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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

  1. $t$ 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.

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  • $\begingroup$ @Berci thanks for note, fixed it. I do suspect it can be solved in a more elementary way, which is part of my question. However, it is strongly implied that I should the Beck mondacitiy theorem, where I got the question from. Please elaborate how you think it can be solved? $\endgroup$ – Mariah Jan 14 '19 at 0:39
  • $\begingroup$ If it's for illustrating Beck's monadicity theorem, that also makes perfect sense, with your way of solution. The elementary approach is just to show that $T$-algebras correspond to vector spaces and $T$-algebra-morphisms to linear maps. $\endgroup$ – Berci Jan 14 '19 at 0:45
  • $\begingroup$ @Berci so I came back to this and realised there is a difficulty in the proof I first attempted. Perhaps you can assist. $\endgroup$ – Mariah May 8 '19 at 18:54
  • $\begingroup$ Do you insist to prove the main claim using Beck's monadicity theorem? For that, we can maybe prove that the coequating relation on $W$ respects the linear structure. $\endgroup$ – Berci May 8 '19 at 20:26
  • $\begingroup$ Another idea might be to use the crude monadicity theorem. Every reflexive pair of vector space maps induces an equivalence relation respecting the operations (although this might be more trouble than it's worth to prove) and coequalizers of such equivalence relations are easily seen to be preserved by the forgetful functor. $\endgroup$ – Kevin Carlson May 9 '19 at 0:57
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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 $s=0$.

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  • $\begingroup$ Eric thank you for the solution. I wonder if there is a simpler way, tell me what you think: it is enough to find a specific splitting of the pair $f,g$ in Set. That is, we know that the coequalizer in Set exists, call it $Z$. Define $q$ to be the quotient map, and $s$ an inclusion. Now the question is, does there exist $t: W \to V$ consistent with $q,s$ so that they constitute a splitting? If so, are all these maps then linear maps? If this was the case we'd have a splitting of $f,g$ in Vect$_{\mathbb{R}}$ by $Z$, and this is preserved by the forgetful functor. $\endgroup$ – Mariah May 8 '19 at 21:04
  • $\begingroup$ I don't think that is a fruitful line of exploration. Before you can make sense of any of it, you need to prove that $Z$ actually is a vector space, and that's essentially equivalent to proving it is the quotient of $W$ by $K$. But if you've done that, you're pretty much already done. $\endgroup$ – Eric Wofsey May 8 '19 at 21:07
  • $\begingroup$ In particular, if you work with more general algebraic structures than vector spaces, it is not true that every $G$-split coequalizer is a split coequalizer (so it may not be possible to have any choice of $s$ and $t$ that are homomorphisms). I'm not sure if there is a counterexample in vector spaces since they are very special (for instance, every epimorphism splits), but this shows that any argument based on making $s$ and $t$ homomorphisms would need to be a somewhat ad hoc argument that uses very special properties of vector spaces and does not work more generally. $\endgroup$ – Eric Wofsey May 8 '19 at 21:12
  • $\begingroup$ MM yup, and also I can't see why $s$ can be forced to be linear in any case... Perhaps the easiest (a bit cheating) solution would be: Vect_$_{\mathbb{R}}$ has all limits, so there is a coequalizer, the cokernel. This is obviously preserved by the forgetful functor.. $\endgroup$ – Mariah May 8 '19 at 21:45
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    $\begingroup$ The forgetful functor does not preserve most coequalizers of vector spaces! For instance, consider if $V=W$ and $f=1$, $g=-1$. You really do need to use the assumption that it is split after applying $G$. $\endgroup$ – Eric Wofsey May 8 '19 at 21:48
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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.)

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