Let $G$ be a locally profinite (i.e., locally compact, Hausdorff, and totally disconnected) topological group, $H \le G$ a closed subgroup, $(\pi, V)$ a smooth representation of $G$, and $(\sigma, W)$ a smooth representation of $H$. Here smoothness of $\pi$ means that for every $v \in V$ there exists a compact open subgroup $K \le G$ such that $\pi(k)v = v$ for every $k \in K$, and likewise for $\sigma$. Also, we take it that $\pi$ and $\sigma$ are $R$-linear representations where $R$ is a commutative ring; that is, $V$ and $W$ are $R$-modules and the actions are $R$-linear. If it makes a difference, feel free to assume below that $R = \mathbb{C}$.


The smooth induction $\mathrm{Ind}\_H^G \sigma$ is the $R$-module of all functions $f\colon G \rightarrow W$ such that $f(hg) = \sigma(h) f(g)$ for all $h \in H$ and $g \in G$ and that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$ for some compact open subgroup $K \le G$ that depends on $f$. The action of $g\in G$ on $\mathrm{Ind}\_H^G \sigma$ is by $f \mapsto f(\*g)$, so $\mathrm{Ind}\_H^G \sigma$ is a smooth $R$-linear representation of $G$. My question is: is the "projection formula map"
$$
\pi \otimes\_R \mathrm{Ind}\_H^G \sigma \rightarrow \mathrm{Ind}\_H^G(\pi|\_H \otimes\_R \sigma)
$$
that sends $\sum v\_i \otimes f\_i$ to $g \mapsto \sum \pi(g)v\_i \otimes f\_i(g)$ an isomorphism? 


I see that it is a well-defined $G$-homomorphism, but I don't see whether it is injective or is surjective. Nor do I see how to write down an inverse.



Projection formula for smooth representations of locally profinite groups
The map is not surjective in general. The basic idea is the induction is something like a product and tensor products and products don't commute. 


Example. $H$ and $\sigma$ trivial, $R$ a field and $\pi$ is an infinite dimensional $R$ vector space with trivial $G$-action. Then if $\psi$ lies in the image of your map then the span set $\{\psi(g): g\in G\}$ is finite dimensional $R$-vector space. 


It is enough to construct a function $\psi$ that doesn't satisfy it. Let 
$\{v\_1, v\_2, \ldots\}$ be an infinite set of linear independent vectors in $\pi$. 
Assume that $G$ is not compact choose an open compact subgroup $K$ of $G$. Then can choose infinitely many distinct cosets $\{ g\_1K, g\_2 K, \ldots\}$. Let 
$\psi: G\rightarrow \pi$ be the function with support $\bigcup\_{n\ge 1} g\_n K$, 
such that $\psi(g)=v\_n$ for all $g\in g\_n K$. This function is fixed by $K$, hence smooth. 


