The left adjoint to the forgetful functor $G\colon\mathsf{Vect}_\mathbb{C}\to\mathsf{Vect}_\mathbb{R}$ and Barr-Beck

Question

Let $G\colon\mathsf{Vect}_\mathbb{C}\to\mathsf{Vect}_\mathbb{R}$ be the forgetful functor from $\mathbb{C}$-vector spaces to $\mathbb{R}$-vector spaces. I am trying to explicitly construct the left adjoint $F$ to $G$.

Any object $V\in\mathsf{Vect}_\mathbb{R}$ is an abelian group $A$ with an action of $\mathbb{R}$. The 'obvious' way to extend this to a $\mathbb{C}$-action is to define $(x+iy)\cdot a=(x\cdot a)+ i\,(y\cdot a)$ which gives $A$ the structure of a $\mathbb{C}$-vector space, and write $F$ to mean the functor given by this construction.

It then seems obvious (which is always cause for alarm) that $GF=\mathrm{id}_{\mathsf{Vect}_\mathbb{R}}$ and reasonably believable (which means nothing really) that $FG\cong\mathrm{id}_{\mathsf{Vect}_\mathbb{R}}$ but this seems far to nice to be true.

In particular, this comes up as an exam question, where we are then asked to describe the explicit monad $GF$ on $\mathsf{Vect}_\mathbb{R}$ and the equivalence (from Barr-Beck) of $\mathsf{Vect}_\mathbb{C}\to\mathsf{Alg}_{GF}(\mathsf{Vect}_\mathbb{R})$.

1. Is my construction of $F$ correct?
2. Are the facts about $GF$ and $FG$ I stated correct?
3. Assuming that some of the above is wrong how can we show that every $G$-spit pair we can construct their coequaliser and show that it is preserved by $G$?

Edit: For the third question, I know that $F$ being a left adjoint implies that it commutes with colimits, and thus that $$F\Big(\mathrm{coeq}\big(G(V)\rightrightarrows G(W)\big)\Big) = \mathrm{coeq}\big(FG(V)\rightrightarrows FG(W)\big)$$ which goes some of the way towards constructing the coequaliser in $\mathsf{Vect}_\mathbb{C}$ given the coequaliser in $\mathsf{Vect}_\mathbb{R}$, but I haven't used the $G$-split property anywhere.

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Edit 2: Thanks to the comments I'm pretty certain that $$F=(-\otimes_\mathbb{R}\mathbb{C})$$ is the left adjoint that we're looking for (and it is exact since $\mathbb{C}$ is a free $\mathbb{R}$-module and thus flat). What I'm having trouble with now is showing that $G$ satisfies the conditions of the Barr-Beck theorem, in particular the fact that every $G$-split pair has a coequaliser and it is preserved by $G$.

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Edit 3: Showing that $G$ preserves the coequaliser (assuming that it exists) follows from the fact that $G$ is exact: since adjoints are unique up to isomorphism, we know that $G\cong\mathrm{Hom}_{\mathsf{Vect}_\mathbb{R}}(\mathbb{C},-)$, but $\mathbb{C}$ is free and thus projective, and thus $G$ is exact. Hence $G$ commutes with colimits.

So all that remains is to explicitly construct the coequaliser of $V\rightrightarrows W$ but I don't know where to begin. How can we use the fact that this pair is $G$-split?

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Final (?) edit: I think that the answer is actually quite trivial: we know that the coequaliser in $\mathsf{Vect}_\mathbb{R}$ always exists since $\mathsf{Vect}_\mathbb{R}=\mathbb{R}\hbox{-}\mathsf{mod}$ has finite colimits (in fact it is bicomplete).

Answer 1

There is no way to turn a real vector space into a complex one in general. Think, for example, about vector spaces of odd dimension. The even dimensional ones can be turned into complex ones but in many ways, and there is no way to do it functorially.

This means that what you wrote does not define a functor (nor anything else, reallythe, as the "definition" does not work: you wrote "this gives the structureof a complex vector space", but it doesn't) so your two first questions have no answer.

I suggest that you try to find the adjoint before embarking in the third one or in any thing else!