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Consider subcategory of $Ab$ of category abelian groups consisting divisible abelian groups (such groups $X$ that for any $x \in X$ and $n \in \mathbb N$ there is $y \in X$ such that $ x = ny$).

I need to show that morphism $\varphi: \mathbb R \to \mathbb T$, $t \mapsto e^{it} $ is monomorhism. It is probably easy so I'm asking for just hint.

P.s. I don't get why $\mathbb T$ is said to be divisible? It is so if we define action of $n$ as $n\cdot e^{it}: = e^{i(t+n)}$, though.

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    What are you calling $\Bbb T$ here? That usually denotes a torus – 2017-01-17
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    @AlexMathers $\mathbb{T}$ is the "1-torus", i.e. $S^1 = \{ z \in \mathbb{C} \mid |z| = 1 \}$. – 2017-01-17

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Concerning the divisibility of ${\mathbb T}$: The action of ${\mathbb Z}$ on ${\mathbb T}$ is uniquely determined by the abelian group structure on ${\mathbb T}$ (it is for any abelian group) by $n.\exp(it)=\exp(it)^n=\exp(itn)$. In particular, ${\mathbb T}$ being divisible means that arbitrary roots exist in ${\mathbb T}$. Can you check that for yourself? Alternatively, note that it suffices to check divisibility for ${\mathbb R}$, because images of divisible groups are again divisible, and $\exp: {\mathbb R}\to{\mathbb T}$ is surjective.

Concerning the actual question: Consider first a morphism $\psi: X\to {\mathbb R}$ of a divisible abelian group $X$ into ${\mathbb R}$ such that $\exp\circ\psi$ is the trivial homomorphism. Then, looking at it as a morphism in the category of all abelian groups, it factors through the kernel of $\exp$. How does that kernel look like? Is it divisible?

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    What do you mean by "trivial homomorphism"? Homomorphism to identity? Then what is the question concerning kernel about? It would be all $X$. – 2017-01-17
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    By the trivial homomorphism I mean the one sending every element to the zero element. When talking about the kernel I meant the kernel of $\exp$, not the one of $\exp\circ\psi$. – 2017-01-17
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    The kernel of $exp$ is $e^{2\pi n}$ – 2017-01-17
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    @Vladislav Where are you getting stuck? – 2017-01-18
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    I've already solved it, thank you. – 2017-01-18
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    of course not, the problem is solved, there are no more questions related. – 2017-01-18
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    @ Hanno I meant, of course the question can be closed. – 2017-01-18
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    @ Hanno sorry, I didn't know that. I've accepted it now. – 2017-01-18
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    @Vladislav Thanks - we can clear up the comment stack then, ok? – 2017-01-18