$\hom(V,W)$ is canonic isomorph to $\hom(W^*, V^*)$ Introduction My Semester just started and we have a new Professor for Linear Algebra II (replacing our former Professor). Apparently we are behind our schedule and thus we only had a brief introduction (5 Minutes on the Blackboard) about the Dual Space and canonical isomorphisms.

In the current problem set there is this optional (not mandatory and not accredited) exercise which I understand as good as nothing about:


  Problem: Let $V$ and $W$ be finite dimensional $\mathbb{K}$-Vectorspaces$^{1)}$. Show that $\hom(V,W)$ is canonical isomorph to $\hom(W^*,V^*)$ and find the canonical isomorphism. 
  
  $^{1)}$ ($\mathbb{K}$ denotes a field here, I presume in english literature they write $\mathbb{F}$)  


Hint (given by my tutor): $$ \Phi: \hom(V,W) \longrightarrow  \hom(W^*,V^*) \\ \Psi \longmapsto (\varphi \longmapsto \varphi\circ \Psi)$$
My problems are vast now:


The literature I am reading does a bad job at explaining this topic
The online literature I find seems to be way beyond my level, they usually include tensor algebra to solve this. So I fall down the math rabbit hole see, On &quot;familiarity&quot; (or How to avoid &quot;going down the Math Rabbit Hole&quot;?)
I don't understand the Hint, I did not encounter such a function before, not even in Analysis and I don't know what's happening. I can only guess that it is a function that somehow completes a circle. 


On the bright side: 


I believe to understand what a canonical isomorphism is. My Professor said it is an isomorphism that does not involve making a choice for a basis. In the above function we do not make such a decision and therefore it would be a candidate for a canonical isomorphism. 
If I would understand the above function, I'd merely have to show that it is injective (trivial Kern) and surjective to complete the task. Unfortunately, as far, we've only done such things using an explicit Matrix and not a function. 
I do know the definition of the Dual Space, also using the Kronecker Delta.  (although I know nothing about it's meaning and geometric intrepreation, if there is any) 
The way my tutors sell this exercise it's supposed to be very easy, once the definitions are clear. 


To-Do-List: 


Understand the function, know what's happening.
Show that the function is surjective   
Show that the function is injective  (trivial Kern)
First notice that if $E$ is a linear space then its dual $E^*$ is the linear space of the  linear forms $$\varphi: E\rightarrow \Bbb K$$

Now if $\Psi\in \operatorname{hom}(V,W)$ i.e. a linear transformation $\Psi:V\rightarrow W$ then $\Phi(\Psi)\in \operatorname{hom}(W^*,V^*)$ since
$$\underbrace{\Phi(\Psi)(\varphi)}_{\in V^*}=\varphi\circ\Psi:V\xrightarrow{\Psi} W\xrightarrow{\varphi} \Bbb K\quad \forall \varphi\in W^*$$
and we prove easily that $\Phi$ is a linear transformation:
$$\Phi(\Psi)(\varphi+\lambda\psi)=\Phi(\Psi)(\varphi)+\lambda\Phi(\Psi)(\psi)$$
and if $\Psi\ne0$ hence there's $x\in V$ such that $W\ni\Psi(x)\ne0$ hence with $\varphi=(\Psi(x))^*$ the linear form such that
$$(\Psi(x))^*(\Psi(x))=1$$
we see that
$$\Phi(\Psi)(\varphi)\ne0$$
hence $\Phi$ is injective, moreover, since 
$$\dim \operatorname{hom}(V,W)=\dim V\times\dim W=\dim V^*\times\dim W^*=\dim\operatorname{hom}(W^*,V^*)$$
we see that $\Phi$ is bijective.
The isomorphism is simply the transpose...

This is as easy as things come, really :-)