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Let $G$ be the topological semigroup whose underlying space is $C(\mathbb{R}^d,\mathbb{R}^d)$ equipped with composition as semigroup operation and let $H$ be the topological group whose underlying space is $C(\mathbb{R}^d,\mathbb{R}^d)$ equipped with pointwise addition as group law; here $C(\mathbb{R}^d,\mathbb{R}^d)$ is equipped with the compact-open topology.

Is there a continuous (non-constant) semigroup homomorphism from $G$ to $H$?

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    $\begingroup$ Nitpick: presumably you wish to exclude the trivial homomorphism $G \to \{0\}$ $\endgroup$
    – Yemon Choi
    Commented Nov 16, 2019 at 20:07
  • $\begingroup$ Of course. I'll add the point. $\endgroup$
    – ABIM
    Commented Nov 16, 2019 at 20:10
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    $\begingroup$ You probably meant to take $G$ as the invertible functions on $\mathbb R^d$? $\endgroup$
    – Christian Remling
    Commented Nov 16, 2019 at 20:14
  • $\begingroup$ @Christian Remling I'm looking for semi-group homomorphisms so there is no need for them to be invertible $\endgroup$
    – ABIM
    Commented Nov 16, 2019 at 22:01

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The new version, for semigroups, is much easier: there is no such homomorphism $\varphi$, for purely algebraic reasons. Consider just the constant functions $c$. Since $cd=c$ in the semigroup, you must map $\varphi(d)=0$. But for any $f$, $fc$ is also constant ($=f(c)$), so $\varphi(f)+\varphi(c)=0$ and thus $\varphi(f)=0$ as well.

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  • $\begingroup$ True, this is really obvious now. Though I feel that the group-theoretic version is not so simple.. where $G\subseteq C(\mathbb{R}^d;\mathbb{R}^d)$ comprised of invertible functions. $\endgroup$
    – ABIM
    Commented Nov 16, 2019 at 23:33
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    $\begingroup$ I don't see the need for the last sentence of this answer. You've already proved $\varphi(d)=0$ for all $d$ (not only constant $d$, because $cd=c$ for all $d$). $\endgroup$
    – Andreas Blass
    Commented Nov 17, 2019 at 0:12
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    $\begingroup$ @AIM_BLB For invertible functions, consider $f(x) = -x \in C(\mathbb{R}^d, \mathbb{R}^d)$. Then $f \circ f = Id$, but there are no torsion points in the additive group. $\endgroup$
    – user44191
    Commented Nov 17, 2019 at 0:19
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    $\begingroup$ @user44191: I don't think it's that easy. This just shows that $f(x)=-x$ will have to be sent to $id=0$ in $H$. (Of course, there are lots of other elements of finite order for which this conclusion holds if $d>1$, as well as a large commutator subgroup that will have to get mapped to $0$, but I still don't think things are very obvious.) $\endgroup$
    – Christian Remling
    Commented Nov 17, 2019 at 0:55
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    $\begingroup$ @AIM_BLB: That's up to you obviously, but I would normally try to avoid making questions a 'moving target' by changing them after posting. $\endgroup$
    – Christian Remling
    Commented Nov 17, 2019 at 2:34

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