A linear transformation $T$ from $V$ to $W$ is an isomorphism iff $\ker(T) = \{0\}$ and $im(T) = W$ This is a result that's proven in my book, and I'm having trouble understanding some parts of the proof, so this is how it goes:

Suppose first that $T$ is an isomorphism. Then the kernel of $T$ is
  found by solving $T(f)=0$. If we apply $T^{-1}$ on both sides, we find
  that $f = T^{-1}(0)=0$, that the $\ker(T)=\{0\}$, as claimed.

My question here is why $T^{-1}(0)$ necessarily equals $0$. Is this a requirement of an invertible or linear transformation? Why does this show the kernel must be $0$?

To see that $im(T)=W$, note that any $g \in V$ can be written as $g =
 T(T^{-1}(g))$.

I don't quite get this either. How does this show that $im(T) = W$?
 A: I hope you can follow this :
Given $T:V\to W$ is an Isomorphism. So T is invertible, which means that $\exists$ S $\ni$ $ST=TS=I$. Now by definition, a Linear Transformation maps $0$ to $0$, ie $T(0)=0$ and since T is invertible so $T^{-1}$ exists and hence $T^{-1}(0)=0$.  
Now note that if $T$ is an isomorphism then so is $T^{-1}$. 
$\therefore $ consider $T(f)=0$. Applying $T^{-1}$ both sides, we get 
$T^{-1}(T(f))=T^{-1}(0)=0\implies f=0$ ,ie $Ker(T)={0}$. 
Since $dim(V)=dim(Ker(T))+dim(Im(T))$ 
So $dim(V)=dim(Im(T))$. But since V and W are isomorphic, so $dim(V)=dim(W)$, and therefore $dim(Im(T))=dim(W)$ and hence $Im(T)=W$. 
Converse is easy. 
Suppose that $ker(T)=\{0\}$ and $Im(T)=W$. So T is one to one. This is easily followed by observing that 
$T(u)=T(v) \implies T(u)-T(v)=0 \implies T(u-v)=0 \implies (u-v)\in Ker(T)$. But since $Ker(T)=\{0\}$, so $u=v$. It easily follows from here that $T$ is invertible. $\blacksquare$
A: First you can prove that $T^{-1}$ is linear. Once we have this 
$$T^{-1}(0)=T^{-1}(0+0)=T^{-}(0)+T^{-1}(0)$$
which implies $T^{-1}(0)=0$.
For the second question. Notice that $T^{-1}(g)$ is an element $h$ of $V$. Then you are showing that (for every $g$) there is $h$ such that $T(h)=g$. This is the definition of being surjective.
Let's prove now the general result that if $T$ is linear then $T^{-1}$ is linear.
We have that for all vectors $u,v$ and all scalar $r,s$ 
$$T(ru+sv)=rT(u)+sT(v)$$
Let's use the particular case in which $u:=T^{-1}(w)$ and $v:=T^{-1}(x)$. We get
$$T(rT^{-1}(w)+sT^{-1}(x))=rT(T^{-1}(w))+sT(T^{-1}(x))=rw+sx$$
Take now $T^{-1}$ on both sides to get
$$rT^{-1}(w)+sT^{-1}(x)=T^{-1}(T(rT^{-1}(w)+sT^{-1}(x)))=T^{-1}(rw+sx)$$
i.e. 
$$rT^{-1}(w)+sT^{-1}(x)=T^{-1}(rw+sx)$$
for all vectors $w,x$ and scalars $r,s$. This is the definition of $T^{-1}$ being linear.
