Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Does the category Monoid of monoids have finite coproducts?

share|improve this question
    
I rolled this question back to its previous version. Someone edited it to simply read aaaaaaaaaaaaaaaaaaaaaaaaaaa. (Which is presumably why it attracted votes to close.) –  David Speyer Mar 13 '10 at 19:04
    
@David: I haven't (yet) voted to close but the question immediately reminded me of mathoverflow.net/questions/18026/… –  Yemon Choi Mar 13 '10 at 19:55
    
I should add that the present question is a bit more interesting than the one I linked to (IMHO). –  Yemon Choi Mar 13 '10 at 20:00
    
There's something a bit strange going on here. After I answered, someone else posted a one-"sentence" answer that sounded like a drunk person yelling nonsense. It must have been multiply flagged then deleted by the moderators. –  Tom Leinster Mar 13 '10 at 20:04
    
Yes, the drunken shouting was deleted. –  David Speyer Mar 13 '10 at 23:28
add comment

1 Answer

Yes. More generally, any category of algebras (in the sense of universal algebra), such as groups, rings, vector space, Lie algebras, ..., has all (small) limits and colimits. See for instance p.210 (end of section IX.1) of Mac Lane's book Categories for the Working Mathematician.

Explicitly, the initial monoid ("0-fold coproduct of monoids") is the one-element monoid. The coproduct $A * B$ of two monoids $A, B$ is constructed similarly to the coproduct of two groups (often called their "free product"). That is, it's the free monoid generated by all the elements of $A$ together with all the elements of $B$, quotiented out by all the relations that hold in $A$ and all the relations that hold in $B$.

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.