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Stream: learning: questions

Topic: On the Kan extension of an object along a group action

Marco Vianello (Jun 01 2025 at 21:51):

Let $G$ be a group acting on an object $X$ of some category $\mathcal C$. For a $c\in \mathcal C$, consider the coproduct $\coprod_G c$ and the natural action of $G$ on it. E. Riehl in Category Theory in Context states that plain morphisms $c\to X$ correspond bijectively to $G$-equivariant maps $\coprod_G c\to X$, so that $\coprod_G c$ is always the left Kan extension of $c$ along the $G$-object $X$ (more formally, the functor $\mathcal BG\xrightarrow{\coprod_G c} \mathcal C$ corresponding to the tautological action of $G$ on $\coprod_G c$ is the left Kan extension of $1\xrightarrow{c} \mathcal C$ along $\mathcal BG\xrightarrow{X} \mathcal C$). I can supply a proof for this silly fact, but I can't make sense of it on a more intuitive ground. Any hint?

Bonus question. We get the bijection

$${\mathcal C}^{\mathcal BG}\bigl(\coprod_G c, X\bigr)\cong \mathcal C(c,X)$$

by restricting a $G$-equivariant map $\coprod_G c \to X$ to the copy of $A$ indexed by the identity element of $G$. It's difficult to overlook the resemblance to something familiar here. Is this fact related to the Yoneda lemma in any way or another?

fosco (Jun 01 2025 at 22:01):

You're generalizing to another category a similar statement for $\bf Set$: the functor $\coprod_G$ in that case sends a set C to $\coprod_{g\in G} C= G\times C$, and you're saying that a function $C\to UX$, where U forgets that X is a G-set, correspond to G-set homomorphisms --so, $G\times C$ is the free G-set on C.

sending C to G x C is a monad, any time $G$ is a monoid, and you're just re-discovering the free-forgetful adjunction.

Sorry I'm being terse, I'm going to bed.

Morgan Rogers (he/him) (Jun 02 2025 at 07:45):

In particular, the "something familiar" is an instance of a free-forgetful adjunction 👍🏻
There is also a right adjoint to the forgetful functor. Of course, these adjoints only exist if $\mathcal{C}$ has the requisite (co)limits

fosco (Jun 02 2025 at 09:30):

Careful, however, that there is a typo in your statement of the Kan extension; in order to Kan extend foo along bar, foo and bar must form a span $X \xleftarrow{foo} Y \xrightarrow{bar} Z$, and $Z \to X$ is the type of the extension $\text{Lan}_{bar}foo$.

fosco (Jun 02 2025 at 09:33):

Instead, the universal property of a (Set-)copower $A\otimes C$ of a set A and an object C is this: there is an everywhere-natural bijection

${\cal C}(A\otimes C,X)\cong {\bf Set}(A,{\cal C}(C,X))$

in your case $A = |G|$ the underlying set of a group $G$. I'll leave you to prove the bijection! How does it relate to the free-forgetul adjunction above?