$SS^\dagger = S^\dagger S = P$.
$S^\dagger(I - P) = 0$.
$S^\dagger P = PS^\dagger = S^\dagger.$
Proof. $\tilde{T}$ has the same range as $T$ as mentioned above, $\overline{R}{\tilde{T}} = V$. Therefore, $\overline{R}{\tilde{S}} = \overline{R}_S$. Also, $S\tilde{S} = \iota_V S P \iota_V \tilde{S} P = \iota_V S \tilde{S} P = \iota_V S S^{-1} P = \iota_V P$. Furthermore, $\ker \tilde{S} = {f : \tilde{S}f = 0} = {f : \iota_V S P f = 0} = {f : Pf = 0} = V^\perp$, as $\iota_V$ and $S$ are injective. Repeat the same argument for the frame sequence ${\tilde{f}_k}$ with the roles of $S$ and $\tilde{S}$ switched and use Lemma 1.1 to arrive at $S^\dagger = \tilde{S}$.
(i) By Property (i) of Lemma 1.1, $SS^{\dagger}$ is the orthogonal projection onto $R_S = R_T$. Therefore, $SS^{\dagger} = P$. Switch the roles of $S$ and $\tilde{S}$ to show the second part.
(ii) $S^{\dagger}(I - P) = S^{\dagger} - S^{\dagger}P = S^{\dagger} - S^{\dagger}SS^{\dagger} = S^{\dagger} - S^{\dagger} = 0$.
(iii) The equality $S^{\dagger}P = S^{\dagger}$ follows form (ii). To show that $PS^{\dagger} = S^{\dagger}$ observe first that $S^{\dagger}P : H \to R_S$. Hence, by 2, Lemma 1.1 and (1.), $S^{\dagger} = S^{\dagger}P = PS^{\dagger}P = PS^{\dagger}SS^{\dagger} = PS^{\dagger}$. $\blacksquare$
Because of the frame property on $V$, among all sequences $c \in \ell^2(\mathbb{N})$ which synthesize an $f \in H$, the sequence $c_0 = (\langle f, S^\dagger f_k \rangle)$ is the one with the minimum norm.
Similarly, among all elements $f \in H$ which analyze to a $c \in \ell^2(\mathbb{N})$, the element $f_0 = S^\dagger T c = \sum_{k=1}^\infty \zeta_k S^\dagger f_k$ is the one with the minimum norm. We have $|f|^2 = |f_0|^2 + |f - f_0|^2$.
Proposition 5.3.5 in [8] can now be restated in terms of $S^\dagger$ which is defined on all of $H$ instead of $S^{-1}$ which is defined only on $V$.
Corollary 2.9 Let ${f_k}_{k=1}^\infty$ be a frame sequence in $H$. For any $f \in H$,
Proposition 2.10 The pseudo-inverse of T is $\tilde{U}$, $T^\dagger = \tilde{U}$. The pseudo-inverse of U is $\tilde{T}$, $U^\dagger = \tilde{T}$. Consequently, we have the following properties
$T^\dagger = T^* S^\dagger$ and $T^\dagger = S^\dagger T^*$.
$(T^\dagger)^* T^\dagger = S^\dagger.$
$(T^\dagger)^* = S^\dagger T.$
$|T|^2 = |S|.$