Showing $F(U) =\operatorname{Hom}(V,U)$ and $G(U)=V \otimes U$ are adjoint functors Following Christian Kassel's "Quantum Groups" page 280
Let for a fixed vector space $V$ $F,G$ be functors from the category of vector spaces (which we will call $A$) to itself defind by: $$F(U) = \operatorname{Hom}(V,U)$$ $$G(U)=V \otimes U$$
The natural transformation $\operatorname{Hom}(U \otimes V,W) \cong \operatorname{Hom}(U,\operatorname{Hom}(V,W))$ shows that $G$ is a left adjoint functor to $F$

So, since $G$ is left adjoint functor to $F$ we have natural transformations $$\nu\colon\operatorname{id}_A \rightarrow FG$$ $$\theta\colon GF \rightarrow\operatorname{id}_A$$
such that the compositions
$$F(V) \rightarrow^{\nu(F(V))} FGF(V) \rightarrow^{F(\theta(V))} F(V)$$
and
$$G(W) \rightarrow^{G\nu(W)} GFG(W) \rightarrow^{(\theta(G(W))} F(V)$$
act as the identity for all vector spaces $V$

Can somebody show me how to define $\nu$ and $\theta$ and help me understand why the compositions are acting as the identity on every vector space? Thanks!
 A: $\DeclareMathOperator{\Hom}{Hom}$When it comes to adjoint functors, the natural transformations involved are often defined in the only possible way you could define them, given very little information that you have.
Starting with the unit $\nu$, for any vector space $U$, we need to produce a linear map $$\nu_U\colon U\to \Hom(V,V\otimes U).$$
$\nu_U$ should take a vector $u\in U$ and produce a linear map $\nu_U(u)\colon V\to V\otimes U$. So, given vector $v\in V$, we know that $\nu_U(u)(v)$ should be a vector in $V\otimes U$. There is really only one thing that we can do and that is to define $\nu_U(u)(v) = v\otimes u$. I prefer to write it like this:
\begin{align}
U &\to \Hom(\,V\ , \,V\otimes U\,)\\
u &\mapsto \quad\quad\,(v\mapsto v\otimes u)
\end{align}
I leave to you to verify that $\nu_U$ and $\nu_U(u)$ are linear maps and that $\nu$ is natural.
Similarly, to define counit $\theta$, for every vector space $U$, we need to produce a linear map
$$\theta_U\colon V\otimes\Hom(V,U)\to U.$$
We can easily define linear maps on simple tensors (by the universal property of tensor product), so given $v\in V$ and $A\colon V\to U$, $\theta_U(v\otimes A)$ should be a vector in $U$. But really, given $A$ and $v$, there is only one way to produce a vector in $U$ and that is to evaluate $A$ at $v$. Thus, $\theta_U(v\otimes A) = Av$ and $\theta_U$ is just evaluation map. (This shows up quite often!) I prefer to write it like this:
\begin{align}
V\otimes \Hom(V,U) & \to U\\
v\otimes  A        &\mapsto Av
\end{align}
I leave to you to show that $\theta_U$ is indeed well defined and $\theta$ is natural.
That the compositions are identities is very straightforward to check.
Finally, a personal remark. I find it quite upsetting that $G$ is left and $F$ is right adjoint and not the other way around.
