> 
> Prove that the category of vector spaces over $\mathbb{R}$,
>  Vect$\_{\mathbb{R}}$, is equivalent to the category of $T$ algebras for
>  some monad $T:$ Set $\to$ Set.
> 
> 
> 

My attempt:

First I know that the forgetful functor $G:$ Vect$\_{\mathbb{R}} \to$ Set has a left adjoint $F$ (which $F$ is the free vector space functor).

Moreover, the functor $T:=G\circ F:$ Set $\to$ Set has the structure of a monad.

Thus, by Beck's monadicity theorem, it suffices to prove that $G$ reflects isomorphisms, and that every $G$ split-pair has a coequalizer in Vect$\_{\mathbb{R}}$, and that $G$ preserves this coequalizer.

---

*Recall that $f,g: V \to W$ in Vect$\_{\mathbb{R}}$ are $G$ split when the diagram on $G(f), G(g)$ has a coequalizer, $(Z,q:G(W) \to Z)$, and there are maps $s:Z \to G(W)$, $t:G(W) \to G(V)$ s.t:*

$q \circ s = Id\_Z$, $f \circ t = Id\_W$, $g \circ t = s \circ q$.

---

So, it is clear that $G$ reflects isomorphisms.

Set is cocomplete, so the pair $f,g$ considered as maps in Set have a coequalizer which is just $W$, as a set, modulo the smallest equivalence relation that contains $\{(f(v), g(v):v\in V\}$. Call this coequalizer in Set $Z$.

We give $Z$ the structure of a vector space:

Let $a,b \in Z$, and define $a +\_{W\_0} b := q(s(a)+\_W s(b))$. Also define $0\_Z = q(0\_Y)$

Now I'm trying to show that $Z$ is a vector space, but the technical details aren't working out for me:

For instance, $z\_1 + 0\_Z = q(s(z\_1) + s(q(0\_Y)))$; but I don't know that $s\circ q(0\_Y) = 0\_Y$. I would know that if either 

1. $s$ is known to be the inclusion map, $q$ is the quotient map.

Or

2. $t$ is a linear map.

How can I resolve this?

I need to be able to show that $Z$ is a vector space, and all $s,t,q$ are linear maps. This would imply that the diagram is split in $Vect\_{\mathbb{R}}$, hence admits the coequalizer vector space $Z$, which $G$ preserves.

I will accept and award the bounty for an answer which takes into account these technical issues and is a formal answer.