# Adjoints to the forgetful functor $U:C^M\to C$

I'm trying to get my head around adjoints to the forgetful functor $U:\bf{C^M}\to \bf {C}$ where $M$ is a monoid interpreted as a category.

My current line of thinking is that the left adjoint $L:\bf C\to C^M$ is given by $L(A)=M\times A$ where $M$ acts on $M\times A$ by $m(M\times A)=(mM)\times A$, the unit $u:A\to M\times A$ is given by the inclusion $u(A)=(1,A)$ and the counit is $\varepsilon:M\times A\to A$ is given by $\varepsilon((m,A))=mA$

It seems to me that this works out quite well though I havent written the down the bijection. One problem that comes to mind is that $M\times A$ might not be an object in $C$, in which case it's not an object in the functor category, and I'm not quite sure how to address this.

Regarding the right adjoint, I'm thinking $R(A)=Hom(M,A)$ but I have almost no idea of how to proceed here. I know that we can interpet the $Hom(M,A)$ sets as objects in many categories, but it does not seem like a general phenomenon.

Edit1:

Ok, i think i got it for $\bf{Set}$. We let the unit $u_A:A\to Hom(M,A)$ be the map $u_A(a)=f_a:M\to A$ given by $f(M)=a$ and the counit $\varepsilon_B:Hom(M,B)\to B$ be given by $\varepsilon(f)=f(1)$. Chasing the diagram seems to work out fine and all so im guessing its alright, ganna write down the bijection in a min. I'm still a bit unsure as to when $Hom_{\bf C}(M,-)$ can be considered an object in $\bf C$ it works out fine in most, (all?), algebraic categories but beyonde this i have no idea.

Anyways, thanks for the replies, they were very helpful, though i have no idea as to why the forgetful functor should be both a tensor functor and a hom functor, some reference would be nice

• Do you know the answer when $C = \text{Set}$? – Qiaochu Yuan May 7 '12 at 18:29
• @user25470: If $\mathbf{C}$ is a nice category (say, a Grothendieck topos) then you can view $U$ as a hom functor and as a tensor functor. Then apply the tensor–hom adjunction. (You can make sense of $M \times A$ in any cocomplete category: it's the coproduct of $M$-many copies of $A$.) – Zhen Lin May 7 '12 at 18:39

First, let's deal with the case where $M$ is an internal monoid in a monoidal category $\mathcal{C}$; then $\mathcal{C}^M$ should be interpreted as the category of left $M$-modules in $\mathcal{C}$. It is not hard to verify that $$\mathcal{C}(X, U A) \cong \mathcal{C}^M (M \otimes X, A)$$ naturally for all objects $X$ in $\mathcal{C}$ and all objects $A$ in $\mathcal{C}^M$, where $M \otimes X$ is regarded as a left $M$-module in the evident way. Thus, we have a left adjoint to $U$: $$M \otimes - \dashv U : \mathcal{C}^M \to \mathcal{C}$$

Morally, this is an instance of the tensor–hom adjunction. Indeed, if $\mathcal{C}$ is monoidally closed (in the sense that $X \otimes -$ has a right adjoint), then we can construct $\mathcal{C}^M (B, A)$ as an equaliser as below: $$\mathcal{C}^M (B, A) \rightarrow \mathcal{C} (B, A) \rightrightarrows \mathcal{C} (M \otimes B, A)$$ The first morphism $\mathcal{C} (B, A) \to \mathcal{C} (M \otimes B, A)$ is the one that corresponds to the morphism $M \otimes B \otimes \mathcal{C} (B, A) \to A$ which evaluates a morphism $B \to A$ on an element of $B$ and then acts on the result by an element of $M$; the second morphism corresponds to the one that acts on an element of $B$ by an element of $M$ and then evaluates a morphism $B \to A$ on the result. But it is clear that $$\mathcal{C} (I, U A) \cong U A \cong \mathcal{C}^M (M, A)$$ so $U$ is enriched-representable by $M$.

The right adjoint is indeed more troublesome. We want $$\mathcal{C}(U A, X) \cong \mathcal{C}^M(A, R X)$$ so, taking advantage of representability and setting $A = M$, we get $$\mathcal{C}(M, X) \cong \mathcal{C}^M(M, R X) \cong U R X$$ Thus, we must find some way of making $\mathcal{C}(M, X)$ into a left $M$-module. This means we have to find a morphism of type $$M \otimes \mathcal{C}(M, X) \to \mathcal{C}(M, X)$$ and by the tensor–hom adjunction this amounts to finding a morphism of type $$\mathcal{C}(M, X) \to \mathcal{C}(M, \mathcal{C}(M, X))$$ but the codomain is $\mathcal{C}(M \otimes M, X)$ by the tensor–hom adjunction again, and so precomposing with the multiplication morphism $M \otimes M \to M$ gives us what we need – but again, all this assuming $\mathcal{C}$ is monoidal closed. Thus, we have a right adjoint of $U$: $$U \dashv \mathcal{C}(M, -) : \mathcal{C} \to \mathcal{C}^M$$

Again, morally this is a tensor–hom adjunction, but it's not as easy to describe. We have to assume $\mathcal{C}$ is a cocomplete category: then we can define the tensor product of a right $M$-module and a left $M$-module and construct a general tensor–hom adjunction; when this is done, $U \cong M \otimes _M -$, so it's no surprise that its right adjoint is a hom functor.

The case where $M$ is an external monoid is a little more complicated. The easiest way to proceed is to assume $\mathcal{C}$ has all small coproducts, so that we can define $M \odot X$ to be the coproduct of $M$-many copies of $X$. In that case, it will be true that $$\mathcal{C} (M \odot X, -) \cong M \times \mathcal{C} (X, -)$$ much like when $\mathcal{C} = \textbf{Set}$. We make the RHS into a right $M$-set, and then Yoneda lemma then turns $M \odot X$ into a left $M$-object in $\mathcal{C}$, in the sense of a functor $\mathcal{B} M \to \mathcal{C}$, where $\mathcal{B} M$ is the delooping of $M$. It is then clear that any $M$-equivariant homomorphism $M \odot X \to A$ must be determined by one of the components $X \to A$, so again we have a left adjoint: $$M \odot - \dashv U : \mathcal{C}^M \to \mathcal{C}$$

Similarly, if we assume $\mathcal{C}$ has all small products, then we can define $X^M$ to be the product of $M$-many copies of $X$, so that we have $$\mathcal{C}(-, X^M) \cong \mathcal{C}(-, X)^M$$ The RHS is automatically a left $M$-set, so the Yoneda lemma makes $X^M$ into a left $M$-object. One sees that any $M$-equivariant homomorphism $A \to X^M$ is determined by one of the components $A \to X$, so we obtain the desired right adjoint: $$U \dashv (-)^M : \mathcal{C} \to \mathcal{C}^M$$

In the case where $\mathcal{C}$ is a Grothendieck topos, both the internal and external approach give the same answer, so I suspect there is a unified way of looking at both. But this is probably enough for now.

• I feel very guilty now, I appreacihiate you took the time to give such a thurough answer, unfortunatly it's way beyond my current level of understanding. But i promise you will take the time to look over again and again untill i get it, ill just have to read up on monoidal categories and representability first. – user25470 May 8 '12 at 13:15
• It is slightly scary that the best reference available online for $BM$ is that... – Mariano Suárez-Álvarez Jul 12 '12 at 16:48