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Let $\mathbb{R}^+$ be the real numbers under addition and let $\mathbb{R}^{*}$ be the nonzero real numbers under multiplication so that $(0, +\infty) \leq \mathbb{R}^{*}$. Suppose there is a group homomorphism $\exp : \mathbb{R}^+ \to (0, +\infty)$, namely $x \mapsto e^x$. With the fact that $\exp$ is strictly increasing, it's easy to see that $\exp$ is injective. To prove that $\mathbb{R}^+ \cong (0, +\infty)$, we could show that $\exp$ is continuous with $\lim_{x \to -\infty} e^x = 0$ and $\lim_{x \to +\infty} e^x = +\infty$ and then apply the intermediate value theorem. Is there an alternative way to prove $\exp$ is surjective using group theory?

I had ideas about using the correspondence theorem (if $G, G'$ are groups and $\varphi : G \to G'$ is a surjective homomorphism, then there is a bijection between the subgroups of $G'$ and the subgroups of $G$ containing $\ker \varphi$) and the fact that $(0, +\infty)$ is the only subgroup of $\mathbb{R}^{*}$ with finite index, but these don't seem to be going anywhere.

jskattt797
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  • This is a good question to ask. My thought is that the answer is probably "no." Surjectivity of the exponential function is an "analytic" statement, and I feel like algebra does not really offer you this sort of information. Maybe someone will prove me wrong though :-). – Alekos Robotis Oct 31 '20 at 05:20
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    @jskattt797 In a context which explicitly involves positive real numbers -- a typical notation for the set of which being $\mathbb{R}{+}\colon=[0, \infty)$ -- the syntax $\mathbb{R}^+$ is not too inspired. Also, your objective is not too clearly formulated. Are you assuming the existence of an injective morphism of totally ordered groups from $(\mathbb{R}, +, \mathrm{O})$ to $\left(\mathbb{R}{+}^{\times}, \cdot, \mathrm{O}{|\mathbb{R}{+}^{\times}}\right)$?, where by $\mathrm{O}$ I am referring to the standard order relation on $\mathbb{R}$ – ΑΘΩ Oct 31 '20 at 05:21
  • @ΑΘΩ I agree, do you have a suggestion for better notation besides $(\mathbb{R}, +)$? I'm assuming that $e^{x + y} = e^x e^y$, $e^x > 0$ for all $x \in \mathbb{R}$, and $x < y \implies e^x < e^y$ when viewing $x \mapsto e^x$ as a map from $\mathbb{R} \to \mathbb{R}$. Then define $\mathbb{R}^+$ and $\mathbb{R}^$ and consider the map $\phi : \mathbb{R}^+ \to \mathbb{R}^$ induced from $\exp$. Since $\exp$ is injective, $\phi$ is also injective. Now I want to prove that $\phi$ is surjective. – jskattt797 Oct 31 '20 at 05:34
  • @jskattt797 As far as referring to the additive group structure $(\mathbb{R}, +)$ by means of solely the support set $\mathbb{R}$ -- according to an abuse of notation and language that is ubiquitous throughout algebra (and not only) -- I would shy away from introducing any dubious abbreviations and would make sure to either 1) explicitly refer to the additive group $\mathbb{R}$ by employing such a syntagm of natural language or 2) to explicitly refer to the entire additive structure $(\mathbb{R}, +)$ in more formal settings. – ΑΘΩ Oct 31 '20 at 05:57
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    @jskattt797 As for your original question -- which is profoundly nontrivial -- one can prove the following result: if $f \colon \mathbb{R} \to \mathbb{R}_+^{\times}$ is a group morphism increasing on an open interval $(a, b)$ with $a<b$, then $f$ is either trivial or an exponential of strictly superunitary basis. The proof will necessarily involve notions of continuity and ultimately the surjectivity of $f$ will follow from the fact that exponentials are surjective, fact at the heart of which lie topological properties, the circumventing of which is not possible and should not be contemplated. – ΑΘΩ Oct 31 '20 at 06:28
  • I doubt that there is a proof using only group theory. Surjectivity must depend on the topological structure of $\mathbb R$ since $\exp : \mathbb Q \to (0,\infty)$ is certainly not surjective. – Paul Frost Nov 01 '20 at 23:51

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