2
$\begingroup$

Background:

A pointed object $X$ in a category $C$ with terminal object $*$ is a map $*\rightarrow X$. Such objects with basepoint-preserving maps form their own category of pointed objects $C^{*/}$. There is a canonical forgetful functor $U:C^{*/}\rightarrow C$ that forgets the basepoint. Furthermore, this has a left adjoint $(-)_{+}:C\rightarrow C^{*/}$ which sends an object $Y$ to the coproduct $Y\coprod *$ equipped with the canonical basepoint inclusion. The adjunction $(-)_{+}\dashv U$ induces a monad on $C$ (think this is called the ``maybe monad"). The category of algebras over this monad is $C^{*/}$.

Question:

There is also a notion of a "multi pointed object:" an object $X$ equipped with a map from a coproduct of the terminal object with itself a bunch of times. The objects with the obvious maps form a category $C_{multi}$ . Does this category arise as a category of algebras over some sort of "maybe" monad?

Edit

To clarify: the objects are objects in $C_{multi}$ with a fixed (we can even assume finite) number of basepoints. The morphisms are maps that preserve those basepoints.

$\endgroup$
  • $\begingroup$ To ask a clarifying question raised in comments below: what are the morphisms between multi-pointed objects? $\endgroup$ – David Roberts Nov 15 '16 at 6:55
  • $\begingroup$ Also, what are the objects? Is the number of points fixed? $\endgroup$ – HeinrichD Nov 16 '16 at 8:29
  • 1
    $\begingroup$ @DavidRoberts The objects are objects in C with a fixed (we can even assume finite) number of basepoints. The morphisms are maps that preserve those basepoints. $\endgroup$ – user84563 Nov 21 '16 at 4:02
  • $\begingroup$ @HeinrichD Yes, sorry for the confusion. I've edited the question to clarify. $\endgroup$ – user84563 Nov 21 '16 at 4:04
6
$\begingroup$

Let $\mathcal{C}$ be a category with coproducts and $S \in \mathcal{C}$. Then we have the slice category $S/\mathcal{C}$. The objects of it are morphisms $S \to X$, where $X$ is an object of $\mathcal{C}$. This generalizes your construction. There is a forgetful functor $S/\mathcal{C} \to \mathcal{S}$ mapping $(S \to X)$ to $X$. It has a left adjoint mapping $X$ to $(S \to S + X)$, where the morphism is the coproduct inclusion. I claim that this adjunction is monadic.

The monad $T$ corresponding to the adjunction sends $X$ to $S+X$, the unit is the coproduct inclusion $X \to S+X$ and the multiplication is the obvious morphism $\mu : S+(S+X) = S+S+X \to S+X$ induced by the codiagonal of $S$. Hence, a $T$-module is an object $X$ together with a morphism $f : S+X \to X$ such that $f|_X = \mathrm{id}_X$ and such that $f \circ (S+f) = f \circ \mu$. Now $f$ is determined by the morphisms $f|_X$ and $f|_S$, but $f|_X=\mathrm{id}$ is fixed, and the equation $f \circ (S+f) = f \circ \mu$ simply says $f|_S = f|_S$ on the first $S$-copy, and $f \circ f|_S = f|_S$ on the second copy; but the latter follows from $f|_X = \mathrm{id}$. Hence, you see that $T$-modules correspond to morphisms $S \to X$. It is also easy to check that this correspondence is compatible with morphisms and clearly it is the canonical one from $S/\mathcal{C}$ to $T$-modules.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But what if the object of possible points is not fixed in advance? Say if C is Set and S is countably infinite? Then we might want higher cardinality sets of points all of a sudden... $\endgroup$ – David Roberts Nov 14 '16 at 11:26
  • 4
    $\begingroup$ If the number of points is not fixed in advance, I don't know what the morphisms are supposed to be. Perhaps the OP should clarify what he/she means. $\endgroup$ – HeinrichD Nov 14 '16 at 14:09
  • $\begingroup$ If $S \to X$ and $T \to Y$ are multipointed objects, then a map is a commutative square? $\endgroup$ – David Roberts Nov 15 '16 at 6:54
  • 1
    $\begingroup$ But this is the arrow category, and the left adjoint to the forgetful functor simply maps $X$ to $0 \to X$ (where $0$ is initial). So this adjunction is definitely not monadic. $\endgroup$ – HeinrichD Nov 15 '16 at 7:45
  • $\begingroup$ OK, but the OP asked for S, T to be coproducts of the terminal object. Morphisms could then be commutative squares with 'top edge' induced by maps of the indexing sets. Not saying this is it, but it makes sense to phrase things this way. $\endgroup$ – David Roberts Nov 15 '16 at 22:59

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.