# “Maybe Monad” for multi-pointed objects?

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.

• To ask a clarifying question raised in comments below: what are the morphisms between multi-pointed objects? – David Roberts Nov 15 '16 at 6:55
• Also, what are the objects? Is the number of points fixed? – HeinrichD Nov 16 '16 at 8:29
• @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. – user84563 Nov 21 '16 at 4:02
• @HeinrichD Yes, sorry for the confusion. I've edited the question to clarify. – user84563 Nov 21 '16 at 4:04

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.
• If $S \to X$ and $T \to Y$ are multipointed objects, then a map is a commutative square? – David Roberts Nov 15 '16 at 6:54
• 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. – HeinrichD Nov 15 '16 at 7:45