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This is a question about the category Grp (groups).

The book "Chapter 0" by Aluffi says that the objects of the category are groups, and the morphisms homomorphisms. He then says that we need not mention that the objects (groups) contain an identity and a unique inverse for every element because all these properties are already contained within the definition of the morphism given: in that $f(1_G)=1_H$ and $f(a^{-1})=(f(a))^{-1}$.

This confuses me. Even if homomorphisms clearly have these properties, do we still not need to mention that a group HAS to contain an identity and inverses for each element? What the above argument suggests is that IF a group has an identity, then it maps to the identity of another group, and so on.

Thank you.

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  • $\begingroup$ What page is this on? $\endgroup$ Commented Jun 10, 2014 at 4:39
  • $\begingroup$ @BrianFitzpatrick- Pg. 60 $\endgroup$
    – user1992
    Commented Jun 10, 2014 at 4:46
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    $\begingroup$ I went down to the library and took a look at the book. Aluffi is not saying that you can remove mention of identities and inverses from the definition of group; he is merely saying that you don't need to mention identities and inverses in the definition of group homomorphism. This is true because any function between groups that preserves multiplication will automatically preserve the identity and inverses. $\endgroup$ Commented Jun 10, 2014 at 15:22

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We can take either point of view: Either groups have identities and inverses that are preserved by homomorphisms, or homomorphisms take identities to identities and inverses to inverses (so there must be identities and inverses in every image).

This perhaps is better motivated by the observation that category theory doesn't actually need the objects. A "just-arrow" category is equivalent. If you go the just-arrow route, then the conserved properties of the morphisms have to be baked into the arrows.

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  • $\begingroup$ Is it possible that no group contains identity or inverses? Or some groups contain identities and inverses whilst others don't? I think both are possible using this definition. $\endgroup$
    – user1992
    Commented Jun 10, 2014 at 4:52
  • $\begingroup$ First, in Grp if any group contains an identity and inverses, and morphisms are enforcing identities and inverses, then every subgroup and every supergroup of that group has identities and inverses. In particular, the trivial group has identities and inverses and then, therefore so does every one of its supergroups (i.e., all groups). In Grp, if morphisms are enforcing identities and inverses and no group has an identity or an inverse, there are no arrows. Since every object has an identity arrow, there are identities and inverses. $\endgroup$ Commented Jun 10, 2014 at 5:02
  • $\begingroup$ Kindly take a look at this. The category theory definition of Grp seems to be equally valid for semigroups as it is for groups. $\endgroup$
    – user1992
    Commented Jun 10, 2014 at 5:18
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    $\begingroup$ No. The same recipe for going from the collection of groups to Grp can be used to go from the collection of semigroups to SemiGrp, but you do not obtain the same object. In particular, even on the same collection, there are vastly more arrows in SemiGrp since those arrows need not respect identities or inverses. $\endgroup$ Commented Jun 10, 2014 at 5:26

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