We consider the unipotent radical $N$ of the Borel subgroup $B$ of $\operatorname{GL}_2(F)$ where $F$ is a local field. Let $\phi$ be a character of the maximal split torus $T$. We inflate $\phi$ to $B$ and consider its smooth induction $\operatorname{Ind}_B^G \phi = (\Sigma, X)$ to $G$ (here $X$ is the space of $G$-smooth functions $f:G \rightarrow \mathbb{C}$ satisfying $f(bg)=\phi(b)f(g)$ for all $b \in B$, $g \in G$, and $\Sigma$ acts on this space via right translation).
Let $V=\ker \alpha_\phi$, where $\alpha_\phi:X \rightarrow \mathbb{C}$ is the surjective map given by $f \mapsto f(1)$. Since $V$ is a representation of $B$, it is also a representation of $N$. We denote the space of locally constant functions $\varphi:N \rightarrow \mathbb{C}$ having compact support by $C_c^\infty(N)$. This space carries a representation of $N$ given by right translation. My question is the following:
Given $f \in V$, define $f_N \in C_c^\infty (N)$ by $f_N(n)=f(\omega n)$. Show that the map $f \mapsto f_N$ is an $N$-isomorphism.
So far, I have shown that indeed $f_N$ as defined above is in $C_c^\infty(N)$, that the map is linear and commutes with the $N$-action. However I've been unable to construct an inverse. How would one do this, or can the bijection be shown in another way?
The key point is that $\operatorname{GL}_2(F)$ equals $B \sqcup B\omega N$, and the multiplication map $B \times N \to B\omega N$ is a homeomorphism in the analytic topology. Thus, given $\varphi \in \operatorname C_\text c^\infty(N)$, we can define $\varphi^N \in V$ by extending $b\omega n \mapsto \phi(b)\varphi(n)$ by $0$ to all of $\operatorname{GL}_2(F)$. The resulting map $\varphi \mapsto \varphi^N$ is inverse to $f \mapsto f_N$.