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
- $s$ is known to be the inclusion map, $q$ is the quotient map.
Or
- $t$ 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.