The inputs are a full column rank matrix $\mathbf{A}^{m\times n}_{n}$ and its singular value decomposition:
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{c}
\mathbf{S}_{n\times n} \\
\mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}
\end{array} \right] \\
%
\end{align}
$$
Colors distinguish $\color{blue}{range}$ spaces from $\color{red}{null}$ spaces.
Think of the input matrix as being column-partitioned like so
$$
\mathbf{A} =
\left[ \begin{array}{c|c}
\mathbf{A}_{1} & \mathbf{A}_{2}
\end{array} \right]
$$
The question is that if we are given $\mathbf{A}$ and its singular value decomposition, can we use that to construct the singular value decomposition of $\mathbf{A}_{1}$?
Special case
The obvious special case is when we miraculously grab all the linearly independent columns in the range space $\color{blue}{\mathcal{R}\left( \mathbf{A} \right)}$:
$$
\mathbf{A} =
\left[ \begin{array}{c|c}
\mathbf{A}_{1} & \mathbf{A}_{2}
\end{array} \right]
%
=
%
\left[ \begin{array}{c|c}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
$$
In this case, we can construct the SVD and pseudoinverse from existing components.
$$
\begin{align}
%
\mathbf{A}_{1} &=
\color{blue}{\mathbf{U}_{\mathcal{R}}} \,
\mathbf{S} \,
\color{blue}{\mathbf{V}^{*}_{\mathcal{R}}}, \\
%
\mathbf{A}_{1}^{+} &=
\color{blue}{\mathbf{V}_{\mathcal{R}}} \,
\mathbf{S}^{-1} \,
\color{blue}{\mathbf{U}^{*}_{\mathcal{R}}}. \\
%
\end{align}
$$
