1
$\begingroup$

I am working on free objects, I am restricting myself primarely to groups, rings and modules (with maybe algebras) so in a sense in the concrete category (if I am not mistaken. This is a thesis work I am doing and as such references is of outmost importans, and I found this note document which gave an interesting definition of free objects which I'll cite

A free object is the mother of all objects of that type, in the sense that every object of that type is a quotient of the free object

This is an extremely good definition for what I am after, especially when you comapare it with this definition which I have seen many times before. I don't include the diagram or that one becuase I do not know how to draw a diagram on here. However I curious, does anyone know a good source for the first definition? Or how to prove it from the more common definition?

$\endgroup$
7
$\begingroup$

This is not a definition, it is a property of free objects. Every object surjecting onto a free object satisfies this too, and these do not have to be free. (Consider the group $S_n \times \mathbb{Z}$ for instance.) I advise you to read any text / book on universal algebra. There, you can find concise definitions and constructions of free objects.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ They state it as a definition. Would you know any recommendation of books where I could find it? $\endgroup$ – Zelos Malum Feb 14 '16 at 11:22
  • $\begingroup$ Well done on giving constructive comments to question :) Well done! $\endgroup$ – Zelos Malum Feb 14 '16 at 11:25
  • 2
    $\begingroup$ No, they don't mean this to be a definition. They write: "We begin with a somewhat cryptic definition, whose meaning will soon become clear." -- in other words, you will have to read further. It is a preliminary "definition". $\endgroup$ – Martin Brandenburg Feb 14 '16 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.