One, (50 points) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying: $f\left(f(x)+f^{2}(y)-1\right)=x+y^{2}-1$.
One, Solution: On the right side of the equation, let $x$ run through the entire set $\mathbf{R}$. It is easy to see that the function $f$ takes all values in $\mathbf{R}$;
If $f\left(x_{1}\right)=f\left(x_{2}\right)$,
then $x_{1}+y^{2}-1=f\left(f\left(x_{1}\right)+f^{2}(y)-1\right)$ $=f\left(f\left(x_{2}\right)+f^{2}(y)-1\right)=x_{2}+y^{2}-1, \therefore x_{1}=x_{2}$.
Therefore, $f$ is injective; combining this, we get $f: \mathbf{R} \rightarrow \mathbf{R}$ is a bijection.
In the original equation, let $y=-y$ to get
$$
\begin{aligned}
& f\left(f(x)+f^{2}(-y)-1\right)=x+(-y)^{2}-1 \\
= & f\left(f(x)+f^{2}(y)-1\right),
\end{aligned}
$$

By the injective condition, we have $f^{2}(-y)=f^{2}(y)$,
$$
\forall y \neq 0, f(-y) \neq f(y) \text {, }
$$

Again, by the injective condition, we get
$$
\therefore \forall y \neq 0, f(-y)=-f(y)(\neq 0) \text {. }
$$

By the surjective condition, we get $f(0)=0$,
In the original equation, let $y=0$ to get $f(f(x)-1)=x-1$, in the above equation, let $x=1$ and use $f^{-1}(0)=0$ to get $f(1)$ $=1$,
In the original equation, let $y=1$ to get $f(f(x))=x$,
In the original equation, let $y=f(y), x=f(x)$ to get
$$
f\left(x+y^{2}-1\right)=f(x)+f^{2}(y)-1 \text {, }
$$

In the above equation, let $y=0$ to get
$$
\begin{array}{l}
f(x-1)=f(x)-1 . \\
\therefore f\left(x+y^{2}\right)=f(x)+f^{2}(y)
\end{array}
$$

From (1), it is easy to see that $f$ is a non-decreasing function.
In (1), let $x=0$ to get
$$
\begin{array}{l}
f\left(y^{2}\right)=f^{2}(y) . \\
\therefore f\left(x+y^{2}\right)=f(x)=f\left(y^{2}\right) . \\
\therefore f\left(x-y^{2}\right)=f(x)-f\left(y^{2}\right),
\end{array}
$$

Combining, we have $f(x+y)=f(x)+f(y)$.
Equation (2), $f(1)=1$ and the monotonicity condition become a classic case, so, $f(x)=x, \forall x \in \mathbf{R}$.