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One sometimes hears about "the forgetful functor $Ab \to Grp$." Given that the image of an object under this functor is still abelian, in what sense is this "forgetful"?

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The fact that it is abelian has been forgotten. –  Ragib Zaman Feb 17 '13 at 1:35
Is there a rigorous definition of ‘forgetful functor’? –  Haskell Curry Feb 17 '13 at 3:14
Ive heard there isnt. –  goblin Feb 17 '13 at 3:19
If $\tau \to \tau'$ is a morphism of algebraic theories, it induces a functor $\mathsf{Mod}(\tau') \to \mathsf{Mod}(\tau)$. Functors of this type are called forgetful functors. This covers all forgetful functors of "algebraic" structure. –  Martin Brandenburg Feb 17 '13 at 4:04
By morphism of algebraic theories, do you mean an interpretation? –  goblin Feb 17 '13 at 4:17

3 Answers 3

up vote 6 down vote accepted

The wikipedia article http://en.wikipedia.org/wiki/Forgetful_functor and the nlab article http://ncatlab.org/nlab/show/forgetful+functor answer your question. In category theory a mathematical object is not regarded as a set with extra structure, but rather as an object of a fixed category. Even if an object stays the same thing set-theoretically after applying a forgetful functor, it is a different object, because the category has changed. For example, an object of the category of groups which happens to be an abelian group is not really an object of the category of abelian groups (perhaps this is true set-theoretically, but this doesn't matter at all); rather it lies in the image of the forgetful functor from abelian groups to groups, and its preimage should not be confused with itsself! This seems to be a bit pedantic, especially when not structure, but only property is forgotten, but there are lots of benefits from this point of view. For example, sometimes I come across papers which write something like $\mathbb{Z} + \mathbb{Z}$. Well, does $+$ refer to the coproduct in the category of groups, or the category of abelian groups? I am always confused. Many students interpret the construction of the complex numbers as $\mathbb{C}=\mathbb{R}^2$ - which is inprecise because actually $U(\mathbb{C})=\mathbb{R}^2$, where $U$ is the forgetful functor from, say, $\mathbb{R}$-algebras to $\mathbb{R}$-modules. There are many more reasons why $\mathbb{C}=\mathbb{R}^2$ is misleading, see math.SE/5108. If only category theory would be more appreciated ...

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A forgetful functor can forget structure but it can also forget properties. So, for instance, the forgetful functor $Ab \to Set$ forgets structure (the binary operation defining the group) and the forgetful functor $Top\to Set$ also forgets structure (the topology on the set). Forgetful functors that forget properties include, besides $Ab \to Grp$, such forgetful functors as $Fld \to Ring$, $ComRing \to Ring$, $ComMon \to Mon$ and so on.

It is important to realize that it is not what makes up a given object that matters but rather how that object relates to the rest of the world. Thus, changing what the rest of the world is, or changing the morphisms (i.e., how the object relates to the rest of the world) makes a big difference.

As an extreme example, consider the category $CLat$ of completes lattices and $CSLat$ of complete semilattices. Now, every complete semilattice is automatically a complete lattice. However, these categories are very different because of how the morphisms are defined. In $CLat$ the morphisms are required to preserve both arbitrary meets and arbitrary joins. In $CSLat$ they are only required to preserve arbitrary meets. So, there is a functor $CLat \to CSLat$ which is the identity on both objects and morphisms. However, this too is a forgetful functor. What is forgotten this time is not structure nor property but rather some of what morphisms should preserve.

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every complete semilattice is automatically a complete lattice - other way around? –  anon Feb 17 '13 at 15:50
@anon if $L$ has all meets than the join of any subset can be computed is the meet of all upper bounds of the set. –  Ittay Weiss Feb 17 '13 at 19:53
Huh. Was not aware of that. (I always thought $\bf N$ (without $0$) under divisibility was complete under meets but not joins, but apparently the meet of an empty set is defined to be the unique maximal element. Now the equivalence of complete semilattices and complete lattices is clear to me.) –  anon Feb 17 '13 at 20:22

Given that "forgetful functor" does not have an agreed upon definition, this question is somewhat soft, and perhaps should be tagged as such. So all this is a matter of individual taste.

As we can read in Wikipedia generally, but not always, "forgetful" functors are faithful and have left adjoints.

However, the functor $Ab \to Grp$ should be called "the inclusion functor", because it is not only faithful - like most forgetful functors - but also injective on objects. And this is a distinguished characteristic of "inclusion functors". You can read this here. Because of this, we say that $Ab$ is a subcategory of $Grp$. Since this particular functor is also full, we say that $Ab$ is a full subcategory of $Grp$.

Also the other functors mentioned by @Ittay: $Fld \to Ring$, $ComRing \to Ring$, $ComMon \to Mon$ are inclusion functors.

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I believe the functor itself , rather than its source, is the "subcategory". True? –  goblin Feb 17 '13 at 22:37
If you incarnate categories as sets (or classes) of objects and morphisms, then the naïve notion of subcategory works well enough. –  Zhen Lin Feb 17 '13 at 23:28

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