$.

Moishe Kohan
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  • With $H$ closed, take $G=\mathbf{R}$ and $H=\mathbf{Z}$. The only homomorphism from $G/H$ to $G$ is constant, because $G/H$ is compact and $G=\mathbf{R}$ has no nontrivial compact subgroup. Hence every continuous homomorphism from $H\times G/H$ to $G$ has a cyclic image. – YCor Apr 26 '19 at 22:14
  • @YCor: OP did not ask about homomorphisms only continuous surjections. – Moishe Kohan Apr 26 '19 at 22:31
  • Indeed. I actually realized this after writing my comment, and actually thought I had erased it. Then I posted an answer with an example not assuming group homomorphism. – YCor Apr 26 '19 at 22:50
  • @YCor: Oh, I see. Nice example! – Moishe Kohan Apr 26 '19 at 22:57
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Here's an example with $H$ closed.

Take $G$ to be the product $\mathbf{R}\times\mathbf{Z}_p$ (here $\mathbf{Z}_p$ is in its usual meaning, i.e., the compact group of $p$-adic integers). Take $H$ to be the infinite cyclic subgroup generated by $(1,1)$.

Claim: $G/H$ is connected.

Granted the claim, let us conclude: given a continuous map $f:H\times G/H\to G$, every coset $\{h\}\times G/H$ being connected, it maps into a single connected component $\mathbf{R}\times\{x_h\}$ of $G$. Since $G$ has uncountably many connected components and $H$ is countable, we deduce that $f$ is not surjective.

To prove the claim, let $\pi:G\to G/H$ be the projection: it is enough to show that $\pi(\mathbf{R}\times\{0\})$ is dense. Indeed, since $\pi(n,n)=0$ for every $n\in\mathbf{Z}$, we have $\pi(\mathbf{R}\times\{0\})=\pi(\mathbf{R}\times\mathbf{Z})$. Since $\mathbf{Z}$ is dense in $\mathbf{Z}_p$, the latter is clearly dense.

YCor
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