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 $G$).


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.


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