Functions in the induced space compactly supported in $PN^-$ modulo $P$

Question

Let $P_0$ be a minimal parabolic subgroup of a connected, reductive group $G$ over a $p$-adic field $k$. Let $P$ be a parabolic subgroup containing $P_0$ with Levi decomposition $P = MN$. Let $N^-$ be a group such that $P^- = MN^-$ is opposite to $P$.

Let $\pi$ be a smooth, irreducible representation of $M$, and consider a function $f$ in the space of the induced representation $\sigma = \operatorname{Ind}_{P}^G \pi $. Then $f$ is locally constant, and there is a compact set $Z$ such that if $f(x) \neq 0$, then $px \in Z$ for some $p \in P$, i.e. $f$ is compactly supported modulo $P$.

I have seen in a few talks and papers the practice of fixing a function $f \in \sigma$ which vanishes outside of the open set $PN^-$. I was wondering a couple of things:

1 . Are are there "many" such functions? How do we know that they exist at all?

2 . If $f$ vanishes outside $PN^-$, since $N^-$ is the union of its open compact subgroups, is it possible to find a compact open subgroup $N_0$ of $N^-$ such that if $f(x) \neq 0$, then $px \in N_0$ for some $p \in P$?

Question 2 seems like a natural thing to expect, but I could not prove it. I thought I might take the image in $P \backslash G$ of the support of $f$, which is compact, and cover it by the images of the open compact subgroups $N_0$. The problem is that I don't think the images of $N_0$ are open in $P \backslash G$, since they are not open in $G$.

Answer 1

Answer moved from the comments.

1. They exist because $P \times N^- \to G$ is an open embedding (even on the level of varieties). This implies that, for $f \in \mathrm C^\infty_{\mathrm c}(N^−; \sigma)$, the function $p n^− \mapsto \pi(p)f(n^−)$ has image in the appropriate function space (for $n^− \in N^−$, the image of $f$ is constant on $\operatorname{stab}_P(f(n^−))N^−_0$, where $N^−_0$ is an open subgroup of $N^−$ such that $f$ is constant on $n^− N^−_0$).

2. Not $N_0$ (which I think it would be better to call $N^−_0$, as above) but $P N_0$ is open, because of (1), so the image of $N_0$ in $P\backslash G$ is open.