**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.



"Maybe Monad" for multi-pointed objects?
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.


