Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Self-explanatory. Just looking for the general tendency for those "algebraic objects," (sorry can't come up with a better term) because no one seems to talk about any such adjoints. Also could use a helpful intuitive description of the "cofree" objects, the way it's clear what indiscrete topology would "look" like in e.g. a digraph.

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

No; the forgetful functors in the categories that I think you're talking about don't preserve coproducts. (In fact they don't even preserve the initial object.)

Just so we're all on the same page, let's write down what it would mean for "cofree" objects to exist in some category $C$ of algebraic objects equipped with a forgetful functor $U : C \to \text{Set}$. This would mean that we had a functor $F : \text{Set} \to C$ together with a natural bijection

$$\text{Hom}_{\text{Set}}(U(c), d) \cong \text{Hom}_C(c, F(d)).$$

On the LHS, we have the set of all maps between the underlying set of some $c \in C$ and some arbitrary set $d \in \text{Set}$. On the RHS, we have the set of all homomorphisms from $c$ to some $F(d) \in C$. Why is this an unreasonable thing to want? Well, if the natural bijection worked in any reasonable way, the image of some element of $U(c)$ in $d$ ought to have something to do with the image of the same element of $c$ in $F(d)$. But on the LHS the images of different elements of $c$ are completely unconstrained, whereas on the RHS they are constrained by the relations that exist in $c$, independent of what $F(d)$ is.

share|improve this answer
    
What an excellent answer! Thank you. –  Adam Ray Sep 29 '12 at 21:58
1  
@AdamRay If you find this answer acceptable, you should click the check mark next to the voting arrows so that other people don't come by to suggest more answers. –  Kevin Carlson Sep 29 '12 at 23:06
1  
@KevinCarlson There is nothing wrong if other people come with more answers, actually someone might come with an answer which sheds a different light to a question. It is indeed a good practice to wait a couple of days before accepting any answer, to give a fair chance to people coming from different time zones to participate in the discussion. –  magma Sep 30 '12 at 0:44
    
Ah, good points, I'll avoid suggesting people accept things so fast in future. –  Kevin Carlson Sep 30 '12 at 0:47
add comment

A right adjoint $\mathcal G$ to the forgetful functor $\mathcal F$ would mean that we have for each (set) map $f\colon \mathcal F(G)\to S$ a suitable group homomorphism $\mathcal G(f)\colon G\to \mathcal G(S)$. Note that $|G|=n$, $|S|=m$ implies that there are exactly $m^n$ maps $f$, hence we must find a group $\mathcal G(S)$ allowing exactly $m^n$ group homomorphisms from $G$. If $n=3$ and $m=2$, this implies that there are exactly 8 such homomorphisms, an even number. However, the number of homomorphisms from $G=\mathbb Z/3\mathbb Z$ to any group is always odd (or infinite): There is the trivial homomorphism and all others can be grouped in pairs via the nontrivial automorphism of $G$, that is we pair $\phi$ with $x\mapsto\phi(-x)$. Therefore, no such right adjoint exists.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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