Let's suppose we a have a $*$-homomorphism $\varphi\colon C(S^{2},M_{n}(\mathbb{C}))\to M_{m}(\mathbb{C})$, where $n,m\in\mathbb{N}$. It is a standard fact that for any projection $p\in C(S^{2},M_{n}(\mathbb{C}))$, the trace $\operatorname{Tr}(p(x))$ is independent of $x\in S^{2}$. My question is:
Suppose I have two projections $p,q\in C(S^{2},M_{n}(\mathbb{C}))$ with $\operatorname{Tr}(p(x))=\operatorname{Tr}(q(x))$. Is it true that $\operatorname{Tr}(\varphi(p))=\operatorname{Tr}(\varphi(q))$?
We have, by the first isomorphism theorem, $$\tag{1} \text{Im}\,\varphi\simeq C(S^2,M_n(\mathbb C))/\ker\varphi. $$ The left-hand-side is simple, so $\ker\varphi$ is maximal. As $M_n(\mathbb C)$ is simple, all ideals are given by compact subsets of $S^2$, so there exists $s_0\in S^2$ such that $$\ker\varphi=\{f:\ f(s_0)=0\}.$$ Then the classes in the quotient are determined by their value on $s_0$, and we deduce that $$\tag{2} C(S^2,M_n(\mathbb C))/\ker\varphi\simeq M_n(\mathbb C). $$ From $(1)$ and $(2)$, the image of $\varphi$ is isomorphic to $M_n(\mathbb C)$.
By the uniqueness of the trace (as a tracial state) in $M_n(\mathbb C)$, we get from the two isomorphisms $(1)$ and $(2)$ that $$ \text{Tr}\,(\varphi(p))=\text{Tr}\,(p(s_0)). $$ Thus $$ \text{Tr}\,(\varphi(q))=\text{Tr}\,(q(s_0))=\text{Tr}\,(p(s_0))=\text{Tr}\,(\varphi(p)). $$