Colimits vs algebra

1. Oct 30, 2005

Hurkyl

Staff Emeritus
Quick version of my question: do all small colimits exist in the category of all groups? Of rings? Of modules? (If not, I expect to be able to take a counterexample and use it to answer by real question)
I'm reading through Jacobson's Basic Algebra II, and have a question about the section on colimits in a category of $\Omega$-algebras.

The theorem Jacobson states is that if you have a diagram in $\mathbf{\Omega}$-Alg where for any two objects A and B in the diagram, the diagram also contains an object C and arrows $A \rightarrow C$ and $B \rightarrow C$, then the colimit exists.

He doesn't say anything about whether a more general colimit can fail to exist, so my question is if all small colimits exist, or if there are examples of diagrams where the colimit fails to exist.

I think I've proven that all colimits exist, but I'm kind of suspicious of my work and if true, curious why Jacobson didn't prove the general case...

Anyways, my proof is as follows:
Suppose we have a diagram indexed by the preorder I. We denote the objects by $A_\alpha$, and the arrows by $\phi_{\alpha \beta}:A_\alpha \rightarrow A_\beta$ for any $\alpha \leq \beta$.

The arrows satisfy $\phi_{\beta \gamma} \circ \phi_{\alpha \beta} = \phi_{\alpha \gamma}$ for any $\alpha \leq \beta \leq \gamma$.

Let X be the disjoint union of the elements of all of the $A_\alpha$, and let F be the free algebra $F(\Omega, X)$.

Let $\Theta$ be the congruence on F generated by the relations $a \equiv \phi_{\alpha, \beta}(a)$ for $a \in A_\alpha$, and by the relations given by the arithmetic of each $A_\alpha$.

I claim that $L = F / \Theta$ is a colimit of the diagram.
Let $i_\alpha : A_\alpha \rightarrow L$ be given by $i_\alpha(a) = [a]_\Theta$ (the equivalence class of a under the congruence $\Theta$).

We have that $i_\beta \circ \phi_{\alpha \beta} = i_\alpha$ because:

$$i_\beta (\phi_{\alpha \beta} (a) ) = [ \phi_{\alpha \beta} (a) ]_\Theta$$
$$i_\alpha (a) = [a]_\Theta$$

and that $a \Theta \phi_{\alpha \beta}(a)$.

Now, let B be any $\Omega$-algebra, and let there be maps $\psi_\alpha:A_\alpha \rightarrow B$ satisfying $\psi_\beta \phi_{\alpha \beta} = \psi_\alpha$.

Then, there exists a unique map $f:F \rightarrow B$ given by $f(a) = \psi_\alpha(a)$ for $a \in A_\alpha$.

All of the relations $a \equiv \phi_{\alpha, \beta}(a)$ for $a \in A_\alpha$ are in the kernel of f, and so are the relations given by the arithmetic of each $A_\alpha$, so f must factor uniquely through $L = F / \Theta$, giving a unique map $p:F / \Theta \rightarrow B$ such that $p([a]_\Theta) = f(a) = \psi_\alpha(a)$ for $a \in A_\alpha$.

We have $p \circ i_\alpha = \psi_\alpha$ because:

$$p( i_\alpha(a)) = p([a]_\Theta) = \psi_\alpha(a)$$

Therefore, L is a colimit of the diagram.

Last edited: Oct 30, 2005
2. Nov 1, 2005

mathwonk

I am an amateur in this stuff compared to you, but I think pretty much any time you have categorical direct sums (i presume this is what you mean by "free"), and quotients, it seems you have direct limits, i.e. colimits, as your construction shows.

i.e. the point is whether the appropriate free objects and their quotients exist, as then your use of them is the standard construction of colimit.

i.e. let Si be any collection of objects and F any direct sum of that family, i.e. an object such that any family of morphisms out of the Si extends uniquely to a morphism out of F.

Then consider some condition on a family morphisms out of the Si that can be stated as saying that some subobject G of F goes to zero. Then taking the quotient of F by G gives an object such that any family of morphisms out of the Si and satisfying the condition, extends to a unique morphism out of F/G.

Last edited: Nov 1, 2005
3. Nov 1, 2005

Hurkyl

Staff Emeritus
A free algebra is a generalization of a free group: for a set X, the free algebra $F(\Omega, X)$ is the unique object in $\mathbf{\Omega}$-alg that contains the elements of X and has the property that any map from X to the algebra A extends uniquely to a map from $F(\Omega, X) \rightarrow A$.

In other words, the free functor $F(\Omega, \_)$ and the forgetful functor U that maps an algebra to its set of elements are adjoint functors, just as they are in Grp. (I still don't understand the ramifications of this fact, but I feel like I'm beginning to learn them for the first time!)

Unfortunately, quotients are a little more subtle in an arbitrary $\mathbf{\Omega}$-alg, and you need to consider congruences on A (subalgebras of AxA that are equivalence relations) to get the quotient algebras of A. I guess the category of monoids is supposed to be a good example of that, though I don't think I've ever even entertained the notion of a quotient monoid before!

