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.
Best Answer
Let $K$ be the subspace of $W$ generated by all elements of the form $f(v)-g(v)$ for $v\in V$. I claim the equivalence relation on $W$ corresponding to $q$ is the same as equivalence mod $K$: that is, $q(w)=q(w')$ iff $w-w'\in K$. The forward direction is immediate from the definition of $K$.
Conversely, suppose $w\in W$ and $k=\sum_{i=1}^n a_i(f(v_i)-g(v_i))\in K$, for scalars $a_i$ and elements $v_i\in K$. We must show that $q(w)=q(w')$ for $w'=w+k$. Using induction on $n$, we can reduce to the case $n=1$, so we must show $q(w+a(f(v)-g(v)))=q(w)$ for $a\in\mathbb{R}$ and $v\in V$. Since $a(f(v)-g(v))=f(av)-g(av)$, we may further assume $a=1$.
We can now verify $q(w')=q(w)$ by a computation where we repeatedly use the various identities relating our functions: \begin{align*} q(w')=q(w+f(v)-g(v))&=q(ft(w)+ftf(v)-ftg(v)) & [ft=1] \\ &=q(gt(w)+gtf(v)-gtg(v)) & [qf=qg] \\ &=q(gt(w)+gtf(v)-gtf(v)) & [gtg=sqg=sqf=gtf] \\ &=q(ft(w)+ftf(v)-ftf(v)) & [qg=qf] \\ &=q(w+f(v)-f(v)) & [ft=1] \\ &=q(w). \end{align*}
(Note that this argument is not specific to vector spaces: a similar argument works for any kind of algebraic structure, to show that if $w'$ and $w$ are words in $W$ that can be obtained by swapping elements of the form $f(v)$ with $g(v)$, then $q(w')=q(w)$: just apply $ft$ to each term of $w'$, use $qf=qg$ to replace $ft$ with $gt$ when computing $q(w')$, and then use $gtg=gtf$ to freely swap terms $gtg(v)$ with terms $gtf(v)$. This shows that the congruence relation that coequalizes $f$ and $g$ with respect to the algebraic structure coincides with the equivalence relation that coequalizes $f$ and $g$.)
Thus we see that $Z$ is just the usual quotient vector space $W/K$, and $q$ is also the coequalizer of $f$ and $g$ in $\mathrm{Vect}_\mathbb{R}$.
Note that it's not necessarily true that $s$ and $t$ are linear. For instance, if $g=0$ then $q=0$ and so $t$ can be any (not necessarily linear) splitting of $f$ and all the required equations will still hold with $s=0$.