2
$\begingroup$

There free functor, L, from Set to Monoid gives the free Monoid on the set. The forgetful functor,R, from Monoid to Set gives the underlying set of the Monoid. These are adjoint. I think the proof amounts to conceiving of the isomorphism

$Hom_{Mon}(L(X), \mathcal{M}) \rightarrow Hom_{Set}(X,R(\mathcal{M}))$.

How do we do this? How do we, in general, prove that a free and forgetful functor are adjoint?

$\endgroup$
5
$\begingroup$

In general, it should be evident from your particular construction of the free object $L (X)$ that any morphism $L (X) \to M$ is determined precisely by the images of the generators of $L (X)$, i.e. by the underlying map $X \to R (X)$. This is the bijection of Hom-sets you are looking for $$\operatorname{Hom} (L (X), M) \xrightarrow{\cong} \operatorname{Hom} (X, R (M)).$$ You should also check that the bijection of Hom-sets is natural in $X$ and $M$, but this also should be evident.


Sometimes this appears as the "universal property of free monoids / groups / modules / etc." One requires for any set $X$ existence of an object $L_X$ together with a universal map of sets $i_X\colon X \to R (L_X)$ ("the inclusion of generators"), such that for any map $f\colon X\to R (M)$ there exists a unique morphism $\widetilde{f}\colon L_X \to M$ such that $f = R (\widetilde{f})\circ i_X$.

The universal property of $i_X$ implies that $L\colon X \rightsquigarrow L_X$ is a functor, it is left adjoint to $R$, and $i_X$ is the adjunction unit.

This is one of the many equivalent ways to specify an adjunction (see e.g. Borceux, Handbook of Categorical Algebra, Volume 1, §3.1):

Let $R\colon \mathcal{D} \to \mathcal{C}$ be a functor. Suppose that for any $X \in \mathcal{C}$ there exists an object $L_X\in \mathcal{D}$ and an arrow $\eta_X\colon X\to R (L_X)$, which is universal in the following sense: for any morphism $f\colon X\to R (M)$ in $\mathcal{C}$ there exists a unique morphism $\widetilde{f}\colon L_X \to M$ in $\mathcal{D}$ such that $f = R (\widetilde{f})\circ \eta_X$.

Then $R$ is right adjoint, its left adjoint is given by $L\colon X \rightsquigarrow L_X$, and $\eta_X$ is the adjunction unit.

enter image description here

$\endgroup$

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.