Proving that $\phi (T) = T^*$ is an isomorphism between vector spaces I am tasked with the following:

I am thus tasked with proving:
$1)$ $\phi(T)$ is linear, so that it respects closure under scalar multiplication and addition.
$2)$ $\phi(T)$ is a bijection.
I only need justification as to whether or not my explanation as to whether $
\phi(T)$ is a bijection is indeed valid or not. First, I will note that linear functionals are isomorphisms, so are bijective themselves and respect closure under addition and scalar multiplication. I work to first prove that $\phi$ is injective.
Injectivity
Suppose $\phi$ is injective.
$$\implies \phi(T_n) = \phi(T_m) \iff T_n = T_m$$
$$\implies T^*_n = T^*_m \iff T_n = T_m$$
$$\implies f \ (T_n) = f \ (T_m) \iff T_n = T_m$$
$$\implies f(T_n - T_m) = 0 \iff T_n = T_m$$
Let's suppose $T_n \ne T_m$.
$$\implies f(T_n-T_m) = 0 \implies T_n - T_m \in \text{Ker} \ f$$
Which is a contradiction, as $f$ is surjective. 
Surjectivity
Assume $\exists \ T \in \mathcal L (V,W)$ such that $\phi(T) = 0$.
$$\implies T^* = 0$$.
$$\implies f(T) = 0$$
$$\implies T \in \text{Ker} \ f$$
Which is a contradiction. Hence, $\phi$ is surjective, making it a bijection and $\phi$ an isomorphism. 
Did I make any mistakes here?
 A: DonAntonio already touched linearity and injectivity questions.  For surjectivity, you seem to be proving injectivity instead.  However, to correctly prove surjectivity, you are going to need to use a dimension-counting argument.  This is because, if $W$ is infinite-dimensional and $V\neq 0$, then $$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)<\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,,$$
where $\mathbb{K}$ is the ground field.  However, the dual map $\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$ is still an injective linear map, regardless of the dimensions of $V$ and $W$.  The proofs of linearity and injectivity are essentially unchanged.
Since $V$ and $W$ in the problem statement are both finite-dimensional, 
$$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)=\dim_\mathbb{K}(V)\,\dim_\mathbb{K}(W)=\dim_\mathbb{K}(W^*)\,\dim_\mathbb{K}(V^*)=\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,.$$
Thus, any injective linear map from $\mathcal{L}(V,W)$ to $\mathcal{L}(W^*,V^*)$ is automatically surjective, whence bijective.
Interestingly, if $W$ is finite-dimensional and $V$ is infinite-dimensional, the map $\phi$ is still an isomorphism.  We are left to show that $\phi$ is surjective.    To show this, let $S:W^*\to V^*$ be a linear map.  Let $n:=\dim_\mathbb{K}(W)$.  Pick a basis $\{w_1,w_2,\ldots,w_n\}$ of $W$, along with the dual basis $\{f_1,f_2,\ldots,f_n\}$ of $W^*$ (i.e., $f_i(w_j)=\delta_{i,j}$ for $i,j=1,2,\ldots,n$, where $\delta$ is the Kronecker delta).  For each $w\in W$, write $w^{**}\in W^{**}$ for its double dual.  Ergo, we see that $S$ takes the form $$S=\sum_{i=1}^n\,e_i\otimes w_i^{**}$$
for some $e_1,e_2,\ldots,e_n\in V^*$ (namely, $e_i:=S(f_i)$ for $i=1,2,\ldots,n$).  Define
$$T:=\sum_{i=1}^n\,w_i\otimes e_i\,.$$
Then, for all $j=1,2,\ldots,n$ and $v\in V$, we have
$$\big(T^*(f_j)\big)(v)=f_j\big(T(v)\big)=f_j\left(\sum_{i=1}^n\,e_i(v)\,w_i\right)=\sum_{i=1}^n\,e_i(v)\,f_j(w_i)=\sum_{i=1}^n\,e_i(v)\,\delta_{i,j}=e_j(v)\,.$$
However, as $e_j=S(f_j)$, we get
$$\big(S(f_j)\big)(v)=e_j(v)$$
for all $j=1,2,\ldots,n$ and $v\in V$.  This proves that $S(f_j)=T^*(f_j)$ for $j=1,2,\ldots,n$.  Because $f_1,f_2,\ldots,f_n$ span $W^*$, we get $S=T^*=\phi(T)$.  Therefore, $\phi$ is surjective whenever $W$ is finite-dimensional.  (Consequently, the dual map $\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$ is an isomorphism if and only if $W$ is finite-dimensional or $V=0$.)
A: Fill in details:
By definition, if $\;T:V\to W\;$ is a linear map, then $\;T^*:W^*\to V^*\;$ is defined as
$\;T^*(g)v:=g(Tv)\;,\;\;v\in V\;,\;\;g\in W^*\;$ , thus
$\;\phi\;$ is linear, because
$$\phi(T+S):=(T+S)^*=T^*+S^*$$
and the last equality follows from
$$(T+S)^*(g)v:=g(T+S)v=g(Tv+Sv)=g(Tv)+g(Sv)=:T^*(g)v+S^*(g)v$$
And also
$$T\in\ker \phi\implies \phi(T):=T^*=0^*\implies\forall\,g\in W^*\,,\,\,T^*(g)v=g(Tv)=0\,,\,\,\forall v\in V\implies$$
$$Tv\in\ker g\,,\,\,\forall g\in W^*\implies Tv=0\,\,\forall\,v\in V\implies T\equiv0.$$
