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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeSep 9th 2012
• (edited Sep 9th 2012)

I added to the “abstract nonsense” section in free monoid a helpful general observation on how to construct free monoids. “Adjoint functor theorem” is overkill for free monoids over $Set$.

• CommentRowNumber2.
• CommentAuthorRodMcGuire
• CommentTimeSep 9th 2012

ummm, shouldn’t

Then a left adjoint to the forgetful functor $Mon(C) \to C$ exists, taking an object $c$ to

$\sum_{n \geq 0} c^{\otimes n},$

which thereby becomes the free monoid on $C$.

really be

Then a left adjoint to the forgetful functor $Mon(C) \to C$ exists, taking $C$ to

$\sum_{n \geq 0} C^{\otimes n},$

which thereby becomes the free monoid on $C$.

an object $c$ of $C$ is not involved.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeSep 9th 2012
• (edited Sep 9th 2012)

ummm… no. There was one typo in what I wrote: that should have been a lower-case $c$ before the period. I’ll go fix that. (Edit: done.)

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeSep 9th 2012

The adjoint functor theorem is useful to do the proofs. (None of the constructions currently come with proofs that they are what we claim they are.)

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeSep 10th 2012

Should I include a proof of the theorem I quoted? (Hm, not sure I really want to put myself out there, but I’ll ask anyway.)

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeSep 10th 2012

I don’t think that it’s necessary now. It might be better to leave the proofs for somebody who doesn’t find the result obvious and wants to write down what they think of. (That’s usually what I do … not that I always find proofs obvious when I leave them out if I’m quoting the results from elsewhere.)