Let $E,F$ be Hilbert $A$-modules, and $T \in \mathcal{L}(E,F)$ with $\lVert T \rVert \leq 1$. Then, $1-T^*T$ has dense range iff $1-TT^*$ has dense range.
This is Lemma 10.3 in Lance's book, but I don't understand the part of the proof where they use the functional calculus to deduce that $\rho((ff^*)^2)=1$. Any explanations would be helpful, or a self-contained proof too.
I don't understand what functional calculus would have to do with this. And I don't have enough familiarity with Hilbert modules; but at least I can show you how it works in the case of Hilbert spaces (I would expect a version of the same ideas to work in the general case, but I don't know enough to tell).
The situation is that you have a state $\rho$ such that $\rho(TT^*)=1$, and $\rho$ is in the predual of $L(F)$. This means that there exists a positive trace-class operator $\rho_0$ such that $\def\tr{\operatorname{Tr}}\rho(X)=\tr(\rho_0X)$ and $\tr(\rho_0)=1$. Then $$ \tr(\rho_0^{1/2}(1-TT^*)\rho_0^{1/2})=\tr(\rho_0)-\tr(\rho_0TT^*)=1-1=0. $$ Since the trace is faithful, $\rho_0^{1/2}(1-TT^*)\rho_0^{1/2}=0$. If $B=(1-TT^*)^{1/2}$ (here we use that $\|T\|\leq1$) this looks like $(B\rho_0^{1/2})B\rho_0^{1/2}=0$, and so $B\rho_0^{1/2}=0$. It follows that $(1-TT^*)\rho_0=0$. That is, $$ TT^*\rho_0=\rho_0. $$ Then $$ \rho((TT^*)^2)=\tr(TT^*TT^*\rho_0)=\tr(TT^*\rho_0)=\tr(\rho_0)=1. $$