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I am trying to think of a category in which not all free objects exists. I thought this might be the case in sets (I thought I might be able to violate the uniqueness ) but I couldn't get anywhere so I was looking for some examples of categories that don't have all free objects?

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  • $\begingroup$ If I understand you well then you are looking for a forgetful functor that has no left-adjoint. Have a look at mathoverflow.net/a/6383/40263 $\endgroup$ – drhab Apr 28 '14 at 10:54
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    $\begingroup$ What do you mean "free objects"? Are you considering a category with some fixed "forgetful functor", and you're looking for a left adjoint? In this case you need to define what a forgetful functor is, because of course not all functors have a left adjoint. $\endgroup$ – Najib Idrissi Apr 28 '14 at 10:55
  • $\begingroup$ @NajibIdrissi I am using this definition en.wikipedia.org/wiki/Free_object Is this not standard? $\endgroup$ – john smith Apr 28 '14 at 11:02
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    $\begingroup$ That is standard, but you have not provided all the data! $\endgroup$ – Zhen Lin Apr 28 '14 at 12:13
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    $\begingroup$ Categories are not the things that have free objects; the things that have free objects are concrete categories, namely categories $C$ equipped with (faithful) functors $F : C \to \text{Set}$. $\endgroup$ – Qiaochu Yuan Apr 28 '14 at 12:55
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The concrete category $(\mathsf{Set},\operatorname{id}_{\mathsf{Set}} : \mathsf{Set} \to \mathsf{Set})$ has all free objects : for any set $S$, the set $S$ itself (with the canonical map $\operatorname{id}_S$) is free on $S$. So your attemp will not get you anywhere.

There is however a very easy (but dumb) example of a concrete category suiting your requirement : take the punctual category $\mathsf e$ (one object, one morphism) and make it concrete with $F \colon \mathsf e \to \mathsf{Set}$ sending the unique object of $\mathsf e$ to $\emptyset$. Then the only set $X$ admitting a free object is $\emptyset$ : indeed,if $X \neq \emptyset$ would admit a free object, it would be the only one of $\mathsf e$ and there should exists some map $i \colon X \to \emptyset$, which is of course absurd.

There must be more natural and less ad hoc examples in the nature, but I don't have the time to think further for now.

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An example that shows up "in nature" is the (concrete) category of fields, equipped with the usual forgetful functor to sets. There isn't a free field on any number of objects, including zero.

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    $\begingroup$ Another example where the failure has a somewhat different flavor is the (concrete) category of finite groups equipped with the usual forgetful functor, which doesn't have a free object on any number of objects except zero. In general there's no reason to expect free objects to exist if the category, say, fails to have coproducts. You can also do silly things like starting with a category that has free objects and then just removing the free objects (this should work most of the time anyway). $\endgroup$ – Qiaochu Yuan Apr 28 '14 at 13:03

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