8
$\begingroup$

Let $f:A\to B$ be a monoid homomorphism. Where can I find an explicit description of the its cokernel? Are there any books on this topic?

If anyone cares, here's my motivation. In the category of groups, the cokernel of the kernel of a group homomorphism $f$ is the quotient of the domain by the kernel, which is comprised of the cosets of the kernel. The first isomorphism theorem says this quotient is isomorphic to the image. This makes sense because the multiplicative kernel action has strongly connected components (because of the existence of inverses), so the cosets of the kernel are the fibers.

For monoids there's no first isomorphism theorem because the kernel is largely uninformative. However, some monoid epimorphisms are known to be the cokernels of their kernels (namely Schreier split monoid epimorphisms), and I would like to see what this means concretely.

Added. By the cokernel of $f$, I mean the universal arrow which postcomposes with $f$ to give the trivial homomorphism. Sorry for not including this from the start, I just thought there's no risk of ambiguity.

$\endgroup$
  • 3
    $\begingroup$ I am not sure what you want here. There is a first isomorphism theorem but one uses congruences not kernels. I don't think any reasonable extension of group extension theory exists for monoids. Things like cohomology seem to work for only very special extensions. Leech's theory seems the best of these. $\endgroup$ – Benjamin Steinberg Dec 12 '16 at 0:51
  • 3
    $\begingroup$ @TheoJohnson-Freyd in the category of groups I understand what quotients mean and how they look like - their elements are cosets. But what does quotient mean for monoids? You can use internal equivalence relations as mentioned in a previous comment, but I'm not sure how to find the cokernel that way. $\endgroup$ – Arrow Dec 12 '16 at 1:13
  • 1
    $\begingroup$ What universal property do you want a cokernel to have? $\endgroup$ – Benjamin Steinberg Dec 12 '16 at 1:57
  • 4
    $\begingroup$ A reformulation of the comment by @BenjaminSteinberg about the desired universal property: What is the definition of "cokernel" in the context of monoid homomorphisms? I would guess that it should mean the coequalizer of the given homomorphism and the constant homomorphism that sends everything to the identity. But it would be good if you'd confirm or correct that guess. $\endgroup$ – Andreas Blass Dec 12 '16 at 2:49
  • 2
    $\begingroup$ I really wonder about that confusion here and why Arrow feels "sorry" for not giving the definition. This definition is standard. Cokernels are defined unambigiously in any pointed category (see Mac Lane or nlab or Wikipedia) and monoids form a pointed category. $\endgroup$ – HeinrichD Dec 12 '16 at 11:20
5
$\begingroup$

First of all, the construction is as for all (pointed) algebraic structures. Let $\sim$ be the congruence relation generated by $f(a) \sim 1$ for $a \in A$. Here, congruence relation means an equivalence relation on the underlying set of $B$ satisfying $b \sim b' \Rightarrow x b \sim x b' \wedge b x \sim b' x$ for all $b,b',x \in B$. Then $\mathrm{coker}(f)$ is the quotient monoid $B/{\sim}$, i.e. the set of equivalene classes of $\sim$ equipped with the monoid structure which is uniquely determined by the property that $B \to B/{\sim}$, $b \mapsto [b]$ is a homomorphism.

It remains to give an explicit description of $\sim$. In the commutative case (more generally, when $f:A \to B$ is central), we have the following:

$b \sim b' \Longleftrightarrow \exists a,a' \in A ( f(a) b = f(a') b')$

In the non-commutative case, the description is much more complicated. You basically have to work with chains of relations and longer products as in the general description of generated congruences. More explicitly, $\sim$ is the transitive closure of the following relation:

$b \approx b' \Longleftrightarrow \exists n \in \mathbb{N} \, \exists a_i,a'_i,c_i,c'_i \in A \, \exists b_i,b'_i \in B: \\ b = b_1 f(a_1) \dotsc b_n f(a_n), ~ b' = b'_1 f(a'_1) \dotsc b'_n f(a'_n),\\ b_1 f(c_1) \dotsc b_n f(c_n) = b'_1 f(c'_1) \dotsc b'_n f(c'_n).$

That being said, it should be pretty clear that you are lost when you want (or have) to use elements instead of the universal property.

Let me mention that the Grothendieck group resp. the universal enveloping group (this is what it's sometimes called in the non-commutative case) of a monoid $A$ is just the cokernel of the diagonal $A \to A \times A$.

$\endgroup$
  • 2
    $\begingroup$ Given the OP was unhappy with the first isomorphism theorem via equivalence relations it wasn't clear this is what the op wanted. $\endgroup$ – Benjamin Steinberg Dec 12 '16 at 13:28

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.