Understanding natural isomorphism of $W\otimes V^*$ and ${\rm Hom}(V,W)$ I just saw in a lecture that if $V$ and $W$ are finite dimensional vector spaces over a field then there is a natural isomorphism from $W\otimes V^*$ and ${\rm Hom}(V,W)$.
If it is asked one to guess for the natural choice of natural isomorphism, it is not hard.
And then, moving to algebraic verifications is also then not hard.
But keeping all these algebraic manipulations, I was unable to understand the isomorphism in some simple examples.
Question: Can one give some concrete examples of $V$ and $W$, where we can easily understand $W\otimes V^*$ and ${\rm Hom}(V,W)$, and then the natural isomorphism between them is clear from it?
 A: Question: "Can one give some concrete examples of $V$ and $W$, where we can easily understand $W\otimes V^*$ and $\operatorname{Hom}_k(V,W)$, and then the natural isomorphism between them is clear from it?"
Answer: There is a natural map
$$\rho: W^*\otimes_k V \rightarrow \operatorname{Hom}_k(W,V)$$
defined as
$$\rho(\phi \otimes v)(w):=\phi(w)v.$$
You may check that $\rho$ is a well defined injective map between $k$-vector spaces and by a dimension count it follows $\rho$ is an isomorphism.
Example: If $V:=W:=k\{e_1,e_2\}, V^*:=k\{x_1,x_2\}$ it follows
$\rho(x_1\otimes e_1)=M$ where $M$ is the matrix
\begin{align*} M= \begin{pmatrix} 1 & 0 \\ 0 & 0  \end{pmatrix} .\end{align*}
This is because
$$x_1\otimes e_1(e_1):=x_1(e_1)e_1=e_1, x_1\otimes e_1(e_2):=x_1(e_2)e_1:=0.$$
Claulating $\rho(x_i\otimes e_j)$ for $i,j=0,1$ you get 4 linearly independent matrices in $\operatorname{End}_k(V)$.
Note: The map $\rho$ is  "functorial" in the following sense: Given maps $f: W_1 \rightarrow W, g: V \rightarrow V_1$, there are maps
$$u: W^*\otimes_k V\rightarrow W_1^*\otimes_k V_1$$
and
$$v: \operatorname{Hom}_k(W,V) \rightarrow \operatorname{Hom}_k(W_1,V_1)$$
such that $v \circ \rho = \rho_1 \circ u$
where
$$\rho_1: W_1^* \otimes_k V_1 \rightarrow \operatorname{Hom}(W_1,V_1)$$
is the canonical map.
Such "natural maps" arise in multilinear algebra, commutative algebra, homological algebra and representation theory. Here is an example where you use such maps to give an explicit isomorphism between representations:
How to construct an isomorphism $\wedge^2 V \xrightarrow{\sim} \wedge^2 V\otimes U^{\prime}$ as representations of $S_5$?
A: Given a basis $\{v_1,v_2,\dots,v_n\}$, a linear map $f\colon V\to W$ is determined once you choose $w_1,w_2,\dots.w_n\in W$, by declaring that
$$
f(v_i)=w_i
$$
and extending by linearity. But of course you get the same map as
$$
f=f_1+f_2+\dots+f_n
$$
where
$$
f_j(v_i)=\begin{cases} w_j & i=j \\[6px] 0 & i\ne j \end{cases}
$$
which can be simplified to $f_j(v_i)=\delta_{ij}w_j=v_j^*(v_i)w_j$, where $\{v_1^*,\dots,v_n^*\}$ is the dual basis.
Can you see now the connection with the tensor product $V^*\otimes W$? Take $v\in V$ and write it as $v=a_1v_1+\dots+a_nv_n$; then
$$
f_j(v)=\sum_{i}a_if_j(v_i)=\sum_i a_i v_j^*(v_i)w_j=v_j^*(v)w_j
$$
and therefore
$$
f(v)=\sum_j v_j^*(v)w_j
$$
But using the basis is not really so important: given any $\xi\in V^*$ and $w\in W$ you can define a linear map $V\to W$ by decreeing that $v\mapsto \xi(v)w$.
In this way you get a map $V^*\times W\to\operatorname{Hom}(V,W)$ which is bilinear (easy check), so this map factors through the tensor product and we have a linear map
$$
\eta\colon V^*\otimes W\to\operatorname{Hom}(V,W)
$$
In the first part, we actually expressed $f$ as
$$
f=\eta\Bigl(\sum_j v_j^*\otimes w_j\Bigr)
$$
which means that when $V$ is finite dimensional, the map $\eta$ is surjective. The map is also injective, in this case. Indeed, any element in $V^*\otimes W$ can be written as
$$
t=\sum_j v_j^*\otimes w_j
$$
and $\eta(t)=0$ implies that, for every $i$,
$$
0=\sum_j v_j^*(v_i)w_j=w_i
$$
This shows that, when $V$ is finite dimensional, an element $t$ in $V^*\otimes W$ is essentially an $n$-tuple of elements of $W$ (but we need to choose a basis of $V$ to make this identification).
What does it mean that the map $\eta$ is natural? In this particular case that it doesn't depend on choice of bases in the two spaces. We used the (finite) basis of $V$ only to prove that when $V$ is finite dimensional then $\eta$ is an isomorphism. No assumption is necessary for $W$.
