3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $f(xg)$.

$\endgroup$
  • $\begingroup$ 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 $f: G \rightarrow S$? $\endgroup$ – Yining Zhang Sep 11 '16 at 1:30
  • 1
    $\begingroup$ 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 $f$. $\endgroup$ – Andreas Blass Sep 11 '16 at 1:41
  • $\begingroup$ Do you have a reference for this that I could use? $\endgroup$ – Morgan Rogers Nov 11 '18 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.) $\endgroup$ – Andreas Blass Nov 11 '18 at 22:54

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.