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I'm new in Category Theory and I just learned about epis and monos. I've been doing some exercises so far, but one of them is confusing me:

Let $i:\mathbb Z \to \mathbb Q$ be the inclusion of integers to rationals (equipped with the usual ring structure):

  1. Show that $i$ an epimorphism
  2. Show that it is not surjective

I tried to assume $g,h: \mathbb{Q} \to R$ and that $g \circ i = h \circ i$. But now I have no idea how to get $g=h$. It is clear for me that it has something to do with the fact $i$ is an inclusion, but I just can't construct any argument. I attempted proof by contrapositive as well, assuming $g\neq h$, but the difficulty persists.

Thanks in advance!

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    What $g\circ i=h\circ i$ means is that $g(n)=h(n)$ for every integer $n$. You need to prove that such a $g$ and $h$ must also agree on all other values -- or in other words that a ring homomorphism $\mathbb Q\to R$ is completely determined by its values on the integers. – hmakholm left over Monica Oct 20 '15 at 09:05
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    A more instructive (and only slightly more involved) approach would be to prove that for any ring $R$ there is at most one ring homomorphism (with unit) $\mathbb Q\to R$. This means that every morphism with $\mathbb Q$ as the codomain is automatically epi. – hmakholm left over Monica Oct 20 '15 at 09:14

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