I'm not seeing how Pederson constructs the following partial isometry in this result.
Lemma: Let $H$ be a Hilbert space. If $T \in F_0(H)$ (Fredholm of index $0$), then there exists a partial isometry $V \in \mathcal{B}_f(H)$ (the finite ranks) such that $T+V$ is invertible.
At the first part of the proof, Pederson just says we can construct by assumption a partial isometry $V$ such that $V^{*}V(H) = \ker(T)$ and $VV^{*}(H) = \ker(T^{*})$, but I'd like to see the explicit construction or existence argument.
I tried just taking the orthogonal projection $P \in \mathcal{B}(\ker(T) + \ker(T^{*}))$ onto $\ker(T)$, using that $P \geq 0$ to write $P=V^{*}V$ for some $V \in \mathcal{B}(\ker(T) + \ker(T^{*}))$ and extend to $H$, which indeed makes $V$ into a partial isometry, but this doesn't actually guarantee that $VV^{*}(H) = \ker(T^{*})$ ($\textit{while it does indeed make $V^{*}V(H) = \ker(T)$}$).
Any help would be appreciated.
I think I see why this is apparent now. The Hilbert spaces $\ker(T)$ and $\ker(T^{*})$ have the same (finite) dimension and thus are isometrically isomorphic. Take $V_0: \ker(T) \cong \ker(T^{*})$ to be such an isometric isomorphism and then extend to $V:H \longrightarrow H$ by taking $V=0$ on $\ker(T)^{\perp}$. Then: $$V^{*}V(H) = V^{*}(\ker(T^{*})) = V^{-1}(\ker(T^{*})) = \ker(T)$$ and similarly: $$VV^{*}(H) = V(\ker(T)) = \ker(T^{*})$$ so $V^{*}V,VV^{*}$ Moreover, it's then clear that $V^{*}V,VV^{*}$ are then the orthogonal projections onto $\ker(T),\ker(T^{*})$, so $V$ is a partial isometry also satisfying $V \in \mathcal{B}_f(H)$.