What's the functor F and what's the morphism from F(A) to A (and what's A)?

closed as unclear what you're asking by David White, Andres Caicedo, Willie Wong, Daniel Moskovich, Noah Stein Aug 5 '13 at 13:52
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
I want to convince you that this is the wrong question to ask. First, a little background. There is a relationship between algebras in the mathematical sense and $F$algebras for $F$ an endofunctor, but in mathematics the endofunctor is usually a monad and we usually want algebras over this monad in the monadic sense, where there are more requirements. The basic relevant example here is that on $\text{Vect}$, the category of vector spaces over a field $k$ (suppressed in the notation), there is a monad whose algebras are the associative $k$algebras. It is a sort of $k$linear version of the list monad. One reason we care about monads so much more than arbitrary endofunctors is because we get them from adjunctions, which are ubiquitous. For example, the monad which gives us associative $k$algebras comes from an adjunction between $\text{Vect}$ and the category $k\text{Alg}$ of $k$algebras. There is a similar monad giving us commutative $k$algebras coming from an adjunction between $\text{Vect}$ and the category of commutative $k$algebras. In both cases here the second category of the adjunction is naturally equivalent to the category of algebras over the corresponding monad; in this situation the adjunction is said to be a monadic adjunction. The monad coming from commutative algebras sends a vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is very close to the exterior algebra functor. Here I want to convince you that the right question to ask, in terms of smashing together the concept of exterior algebras with the concept of algebras over an endofunctor, is actually "is the exterior algebra functor a monad, and if so, does it come from a monadic adjunction?" The answer is yes in the following sense. We need to replace $\text{Vect}$ with the category of super vector spaces (the same super as in supersymmetry). The objects of this category are direct sums $V \cong V_0 \oplus V_1$ where $V_0$ is the even or bosonic part and $V_1$ is the odd or fermionic part, and the morphisms are linear maps preserving parity. There is a monadic adjunction between this category and the category of supercommutative algebras (I'm suppressing the underlying field here again), and the corresponding monad is a super version of the symmetric algebra monad which, on purely odd super vector spaces, reproduces the usual exterior algebra. 

