Consider the strictly convex quadratic function $f(x) = \frac{1}{2}x^tPx - q^tx + r,$ where $P \in \mathbb{R}^{n \times n}$ is a positive definite matrix, $q \in \mathbb{R}^n$ and $r \in \mathbb{R}.$ Let $\mathcal{H} := \{H: H \text{ is a }k- \text{dimensional subspace in } \mathbb{R}^n\}.$ Clearly, the restriction of $f$ to any $H \in \mathcal{H}$ is again a strictly convex function. For any $H \in \mathcal{H},$ we will use $x_H$ to denote the unique optimal point of the following problem
\begin{equation*} \underset{x \in H}{\text{min.}} \; f(x). \end{equation*}
Now consider the map, $\psi(H) = x_H.$
Prove / Disprove: The map $\psi$ is bijective.
Remark: It is assumed that $P$ is invertible and $q \neq \mathbf{0}.$
Edit: Whether $\psi$ is surjective depends on the definition of its codomain. For example, if $n=2$ and $k=1$, the image of $\psi$ is a curve on $\mathbb{R}^2$. So, if the codomain of $\psi$ is $\mathbb{R}^2$, $\psi$ is certainly not surjective.
At any rate, $\psi$ may not be injective even if the origin is not a global minimum. Consider $a=(1,0,0)^T$ and $f(x)=\|x-a\|^2$. Let $\{e_1,e_2,e_3\}$ be the canonical basis of $\mathbb{R}^3$. Then $x_H=a$ when $H=\operatorname{span}\{e_1,e_2\}$ or $\operatorname{span}\{e_1,e_3\}$.