1
$\begingroup$

The right adjoint to the forgetful functor $U:\mathbf{Group}\rightarrow\mathbf{Monoid}$ is the functor $I:\mathbf{Monoid}\rightarrow\mathbf{Group}$ that sends a monoid to its group of units, i.e., the invertible elements. Is $I$ full?

By basic results in category theory, this is equivalent to the question: is the counit split monic? (The counit is the inclusion map, so it's monic, even injective.)

Just to spell it out: is there a monoid $M$, with group of units $G\subseteq M$, such that there is no monoid homomorphism $\varphi:M\rightarrow G$ that restricts to the identity on $G$ (in the jargon, no retract of $M$ onto $G$).

$\endgroup$
2
$\begingroup$

The answer to the title question is no and the answer to the last body question is yes.

For example, you can take $M$ to be the multiplicative monoid of a ring with at least two units. A monoid homomorphism $\varphi : M \to M^{\times}$ must send the zero element $0 \in M$ to an idempotent, and since $M^{\times}$ is a group that idempotent is the identity. But since $0$ is absorbing, everything else must also be sent to the identity. So there is only one homomorphism $M \to M^{\times}$ and it doesn't restrict to the identity on $M^{\times}$ unless $M^{\times}$ is trivial.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.