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It is well known that in many cases, the forgetful functor has a left adjoint functor. For example, the free group functor, abelianization functor, universal enveloping algebra functor and so on.

I want to know some examples that the forgetful functor has a right adjunction.

One example I know is that the forgetful functor $U:\mathbf{Top} \rightarrow \mathbf{Sets}$ has a right adjoint functor which equips any set the indiscrete topology.

Another example I know is that the forgetful functor $U:\mathbf{C-bicomod} \rightarrow \mathbf{Vect}{_k}$ has free bicomodule as right adjunction. Here $C$ is a coalgebra over $k$, and $\mathbf{C-bicomod} $ denotes the category of $C$-bicomodules.

Can some one give me other examples? Thanks a lot.

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2 Answers 2

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Let $G$ be a group and $G$-Set the category of left $G$-sets, i.e., sets equipped $X$ with a left action of $G$ on $X$. The forgetful functor from $G$-Set to Set has a right adjoint, which sends each set $S$ to the set of all functions $f:G\to S$, equipped with the $G$-action that sends $(g,f)$ (where $g\in G$ and $f:G\to S$) to the function $(gf):G\to S$ sending any $x\in G$ to $f(xg)$.

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  • $\begingroup$ Thanks for you answer. Can you explain a little bit more why this is true. If I have a set morphism from a $G$-set to a set $S$, how can I get a $G$-morphism from this $G$-set to the set of functions $f: G \rightarrow S$? $\endgroup$ Commented Sep 11, 2016 at 1:30
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    $\begingroup$ Suppose $(X,*)$ is a $G$-set (where $X$ is a set and $*$ is an action of $G$ on $X$) and you have a function $f:X\to S$. Then there is an induced morphism (i.e., $G$-equivariant function) $\hat f$ from $(X,*)$ to the set of functions $G\to S$ that sends each $x\in X$ to the function sending any $g\in G$ to $f(gx)$. One needs to check that this respects the $G$-actions and that every $G$-equivariant map from $(X,*)$ to my alleged right adjoint at $S$ is of the form $\hat f$ for a unique $f$. $\endgroup$ Commented Sep 11, 2016 at 1:41
  • $\begingroup$ Do you have a reference for this that I could use? $\endgroup$ Commented Nov 11, 2018 at 22:46
  • $\begingroup$ @MorganRogers This is a special case of the general notion of (right) Kan extension, and if I remember correctly there's a treatment of these in Mac Lane's book "Categories for the Working Mathematician". (To apply the notion of Kan extension to the present situation, think of a group $G$ as a one-object category, with the elements of $G$ as its morphisms, and note that $G$-sets are just set-valued functors on this category.) $\endgroup$ Commented Nov 11, 2018 at 22:54
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Fix a ring $R$. There is a forgetful functor from the category of (right) $R$-modules to abelian group, called the restriction of scalars, by forgetting the multiplicative structure $$U : \, \mathbf{Mod}_R \rightarrow \mathbf{Ab} \, .$$ The way you should think about this is that there is a unique ring-homomorphism $f: \mathbb{Z} \rightarrow R$. This turns an $R$-module $M$ into a $\mathbb{Z}$-module, i.e. an abelian group.

On the other side, for all abelian groups $A$, the set of all group homomorphisms, $\operatorname{Hom}_{\mathbb{Z}}(R, A)$ has the structure of a (right) $R$-module via $(\varphi\cdot r)(x) = \varphi(rx)$. This defines a functor, the co-induction functor (co-induced by $\mathbb{Z}$ to be precise), $$\operatorname{Hom}_{\mathbb{Z}}(R, -) : \mathbf{Ab} \rightarrow \, \mathbf{Mod}_R \, .$$

It turns out that co-inducing is right-adjoint to restricting scalars and we get $U \dashv \operatorname{Hom}_{\mathbb{Z}}(R, -)$.

See Change of Rings for more details.

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