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I have two different Lie groups, one is an arcwise connected Lie subgroup

$G_1 \subseteq \mathbb{R}^m \times (S^1)^n$

and the other is

$G_2 = \mathbb{R}^k \times (S^1)^l$

While $G_1$, $m$ and $n$ are fixed, I can change $G_2$ by varying $k$ and $l$ as I please. I'm interested in continuous

$f : G_1 \to G_2$

$g : G_2 \to G_1$

and was wondering if anything at all can be said about $f$ or $g$? Ideally, I would like to find any condition for $l$ and $k$, so that $f$ or $g$ become "almost" homeomorphic or that both are surjective. By "almost" I'm trying to express that there exists an $f$ and $g$ for which

$(g \circ f)(p) = p$

is true except for a tiny finite subset. Is that possible?

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  • $\begingroup$ Conditions on $f,g$ --- any function? smooth map? Abstract group homomorphisms? Lie group homomorphisms? ... $\endgroup$
    – user10354138
    Jun 3, 2019 at 10:53
  • $\begingroup$ I've added the condition that $f,g$ are continuous. $\endgroup$
    – Mike
    Jun 3, 2019 at 10:58
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    $\begingroup$ There is a continuous surjection $G_1\to G_2$ regardless of what $m, n, k,l$ are as long as at least $m\geq 1$. It is a variation of a space filling curve. Now for the stronger condition $g\circ f=id$ it means that $G_1$ can be embedded as a retract of $G_2$. And I think that this cannot be done when $n>l$ or $m>k$ by careful homology analysis and/or invariance of domain. BTW how is that question related to the Lie theory? How is the group/differential structure of $\mathbb{R}^n$, $S^1$ relevant here? Are you assuming that $f,g$ are Lie homomorphisms (which makes that really simple)? $\endgroup$
    – freakish
    Jun 3, 2019 at 13:18
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    $\begingroup$ Thank you for pointing out space filling curves, hadn't thought of those yet! I was hoping that restricting $G_1$ to be a Lie subgroup would make things easier. In general I am interested in all connected differentiable submanifolds of $\mathbb{R}^m \times (S^1)^n$. $\endgroup$
    – Mike
    Jun 3, 2019 at 16:23

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