But! Fortunately, since I wrote this, I came across two useful pieces of information:

Jacobson does give a proof that arbitrary coproducts exist in $\mathbf{\Omega}$-alg.

The Wikipedia page on limits says what you're saying, but in the most general context: a category has all colimits iff it has all coproducts (i.e. direct sums) and coequalizers! (In Grp, the coequalizer of the inclusion K->G and the zero map K->G is the projection G->G/K. In general, it's the colimit of a pair of arrows between a pair of objects)

I'm somewhat more confident in my proof that coequalizers exist than in the proof that I wrote here that colimits exist! (In fact, if it's at all valid, I think I only proved it for one case of colimits) So, I'm reasonably happy.

(P.S. I'm glad I at least sound like I know what I'm talking about, because I feel like I only kinda sorta have an idea what's going on!)

Last edited: Nov 1, 2005
4. Nov 1, 2005

mathwonk

well i am not up on your question or proof yet but it seems you do not quite want a free algebra, but a coproduct, i.e. direct sum of algebras.

i.e. that a free algebra generated by an arbitrary set exists is fairly clear, since you just take the polynomial algebra generated by those variables in the set, i.e. all polynomials in those elements with coefficients in the ring of coefficients.

i guess free objects are always adjoint to forgetful objects, that is the
meaning of "free".

it is not so obvious to me that an arbitrary (infinite) direct sum of algebras exists although for two algebras one presumably takes the tensor product as algebras, as that has the mapping property one wants.

indeed this is the proof that products exist in the category of affine varieties, since the functor from rings to varieties is contravariant.

5. Nov 1, 2005

Hurkyl

Staff Emeritus
In the case of a coproduct, I would be trying to construct the coproduct by starting with a free algebra over the generators of the individual algebras, and modding out by the relations generated by the relations on the individual algebras.

For a more general colimit, I was also trying to throw into the mix all of the relations implied by the diagram under question.

Incidentally, a $\Omega$-algebra doesn't have to be so nice as to be a module over a ring! All it is is a set equipped with operators. ($\Omega$ is the type of the algebra -- e.g. it could be "one binary operation, one unary operation, and one nullary operation") You can define a variety (in the universal-algebra sense) by also specifying equational relations.

So while you can get thinks like groups and rings out of it, you can also get lattices (boolean or otherwise), semigroups, and even uglier things. (You can't get fields, because the inverse operation is only a partial operation)

The existance of direct sums, etc, seems clear to me too, even for the above general kind of algebra, but I've been wrong on a lot of things that seem clear. (While trying to process the things I got out of the chapter, I think I spent half a day trying to prove something for which the map Z-->Q is an easy counterexample)

6. Nov 2, 2005

matt grime

Your categories are abelian and have small coproducts (which are themselves colimits) so you're all ok. Examples where colimits fail to exist are plentiful and a simple way to make them is to limit to, say, finite groups. Then the colimit may well be an infinite group. I did skim only, and haven't paid attention to the sigma alg part though.

colimits may also fail to exist in categories that are not abelian, even if they have arbitrary coproducts. In many interesting examples kernels and cokernels do not exits (eg derived categories, stable categories, triangulated categories in general). Here is my favourite example again of when colimits fail to exist in a nice situation with a finite diagram.

Consider the diagram

$$\bullet \leftarrow \bullet \rightarrow \bullet$$

then in the category of homotopy classes of topological spaces where morphisms are cont maps modulo homotopy this diagram has no colimit.

Proof. Consider replacing the dots with the disc, circle and disc with the morphisms identifying the edge of the discs with the circle, then the colimit ought to be the sphere. However we ought to be able to change things up to homotopy eg replace discs with points, but then the colimit is just another point and the point and the sphere are not homotopic.

In general the only way to see if colimits exist is to construct, as you did, what the colimit *ought* to b, and it is the disjoint union of all the objects in the colimit modulo the relations imposed by the morphisms, and then to see if that object can be realized in such a way as to be in the category, and moreover that any realizations are isomorphic in the category (ie are independent of the choices made). The finite groups will have cases where the realization is necessarily not in the category of finite groups, and the homotopy category is not independent of the choices made.

Last edited: Nov 2, 2005
7. Nov 2, 2005

mathwonk

matt, i thought all abelian categories were subcategories of abelian groups, so i so, then there is nothing to do right? i.e. anything true there is true in any abelian category, ehh?

8. Nov 2, 2005

Hurkyl

Staff Emeritus
One reference I have says that any Abelian category can be faithfully embedded into the category of abelian groups.

Apparently, you can do even better and say that it's isomorphic to a full subcategory of modules over some ring.

9. Nov 3, 2005

matt grime

Yes, any abelian category is equivalent to a full subcategory of some module category, and that is how one should always treat abelian categories. I forget whose theory that is. oh, now i recall, it's the freyd mitchell embedding theorem.

However, taking small colimits can still take you outside the subcategory since it is not guaranteed to be closed under arbitrary direct sums.