We denote $V \subset \mathbb{R}^n$ to be an open set.
Definition 1: Let $\text{Alt}^p(\Bbb R^n)$ denote the set of all alternating multlinear functionals on $\Bbb R^n$.
Definition 2: Let $\Omega^p(V)$ denote the set of all differential $p-$forms where $\omega: V \subset \Bbb R^n \to \text{Alt}^p(\Bbb R^n) $
Definition 3: Let $C^\infty(V,\Bbb R)$ denote the set of all smooth maps from $V$ to $\Bbb R.$
I am reading notes (prob 3.1) that say that $\Omega^2(\Bbb R^2) \cong C^\infty (V, \Bbb R)$ in the following manner. Don't worry, this small part of the question only requires a good understanding of linear algebra to explain to me.
For any $\tau \in \text{Alt}^3(\Bbb R^3)$, $\tau = \tilde{\tau} e^1 \wedge e^2 \wedge e^3$ where $\tilde{\tau} \in \Bbb R.$ So we regard an $\omega \in \Omega^3(V)$ as a smooth map $\omega : V \to \text{Alt}^3(\Bbb R^3)\cong \Bbb R$, that is a smooth function on $V$.
So what I am not understanding is that, what is the isomorphism map between $\text{Alt}^3(\Bbb R^3)\cong \Bbb R$? All I see is that they have the same dimension, so isomorphism exists. What is the point of introducing $\tau = \tilde{\tau}e^1 \wedge e^2 \wedge e^3$?. He repeats the same thing again for $\text{Alt}^2(\Bbb R^2)\cong \Bbb R$ later.
The isomorphism between the two spaces is more or less obvious. If $V$ is a real vector space with basis $e_1,\ldots ,e_k$, then the map $$ \Phi\colon Alt^k(V)\rightarrow V, \; \omega\mapsto \omega(e_1,\ldots ,e_k) $$ is a linear isomorphism. For a proof, see for example Corollary $B.2.12$ here.