I'm trying to understand the idea of an [Albanese variety](https://en.wikipedia.org/wiki/Albanese_variety).  It reminds me of something simpler:

Given a set $X$ with a chosen point $x \in X$, we can form the free abelian group on the pointed set $(X,x)$, which is just the free abelian group on $X$ modulo a relation saying $x = 1$.    

If we call this group $A(X,x)$, it has a nice universal property: any map of pointed sets $f$ from $X$ into an abelian group $A$ factors uniquely as the obvious inclusion 

$$i_X \colon X \to A(X,x)$$ 

followed by some homomorphism

$$\overline{f} \colon A(X,x) \to A.$$

So:

$$  f = \overline{f} \circ i_X $$

The process of taking the free abelian group on a pointed set defines a functor 

$$  A \colon \mathrm{Set}_* \to \mathrm{AbGp} $$

which has a right adjoint

$$  U\colon \mathrm{AbGp} \to \mathrm{Set}_* $$

sending any abelian group $A$ to its underlying pointed set $(A,1)$.  The composite 

$$ U A \colon \mathrm{Set}_* \to \mathrm{Set}_* $$

is thus a [monad](https://en.wikipedia.org/wiki/Monad_(category_theory)), and if I'm not mistaken, the [algebras](https://en.wikipedia.org/wiki/Monad_(category_theory)#Algebras_for_a_monad) of this monad are just abelian groups.

The idea of an Albanese variety seems to be similar, but working with projective algebraic varieties instead of sets.  Namely:

Given any such variety $X$ with a chosen point $x$ there is an [abelian variety](https://en.wikipedia.org/wiki/Abelian_variety) called the **Albanese variety** $A(X,x)$, apparently defined by the following universal property: there is a map of varieties 

$$i_X \colon X \to A(X,x)$$ 

such that any map of pointed varieties $f$ from $X$ into an abelian variety $A$ factors uniquely as $i_X$ followed by some map of abelian varieties 

$$\overline{f} \colon A(X,x) \to A.$$  

So:

$$  f = \overline{f} \circ i_X $$

(The [Wikipedia article on Albanese varieties](https://en.wikipedia.org/wiki/Albanese_variety) doesn't clearly require that $\overline{f}$ be a map of abelian varieties, but [Ravi Vakil's lecture notes](http://math.stanford.edu/~vakil/02-245/sclass18A.pdf) do, so I'm going with that.)

So, naturally, I'm wondering: **does taking the Albanese variety define a functor from pointed varieties to abelian varieties, which has a right adjoint, which together define a monad on the category of pointed varieties whose algebras are the abelian varieties?**

The map $i_X \colon X \to A(X,x) $ would then be the unit of this monad, while the multiplication of the monad would be the map of abelian varieties $\overline{1} \colon A(A(X,x),x) \to A(X,x) $ obtained from the identity map of varieties $1 \colon A(X,x) \to A(X,x)$.