Is the following Proof Correct ? If so please do point out any extraneous details any remarks on how to shorten the argument would also be appreciated.
Theorem. The vector spaces $V$ and $\mathcal{L}(\mathbf{F},V)$ are isomorphic, where $\mathcal{L}(\mathbf{F},V)$ denotes the set of all linear transformations from $\mathbf{F}$ to $V$.
Proof. Consider the Transformation $\phi:V\to\mathcal{L}(\mathbf{F},V)$ defined as follows.
$$\phi(v) = T:\mathbf{F}\to V\ \ \operatorname{where}\ \overset{\alpha}T = T(1) = v$$
We now demonstrate that with the above definition $\phi$ is an isomorphism from $V$ to $\mathcal{L}(\mathbf{F},V)$. To show Linearity let $w = av_1+bv_2$ where $a,b\in\mathbf{F}$ then $\phi(w)$ equals $T\in\mathcal{L}(\mathbf{F},V)$ such that $\overset{\alpha}T = w$, furthermore $\phi(av_1) = T_1$ such that $\overset{\alpha}T_1 = aT_1v_1 = a\phi(v_1)$ and similary for $bv_2$ we have $\phi(bv_2) = T_2$ such that $\overset{\alpha}T_2 = bT_1v_2 = b\phi(v_2)$ consequently $\phi(av_1+bv_2) = a\phi(v_1)+b\phi(v_2)$.
Now assume for $v_1,v_2\in V$ , $T_1 = \phi(v_1) = \phi(v_2) = T_2$ then from defintion $v_1 = \overset {\alpha}T_1 = \overset{\alpha}T_2 = v_2$ implying that $\phi$ is injective.
Now assume that $T\in\mathcal{L}(\mathbf{F},V)$ from definition $T(1)=w$ for some $w\in V$ consequently $\phi(w)=T$.
$\blacksquare$