Constructing the left adjoint to the forgetful functor $G:\textbf{Vect}_\mathbb{C}\to\textbf{Vect}_\mathbb{R}$ In general if $R$ is a commutative ring and $S$ is a commutative $R$-algebra, then the forgetful functor $S\mathrm{-mod} \to R\mathrm{-mod}$ has a left adjoint $S\otimes_R -: R\mathrm{-mod}\to S\mathrm{-mod}$. It is pretty straightforward to show that these functors are adjoint, and all is well.
However, I wonder if there is a way to "figure out" what the left adjoint must be, without the insight that it is the tensor product. In particular, I am interested in the case where $R = \mathbb{R}$ and $S = \mathbb{C}$. That is, let $G:\textbf{Vect}_\mathbb{C}\to\textbf{Vect}_\mathbb{R}$ be forgetful. The General Adjoint Functor Theorem tells us that $G$ has a left adjoint, and I would like to construct it without knowing in advance what it is.
My Attempt
First of all, at least as sets, we have
$$
G(\mathbb{C}) \cong\operatorname{Hom}_\mathbb{R}(\mathbb{R}, G(\mathbb{C})) \cong \operatorname{Hom}_\mathbb{C}(F(\mathbb{R}), \mathbb{C}).
$$
Now, assuming that this isomorphism is actually of $\mathbb{C}$-vector spaces, we see that $F(\mathbb{R}) \cong \mathbb{C}$. Since the left adjoint preserves colimits, for any real vector space $W$, we have
$$
F(W) \cong F(\bigoplus_{\operatorname{dim}_\mathbb{C} W}\mathbb{R}) \cong \bigoplus_{\operatorname{dim}_\mathbb{C}W}F(\mathbb{R}) \cong \bigoplus_{\operatorname{dim}_\mathbb{C} W}\mathbb{C}.
$$
This is correct, since tensoring with $\mathbb{C}$ (which we know is actually the left adjoint) replaces each copy of $\mathbb{R}$ with a copy of $\mathbb{C}$. However, I don't see how to deduce the action of this functor on morphisms. Is there an elegant way to do this?
 A: Understanding how $F$ acts on morphisms requires understanding the correspondence $\def\Hom{\operatorname{Hom}}\def\R{\mathbb{R}}\def\C{\mathbb{C}}\def\id{\operatorname{id}}$
$$
\Hom_\R(V,G(W)) \cong \Hom_\C(F(V),W)
$$
Since $\Hom_\R(-,-)$ is continuous in both variables, we can identify $V\cong\bigoplus_{\dim_\R V}\R$ and $F(V)\cong\bigoplus_{\dim_\C F(V)}\C$ and view the above as
$$
\prod_{\dim_\R V}\underbrace{\Hom_\R(\R,G(W))}_{G(W)} \cong \prod_{\dim_\C F(V)}\underbrace{\Hom_\C(\C,W)}_W
$$
Finding such a correspondence is trivial: $\dim_\R V = \dim_\C F(V)$, and $G(W)=W$ as sets, so this is really just a dolled-up identity map.
Tracing through all the identifications made above, we find that a linear map $\phi:V\to G(W)$ corresponds to the linear map $F(V)\to W$ just obtained by extending $\phi$ by $\C$-linearity (as epxected), and the two morphisms are called adjuncts.
Now, given an $\R$-linear map $f:U\to V$, then naturality of the above implies that
$\require{AMScd}$
\begin{CD}
\Hom_\R(V,G(W)) @<\cong<< \Hom_\C(F(V),W) \\
@Vf^*VV @VVF(f)^*V \\
\Hom_\R(U,G(W)) @>\cong>> \Hom_\C(F(U),W)
\end{CD}
must commute. Take $W := F(V)$, then following $\id_{F(V)}$ along the two paths from $\Hom_\C(F(V),F(V))$ to $\Hom_\C(F(U),F(V))$ determines $F(f)$.
More precisely, $F(f)$ is the (right) adjunct of $\eta_V\circ f$, where $\eta_V$ is the (left) adjunct of $\id_{F(V)}$ (that is, the adjunction unit).
