I have the following exercise:
Let $\pi:\mathcal A \rightarrow \mathcal B$ be a *-homomorphism between two unital $C^*$ algebras $\mathcal A$ and $\mathcal B$ which maps the unit to the unit. Assume $\forall A \in \mathcal A: A > 0 \Rightarrow \pi(A) > 0$. Proof, that $\pi$ is isometric.
My approach: I know $\pi \text{ isometric} \Leftrightarrow \pi \text{ injective}$, so I tried to prove, that there is an $A>0$ with $\pi(A) \not> 0$, if $\pi$ is not injective.
If $\pi$ is not injective, $\mathrm{ker}\ \pi \ne \{0\}$, so that I can take an $B\in \mathcal A$ with $B\ne 0$ and $\pi(B) = 0$. So $\tilde B = BB^*$ is self-adjoint and $\tilde B^2 \ge 0$. If $\tilde B$ is invertible, one has $\tilde B^2 > 0$ and $\pi(\tilde B^2) = 0 \not > 0$, which would be a contradiction to $\forall A \in \mathcal A: A > 0 \Rightarrow \pi(A) > 0$.
But want can I do, if $\tilde B^2$ is not invertible. Is there a way to construct an invertible element from $\tilde B^2$ with which I can do the proof? Thanks for any help!