A unitary representation of a topological group $\rho:G\to\mathcal{U}(H)$ on an Hilbert space it is said to be irreducible if the only closed subspaces $H'\subset H$ such that $\rho(g)H'\subset H'$ are $H'= \{0\} $ and $H'=H$.

This question is linked to Definition of irreducible representation requires the invariant subspace to be closed?

Since I don't manage to prove that the classical definition (that do not assume the subspace to be closed) and the def. above are equivalent I need to prove again the Schur's Lemma.

Schur's Lemma

If $\rho_i:G\to U(H_i) $ for $i=1,2$ are irreducible unitary representations and $\varphi\in Hom_{\mathbb{C}}(H_1,H_2)$ is a G-module homomorphisms i.e. $\rho_2(g)\cdot\varphi(h) = \varphi(\rho_1(g)\cdot h)$ for any $h\in H_1,g\in G$. Then $\varphi$ is either the null homomorphism or an isomorphism.

My try. $Ker\varphi$ is closed and invariant subspace of $H_1$ therefore it must be either trivial or the whole $H_1$. Thus $\varphi$ is the null homomorphism or it is injective. Suppose $\varphi$ is injective, then $Im\varphi$ is invariant, but maybe it is not closed, $\overline{Im\varphi}$ is closed and invariant but maybe it is not proper so there is no contradiction.

How can we conclude that $\varphi$ is an isomorphism?


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