3
$\begingroup$

The following is reflective in the category $\mathbf{Sem}$ of semigroups and their homomorphisms: we add a new neutral element to each semigroup, even monoids; then the reflections are identity mappings on the old structures, and homomorphisms in this reflective subcategory will preserve the neutral element.

What if we add neutral elements only to semigroups that are not already monoids? Why won't this be a reflective subcategory in $\mathbf{Sem}$? What is the easiest proof of this?

$\endgroup$
  • $\begingroup$ Please share your thoughts so far :) $\endgroup$ – Shaun Sep 23 '14 at 12:44
  • $\begingroup$ I really cannot imagine.If reflections are still identities on old structures, then something will fail,and other choice should not be possible,but I cannot see this precisely :( $\endgroup$ – user3357120 Sep 23 '14 at 15:21
  • 1
    $\begingroup$ It's not a functor. $\endgroup$ – Martin Brandenburg Sep 23 '14 at 16:19
  • $\begingroup$ Unfortunately,after two days,I still cannot see why...Any further hint?Composition fails--why? $\endgroup$ – user3357120 Sep 26 '14 at 17:51
  • $\begingroup$ I've got the point now.My problem was that I draw only one commutative diagram for the adjunction from the two as given in the book CWM. $\endgroup$ – user3357120 Sep 27 '14 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.