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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 $\mathbb Q \to (\mathbb Q/\mathbb Z)\oplus(\mathbb Q/\mathbb Z)$.

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2 Answers 2

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.

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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.

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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

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