Let $A$ have reduced SVD $A = \tilde{U} \tilde{\Sigma} \tilde {V}^*$, and define the Moore-Penrose pseudoinverse of $A$ as $A^+ = \tilde{V} \tilde{\Sigma} \tilde{U}^*$. In the $l^2$ norm $\| \cdot \|$, is $A \mapsto A^+$ well-conditioned? That is, given a perturbation matrix $B$, is $\|(A + B)^+ - A^+\|/\|A^+\|$ controllable by $\|B\|/\|A\|$?
I think the answer is yes, but I have no idea how to prove it.
No, take for example $T(\epsilon) = \begin{bmatrix} 1 & 0 \\ 0 & \epsilon\end{bmatrix}$. Then $T^{+}(\epsilon) = \begin{bmatrix} 1 & 0 \\ 0 & \eta\end{bmatrix}$, where $\eta = 0$ for $\epsilon = 0$ and $\eta = 1/\epsilon$ for $\epsilon\neq 0$.
Now, for $A = T(0)$ and $A + B = T(\epsilon)$ we see, that $\|(A+B)^{+} - A^{+}\|/\|A^{+}\| = \eta\rightarrow \infty$ and $\|B\|/\|A\|\rightarrow 0$ for $\epsilon\rightarrow 0$.