Let $F$ be a number field and $G$ (resp. $H$) an odd orthogonal (resp. metaplectic group) over $F$.

Let $v$ be a finite place of $F$ and $\sigma_v$ a supercuspidal representation of $G_v(F_v)$. Let $\pi_v=\theta_v(\sigma_v)$ be the local theta lift of $\sigma_v$ to $H_v(F_v)$ and $\pi$ be a irreducible cuspidal representation of $H(\mathbb{A})$ such that the localization of $\pi$ at $v$ is $\pi_v$.

Consider the global theta lift $\Theta(\pi)$ of $\pi$ to $G(\mathbb{A})$ and assume that it is nonzero.

In some literature, it is written that $\Theta(\pi)$ is cuspidal because $\sigma_v$ is supercuspidal.

I can't understand this because we don't know that $\Theta(\pi)$ is irreducible.

I am wondering whether the statement in the literature is really true.

(ps: My guess is that suppose that $\rho$ is an irreducible constituent of $\Theta(\pi)$. Then there is a nonzero linear map $l:\omega_v \to \pi_v \otimes \rho_v$, where $\omega_v$ is the local Weil representation. By the maximal semi-simplicity of the local theta lift $\theta(\pi_v)$, $\theta(\pi_v) \simeq \rho_v$. Therefore $\rho_v=\sigma_v$ and so $\rho$ is cuspidal.

But I am not sure $\Theta(\pi)$ is cuspidal if all of its irreducible constituents are cuspidal.