Closeness of points in the irreducible decomposition of a C$^{*}$-algebra representation

Question

Suppose $X$ and $Y$ are compact metric spaces. Let $\varphi\colon C(X)\to M_{n}(C(Y))$ be any $*$-homomorphism. If $\pi$ is an irreducible representation of $M_{n}(C(Y))$, then $\pi$ is unitarily equivalent to a point evaluation $\textrm{ev}_{y}$. The $*$-homomorphism $\textrm{ev}_{y}\circ\varphi\colon C(X)\to M_{n}(\mathbb{C})$ is a representation of $C(X)$. Since it's a finite-dimensional representation, we can find a unitary $u_{y}\in M_{n}(\mathbb{C})$ and a set of points $X_{y}=\{x^{y}_{1},\ldots,x^{y}_{n}\}\subset X$ such that for all $f\in C(X)$, $$ (\varphi\circ f)(y)=(\textrm{ev}_{y}\circ\varphi)(f)=u_{y} \begin{pmatrix} f(x^{y}_{1}) & 0 & \cdots & 0\\ 0 & f(x^{y}_{2}) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & f(x^{y}_{n}) \end{pmatrix} u_{y}^{*}. $$

My question is:

Does the Hausdorff distance of the set $\{x_{1}^{y},\ldots,x_{n}^{y}\}$ vary continuously in $y$? More precisely, if $y_{m}\to y$ in $Y$, does $D_{m}:=\max_{1\leq i\leq n}\min_{1\leq j\leq n}d_{X}(x^{y}_{i},x^{y_{m}}_{j})\to 0$ as $m\to \infty$?

This question is motivated by the simpler case where $\varphi$ gives rise to continuous functions $\lambda_{1},\ldots,\lambda_{n}:Y\to X$ satisfying $$ \varphi(f)=\begin{pmatrix} f\circ\lambda_{1} & 0 & \cdots & 0\\ 0 & f\circ\lambda_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & f\circ\lambda_{n} \end{pmatrix}. $$

In this case for each $y\in Y$, $X_{y}=\{\lambda_{1}(y),\ldots,\lambda_{n}(y)\}$, and the result holds.

I asked this question some weeks ago on stack exchange, but did not get an answer: https://math.stackexchange.com/questions/2787579/closeness-of-points-in-the-irreducible-decomposition-of-a-c-algebra-repres

Answer 1

Yes, it varies continuously. To see this, suppose it did not, i.e., ${\rm max}_i {\rm min}_j d(x^y_i, x^{y_m}_j) \not\to 0$. Passing to a subsequence, we can assume that ${\rm min}_j d(x^y_i, x^{y_m}_j)$ is bounded away from $0$ for some fixed value of $i$. Now find $f \in C(X)$ such that $f(x_i) = 1$ and $f(x^{y_m}_j) = 0$ for all $m$ and $j$. Then ${\rm ev}_{y_m}\circ \phi(f) = 0$ for all $m$ but ${\rm ev}_y\circ \phi(f) \neq 0$. But $\phi(f) \in M_n(C(Y)) \cong C(Y; M_n)$ so that $y \mapsto {\rm ev}_y(\phi(f))$ must be a continuous map from $Y$ into $M_n$, which is a contradiction.