Let $V$ a vector space over $\mathbb{F}$ and let the vector space $\mathscr{L}(\mathbb{F},V) = \{f: \mathbb{F} \to V: f$ is linear $\}$. I wanna show that $V \cong \mathscr{L}(\mathbb{F},V)$.
My solution:
For each $v \in V$, define $f_v: \mathbb{F}: \to V$ s.t $f(\alpha) = \alpha v$. It's ovious that $f$ is linear.
Now define $\psi: V \to \mathscr{L}(\mathbb{F}, V)$ given by $\psi(v) = f_v$.
$\psi$ is linear:
If fact, lets $v_1,v_2 \in \mathbb{V}$ and $\lambda \in \mathbb{F}$, then $\psi(v_1 + \lambda v_2) = f_{v_1+ \lambda v_2}$. Futhermore, for each $\alpha \in \mathbb{F}$ we have $f_{v_1+ \lambda v_2}(\alpha) = \alpha(v_1+ \lambda v_2) = \alpha v_1 + \lambda \alpha v_2 = f_{v_1}(\alpha) + \lambda f_{v_2}(\alpha).$
Therefore $\psi(v_1 + \lambda v_2) = \psi(v_1) + \lambda \psi(v_2).$
$\psi$ is injective:
Let $v \in \ker \psi$ then $\psi(v) = 0_f$, where $0_f: \mathbb{F} \to V$ is given by $0_f(\alpha) = 0, \forall \alpha \in \mathbb{F}. $ However, $\psi(v) = f_v$, Then $f_v = 0_f$, for all $\alpha \in \mathbb{F}$. Then, for $\alpha=1$ we descover $v=0$, that is, $\ker \psi = \{0\}$.
$\psi$ is surjective:
Let $T \in \mathscr{L}(\mathbb{F}, V)$, and take the vector $T(1) \in V$. Note that, $\psi(T(1)) = f_{T(1)}$, but $f_{T(1)} : \mathbb{F} \to V$ is given by $f_{T(1)}(\alpha) = \alpha T(1) = T(\alpha),$ $ \forall \alpha \in \mathbb{F}$.
Then $T = \psi(T(1)).$
Is that correct?
Sounds alright to me. However, at some point in the proof of injectivity, you say that $f_v = 0_f$ for all $\alpha$, didn't you mean $f_v(\alpha) = \vec{0}$ for all $\alpha$ ?
Otherwise, looks correct :)