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We just started learning about Categories in an Algebra Class. I was reading ahead a little bit, and reached functors. A functor from a category $B \to C$ is a map that acts on objects on $B$ AND on the morphisms between them, or just on the objects? There's also a small exercise, and is to show that there are no functors from the category of group to the category of abelian group, sending each group to its center. I will post the answer down, in order to avoid long questions (they already gave a hint).

he example they gave is $S_{2} \to S_{3} \to S_{2}$. First, if I call $g$ the inclusion of $S_{2}$ into $S_{3}$ and $\pi$ the projection of $S_{3}$ into $S_{2} \cong S_{3}/A_{3}$, we can see that $\pi \circ g$ is the identity on $S_{2}$. Now $Z(S_{3})$ is trivial, so any homomorphism $Z(S_{2})\to Z(S_{3}) \to Z(S_{2})$ are all trivial. Now, if such a functor existed, then $Tg$ is trivial and $T \pi$ then so is $T \pi \circ Tg$, but $T(\pi \circ g)$ must be the identity on $Z(S_{2}) \cong S_{2}$. Does this seem reasonnable ?

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    $\begingroup$ Yes, a functor also has to do things to morphisms; otherwise it would just be a function between sets of objects. $\endgroup$ – Qiaochu Yuan Sep 4 '15 at 20:38
  • $\begingroup$ Your example is fine. Its conclusion follows from the fact that a functor's action on mophisms is a "homomorphism" of categories in that it preserves the composition and carry identities to identites. $\endgroup$ – Matematleta Sep 4 '15 at 21:10

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