TITLE: Understanding the tensor-hom adjunction intuitively
QUESTION [14 upvotes]: I'm currently trying to teach myself some category theory. Recently, I learned that the tensor product is left adjoint to the hom functor in suitable categories, e.g. vector spaces with linear maps, i.e. for vector spaces $U$, $V$, and $W$,
$$\mathrm{Hom}(U \otimes V, W) \cong \mathrm{Hom}(U, \mathrm{Hom}(V,W))$$
I am trying to gain some intuition for adjoints by looking at special cases. This one is puzzling me a bit. Here is one of my attempts to get a grasp of it: 
If my understanding is correct, a (covariant) $k$-tensor $\omega$ on a vector space $V$ can be defined equivalently as either an element of the $k$-fold tensor product of the dual space $V^*$, i.e. $\omega \in \bigotimes_{i=1}^k V^*$, or as a $k$-form on $V$, i.e. a $k$-linear map $\omega: \bigoplus_{i=1}^k V \to \mathbb{R}$. Thus, we have a bijection between $\bigotimes_{i=1}^k V^*$ and the set of $k$-forms on $V$. Furthermore, if $V$ is finite dimensional, then we also have the well-known isomorphism $V^* \otimes V^* \cong \mathrm{Hom}(V,V^*)$. 
It seems to me that these observations can be interpreted as special cases of the adjunction above, or are at least related in some way. In particular, we have 
$$\mathrm{Hom}(V^* \otimes V^*, V) \cong \mathrm{Hom}(V^*, \mathrm{Hom}(V^*,V))$$
What is the proper way to interpret these facts in the language of category theory? Can this be generalized at all to help give more intuition for the adjunction in terms of the properties of tensor products? 
Thank you!

REPLY [2 votes]: $\DeclareMathOperator{Hom}{Hom}$
I just want to give an explicit example, where I used this tensor-hom adjunction.
Say we have finite dimensional inner product space $(V,\langle,\rangle)$ and a subspace $W\subset V.$ Then we define orthogonal complement of $W$ as
$$W^\perp=\{v:(\forall w\in W)\langle v,w\rangle=0\}.$$
However I would like to see $W^\perp$ in more "algebraic" form.
So as you noticed, due to the adjunction, my inner product space $\langle,\rangle:V\otimes V\to\mathbb{R}$ can be seen as element of $\Hom(V,V^*).$ Let denote this as $L_{\langle,\rangle}:V\to V^*.$ Explicitly we have that
$$(L_{\langle,\rangle}(v_1))(v_2)=\langle v_1,v_2\rangle.$$
Now consider inclusion $i:W\hookrightarrow V$ and associated dual map $i^*:V^*\to W^*.$ Let us compute the kernel of composition $i^*\circ L_{\langle,\rangle}:V\to V^*\to W^*.$ We have that
$$\ker(i^*\circ L_{\langle,\rangle})=\{v:i^*(L_{\langle,\rangle}(v))=0\}=\{v:\forall(w\in W)L_{\langle,\rangle}(v)(w)=0\}=\{v:\forall(w\in W)\langle v,w\rangle=0\}=W^\perp.$$
Hence we presented $W^\perp$ as a kernel of some linear mapping.
Such approach also allows you to define non-degeneracy, reflexivity of bilinear mapping in the language of kernels and images. One can also show that two maps $F:A\to B, G:B\to A$ are adjoint if and only if $L_{\langle,\rangle_B}\circ F=G^*\circ L_{\langle,\rangle_A}.$
More compactly. tensor-hom adjuntion allows you to translate "set-theoretic" conditions into "algebraic" ones. It works both ways by the way.