Below you find my attempt to solve the following exercise:
Show that $V$ and $\mathcal{L}(\mathbb{F},V)$ are isomorphic vector spaces.
If $V$ were a finite-dimensional vector space, then I would be confident that my proof below (with minor changes) would be correct. However, $V$ can be infinite-dimensional in the exercise above. While I know that some results for finite-dimensional vector spaces remain true in the infinite-dimensional case, other results do not. For this reason, I am not 100% confident that my proof is correct. Could you help me see if there are any mistakes? If any, could you explain to me how to fix them?
Proof. Suppose $\{\alpha\}$ is a basis of $\mathbb{F}$ and $\{v_\lambda|\lambda\in \Lambda\}$ is a basis of $V$, where $\Lambda$ is an indexing set (which can be uncountable). Let $T_\lambda:\mathbb{F}\to V$ be a linear map such that $T_\lambda(\alpha)=v_\lambda,$ where $\lambda\in \Lambda$. We will prove that the set $\{T_\lambda|\lambda\in \Lambda\}$ is a basis of $\mathcal{L}(\mathbb{F},V)$.
Let $\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda$ denote a finite linear combination of the $T_\lambda$'s.
Linear independence: If $\alpha_\lambda\in\mathbb{F}$ are scalars such that $\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda=0$, then
$$\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda(\alpha)=0\Rightarrow\sum_{\lambda\in\Lambda}\alpha_\lambda v_\lambda=0\Rightarrow \alpha_\lambda=0.$$
Linear span: Let $\alpha\in\mathbb{F}$ and $T\in\mathcal{L}(\mathbb{F},V)$ be arbitrary. Since $T(\alpha)\in V$, it follows that
$$T(\alpha)=\sum_{\lambda\in\Lambda}\alpha_\lambda v_\lambda=\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda(\alpha)=\left(\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda\right)(\alpha)\Rightarrow T=\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda.$$
Let $\varphi:\mathcal{L}(\mathbb{F},V)\to V$ be defined as
$$\varphi\left(\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda\right)=\sum_{\lambda\in\Lambda}\alpha_\lambda v_\lambda.$$
We will prove that $\varphi$ is an isomorphism.
Linearity: Let $\alpha\in\mathbb{F}$ and $S,R\in\mathcal{L}(\mathbb{F},V)$. Then
\begin{align}
\varphi(\alpha S+R)&=\varphi\left(\alpha\sum_{\lambda\in\Lambda}\beta_\lambda T_\lambda+\sum_{\lambda\in\Lambda}\gamma_\lambda T_\lambda\right)=\varphi\left(\sum_{\lambda\in\Lambda}\left(\alpha\beta_\lambda+\gamma_\lambda\right) T_\lambda\right)\\
&=\sum_{\lambda\in\Lambda}\left(\alpha\beta_\lambda+\gamma_\lambda\right)v_\lambda=\alpha\sum_{\lambda\in\Lambda}\beta_\lambda v_\lambda+\sum_{\lambda\in\Lambda}\gamma_\lambda v_\lambda=\alpha\varphi(S)+\varphi(R).
\end{align}
Injectivity: Suppose $S,R\in\mathcal{L}(\mathbb{F},V)$. Then
$$\varphi(S)=\varphi(R)\Rightarrow\varphi\left(\sum_{\lambda\in\Lambda}\beta_\lambda T_\lambda\right)=\varphi\left(\sum_{\lambda\in\Lambda}\gamma_\lambda T_\lambda\right)\Rightarrow\sum_{\lambda\in\Lambda}\beta_\lambda v_\lambda=\sum_{\lambda\in\Lambda}\gamma_\lambda v_\lambda\Rightarrow\beta_\lambda=\gamma_\lambda.$$
Therefore, $S=R$.
Surjectivity: Consider an arbitrary $v\in V$ such that $v=\sum_{\lambda\in\Lambda}\alpha_\lambda v_\lambda$. Choose $T=\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda$. Then $T\in\mathcal{L}(\mathbb{F},V)$ and
$$\varphi(T)=\varphi\left(\sum_{\lambda\in\Lambda}\alpha_\lambda T_\lambda\right)=\sum_{\lambda\in\Lambda}\alpha_\lambda v_\lambda=v.$$
Best Answer
Using notation from OP, given $v\in V$ we can define a map $T_v\colon\mathbb{F}\to V$ as $T_v(a)=a{\cdot} v$.
The map $T_v\colon\mathbb{F}\to V$ is linear i.e., $T_v\in\mathcal{L}(\mathbb{F},V)$ and we can consider the map $\psi : V\to\mathcal{L}(\mathbb{F},V)$ defined as \begin{gather} \psi : V\to\mathcal{L}(\mathbb{F},V)\\ \psi(v)=T_v \end{gather}
The map $\psi$ is linear.
The one-dimensional vector space $\mathbb{F}$ is a field, hence it is provided with the canonical basis $\{1\}$. Let us define the map $\phi\colon\mathcal{L}(\mathbb{F},V)\to V$ as $\phi(T)=T(1)$, the map $\phi$ is linear too.
The maps $\psi$ and $\phi$ are inverse of each other, i.e. $\psi\circ\phi=\operatorname{Id}_{\mathcal{L}(\mathbb{F},V)}$ and $\phi\circ\psi=\operatorname{Id}_{V}$.
It remains to prove the preceding claims, which should be standard. For the sake of example: $\psi\circ\phi(T)=\psi(T(1))=T_{T(1)}$, let us calculate $T_{T(1)}(a)=a{\cdot}T(1)=T(a1)=T(a)$ $\forall a\in\mathbb{F}$ i.e., $T\equiv T_{T(1)}$.