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Was wondering if someone could help. I am just about getting to grips with cateogry theory and adjoints. I have a query though, reading through some material.

Does the forgetful functor from the category Group of groups to Set have a left adjoint?

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Tom, which introductory material on category theory have you been reading? My experience is that almost all introductions to the subject would include this as a basic example of an adjoint functor. But maybe you're reading something more aimed at computer scientists, for instance? Anyway, it sounds like you could do with some extra reading material, and you're in luck: category theory is especially well-represented on the web, and if you google a phrase such as "introduction to category theory" you'll turn up some good stuff. –  Tom Leinster Feb 28 '10 at 1:42
    
Yea Im a computer scientist so need to get to grips with this! –  Tom Bath Feb 28 '10 at 17:06
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1 Answer

Yes, the functor that takes a set to the free group generated by that set.

http://en.wikipedia.org/wiki/Free_group

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Wow quick! Thanks! I need to start proving these types of questions. Do you know where I could get a model proof for this? It would be great to see a perfect answer before I try to do similar on other functors :) –  Tom Bath Feb 27 '10 at 14:32
    
usually you should try wikipedia first and if this doesn't help u can still ask en.wikipedia.org/wiki/… –  HenrikRüping Feb 27 '10 at 14:46
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Most abstract algebra books that cover categories at all prove this. It's actually an example of a more general phenomenon, that's come up here many times: the adjoint to a forgetful functor is a free object on a set. If you're in a theory where the free X is defined, it's probably the adjoint of the forgetful functor in the other direction. –  Charles Siegel Feb 27 '10 at 15:03
    
In general, if you want to understand why things like "free group" give rise to adjoint functors, you may want to look at Theorem IV.2.(ii) of Mac Lane's "Categories for the Working Mathematician." You will see there that if G:A->X is a functor, and for all x in X you have an object F_0(x) of A and a "universal arrow" from x to G(F_0(x)), then you can define a functor F:X->A with object function F_0 which is left adjoint to G. So, if you have something similar to a "free object" for every object of X, you get an adjunction. –  user2734 Feb 27 '10 at 16:37
    
@Tom: The adjoint functor theorem should apply to give free constructions in many such cases (by proving the existence of the adjoint to the forgetful functor, cf. Charles's comment). –  Akhil Mathew Feb 28 '10 at 1:02
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