Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am sorry, maybe i am just confused or have not understood some definition, but i do not understand the following remark in Abstract and Concrete Categories -- The Joy of Cats on page 117:

$\mathbb Z \hookrightarrow \mathbb Q$ is a monomorphism in Sgr (the category of semigroups) that is not regular (not an equalizer of two morphisms).

It seems to me it is an equalizer of the two natural morphisms $\mathbb Q \to \mathbb Q *_{\mathbb Z} \mathbb Q$ (the free product of two copies of the group $\mathbb Q$ with amalgamated copies of $\mathbb Z$).

Where am i mistaken? Thanks.

Update. There are simpler pairs of morphisms for which $\mathbb Z \hookrightarrow \mathbb Q$ would be an equalizer: consider different morphisms $\mathbb Q \to (\mathbb Q/\mathbb Z)\oplus(\mathbb Q/\mathbb Z)$.

share|cite|improve this question

It looks like an equalizer to me. Perhaps the authors meant the inclusion homomorphism of $\mathbb N$ (rather than $\mathbb Z$) into $\mathbb Q$. This is not an equalizer. Indeed, if two homomorphisms from $\mathbb Q$ into any semigroup $S$ agree on $\mathbb N$, then they must also agree on $\mathbb Z$, because inverses in $S$ are unique when they exist.

share|cite|improve this answer
Thanks for the confirmation, maybe i will write to the authors. – Alexey Jul 20 '13 at 21:58
It would be simpler then to give $\mathbb N \hookrightarrow \mathbb Z$ as an example. – Alexey Jul 21 '13 at 10:02
@Alexey Yes; my answer was based not on making the example as simple as possible but on keeping it as close as possible to what you quoted. In other words, assume just a single typo rather than two. – Andreas Blass Jul 21 '13 at 15:31
up vote 1 down vote accepted

It turns out the authors meant $\mathbb Z$ and $\mathbb Q$ as semigroups with respect to the multiplication and not the addition.

share|cite|improve this answer
did they answer to your mail? – magma Jul 24 '13 at 20:07
Yes, from their answers i have understood that it was about multiplication and not addition. Maybe it was mentioned somewhere in the book, but i did not read all. – Alexey Jul 24 '13 at 21:43

Your Answer


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.