Well-defineness of Chern roots?

Question

Let $\mathcal{E} \to X$ be a rank $n$ (complex) vector bundle on a space $X$ (possibly with some other mild conditions to make the splitting principle holds). According to the splitting principle, there is a space $X'$ and a continuous function $f: X' \to X$ inducing an injection on the cohomology groups $H^k(X;\mathbb{Z}) \to H^k(X';\mathbb{Z})$ and such that the pullback bundle $f^*\mathcal{E} \to X'$ is a direct sum of line bundles. In that case, the total Chern class $c_\bullet(f^*\mathcal{E})$ can be written as $$(1+x_1) \cdots (1+x_n) \in H^\bullet(X';\mathbb{Z})$$ for some $x_i \in H^2(X';\mathbb{Z})$ (and the product is the cup product). Then we call $x_1, \cdots, x_n$ the Chern roots of $\mathcal{E}$. Here is the question I can't understand: aren't Chern roots, by their definition, dependent on the space $X'$ that is chosen in the splitting principle? If that's the case, then why do people only talk about "the Chern roots" of $\mathcal{E}$? Are Chern roots unique in some sense?