4. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
(1) For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1}, x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right) \neq f\left(x_{2}\right)$.

Then the value of $f(0)+f(1)+f(-1)$ is
A. 0
B. 1
C. -1
D. Uncertain
4. Note that $f(0)=f^{3}(0)$, so $f(0)[f(0)-1] \cdot[f(0)+1]=0$. Therefore, $f(0)=0,1$ or -1. Also, $f(1)=f^{3}(1)$, similarly we know $f(1)=0,1$ or -1. Again, by $f(-1)=f^{3}(-1)$, we also know $f(-1)=0,1$ or -1. But $f(-1), f(0)$, and $f(1)$ are all distinct, so $\{f(-1), f(0), f(1)\}=\{-1,0,1\}$, which means $f(-1)+f(0)+f(1)=0$. Hence, the answer is A.