One, (20 points) In the same coordinate plane, the quadratic function $y=x^{2}-x+a$ intersects with the linear function $y=x$ at points $A$ and $B$. The function $y_{1}=y^{2}-y+a$ intersects with the linear function $y_{1}=x$ at the same points $A$ and $B$, and there are no other intersection points besides these two. Find the range of values for $a$.
Given the quadratic function $y=x^{2}-x+a$ and the linear function $y=x$ intersect at points $A$ and $B$, we know that $x^{2}-x+a=x$ has two distinct real roots. Therefore, $\Delta=4-4a>0$, hence $a<1$.
(2) $x^{2}-x+a=-x$ and $x^{2}-x+a=x$ have the same real solutions, i.e., $-x=x$ solves to $x=0$. When $x=0$, substituting into either of the above equations gives $a=0$. When $a=0$, the two equations are the same, i.e., $x^{2}=0$, which of course has the same real solution.
In summary, the range of values for $a$ is $0 \leqslant a<1$.