Vect$_{\mathbb{R}}$ is equivalent to the category of $T$ algebras for a monad Set $\to$ Set.

Question

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.

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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:

$q \circ s = Id_Z$, $f \circ t = Id_W$, $g \circ t = s \circ q$.

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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.

Answer 1

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 $s=0$.

Answer 2

As an outline of the direct proof not using Beck's monadicity theorem: let us view $T(I)$ as giving equivalence classes of formal (non-flattened) expressions in the vector space language with atoms $e_i$ for $i \in I$. This has an obvious functor structure; and the monad structure is given by $\eta_I : I \to T(I)$, $i \mapsto e_i$ and $\mu_I : T(T(I)) \to T(I)$ is the "flattening map" e.g. $\mu(4 e_{2e_i - 3e_j} - e_{e_k}) = 4 (2 e_i - 3 e_j) - e_k$.

Now, let us see what a $T$-algebra looks like: we need to have a set $X$ with a map $\alpha : T(X) \to X$ satisfying the algebra conditions. Given such a map $\alpha$, we can define vector space operations by $x + y := \alpha(e_x + e_y)$ and $\lambda x := \alpha(\lambda e_x)$. Now, the condition $\alpha \circ \eta = \mathrm{id}$ just means that $\alpha(e_x) = x$. For the condition $\alpha \circ T\alpha = \alpha \circ \mu$, let us apply this for example to the element $e_f + e_g$ where $f,g \in T(X)$. We see that $\mu(e_f + e_g) = f + g$ whereas $T\alpha(e_f + e_g) = e_{\alpha(f)} + e_{\alpha(g)}$; so $\alpha(f + g) = \alpha(e_{\alpha(f)} + e_{\alpha(g)})$ which by definition is equal to $\alpha(f) + \alpha(g)$. Similarly, $\alpha(\lambda f) = \lambda \alpha(f)$.

Now, to check the vector space axioms, we have for example that $\alpha[\lambda (e_x + e_y)] = \lambda \alpha(e_x + e_y) = \lambda [\alpha(e_x) + \alpha(e_y)] = \lambda (x+y)$ whereas $\alpha[\lambda e_x + \lambda e_y] = \alpha(\lambda e_x) + \alpha(\lambda e_y) = \lambda \alpha(e_x) + \lambda \alpha(e_y) = \lambda x + \lambda y$. Since $\lambda (e_x + e_y) = \lambda e_x + \lambda e_y$ are in the same equivalence class in $T(X)$, it follows that $\lambda (x+y) = \lambda x + \lambda y$. The other vector space axioms can be proven similarly.

Furthermore, the vector space operations uniquely determine $\alpha$, as we can prove by structural induction on elements of $T(X)$ using the linearity of $\alpha$ that we showed above.

Conversely, given any vector space $X$, we can define $\alpha : T(X) \to X$ to be the "evaluation map" e.g. $\alpha(4 (e_x + e_y) - 3 e_z) = 4 (x+y) - 3z$ and it should be straightforward to see that this $\alpha$ defines a $T$-algebra structure on $X$. These two constructions will now give the two functors between $\mathrm{Vect}_{\mathbb{R}}$ and $\mathrm{Set}_T$, and it remains to check that they define an equivalence of categories. In one direction, the fact mentioned above that the vector space operations derived from $\alpha$ uniquely determine $\alpha$ will figure prominently.

(It should also be possible to adapt this proof to the equivalent formulation that $T(I)$ is the set of functions $I \to \mathbb{R}$ with finite support, and functoriality is $T(\phi : I \to J) : f \mapsto (j \mapsto \sum_{i \in \phi^{-1}(j)} f(i))$. However, then the algebra gets obscured -- e.g. in considering $e_f + e_g$ above we might end up needing to consider two separate cases for $f = g$ vs. $f \ne g$ -- and in the result it might not be as easy to see how you would generalize the argument to other varieties of algebras.)