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I have a couple examples of functors that were given to me in class that are faithful, but not full. However, I'd like an actual proof of these facts in case I have to explain myself on an exam.

Example 1: $F:Rings\rightarrow Abelian$ $Groups$ defined by $F(R)=(R,+)$. If $(A,+)$ is an abelian group, define $xy=0$ for $x,y\in A$. The $(A,+)$ is a ring (Could someone please explain to me why this is true?) and $F(A,+,*)=(A,+)$. I'm told this is essentially surjective, faithful, and not full. I'd like an explanation as to why. The more detail the better in case I have to justify myself on an exam.

Example 2: The forgetful functor $F: Groups\rightarrow Sets$ is faithful, but not full. My notes say that each group maps to a unique set and group homomorphism are subsets of the functors. But I don't really understand what that means. Also, my notes claim that $F$ is not full since there exists functions between groups which are not group homomorphisms, but I'd like more explanation on that fact as well.

Please be as detailed as possible with your explanations. I'm very new at Category Theory and I need things broken down for me.

If you have any other examples of faithful, but not full functors please share!

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    $\begingroup$ Faithful (resp. Full) refers to injectivity (resp. surjectivity) only on the induced map between Hom(X,Y). In particular, a faithful functor need not be injective on the underlying objects. Thus the forgetful functor from Groups to Sets is faithful because two different group homomorphisms are also different as set theoretic functions. This functor is not full because, as your reference says, there are set-theoretic functions between groups which are not homomorphisms (eg constant functions not mapping to the identity element). $\endgroup$ – lulu Dec 13 '15 at 19:29
  • $\begingroup$ That was really helpful! Could you give me an example of a group homomorphism that maps to a set function and a set function that doesn't get mapped to by a homomorphism? Just so I have a better idea of the morphisms we're working with $\endgroup$ – Michael Dec 13 '15 at 19:37
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    $\begingroup$ Any group homomorphism starts out as a set-theoretic function! That is, it is a set-theoretic function that has other nice properties. The forgetful functor literally just forgets the other nice properties. I mentioned in my earlier comment that constant functions are not homomorphisms (unless the constant happens to equal the identity). For another example, take any permutation of G which doesn't take $e$ to itself. $\endgroup$ – lulu Dec 13 '15 at 19:40

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